solutions manual of geometry and discrete mathematics

Solutions for Selected Problems

Exercise 1.1

4. c. Clearly if n � 41 the expression becomes 412 � 41� 41 or 41 � 43, which is composite.Actually, if n � 40, we have 402 � 40 � 41 � 40 (40 � 1) � 41 � 412.The expression gives a prime number for n � 1, 2,3, … , 39, but not for 40.

6. The expression to be tested is n2 � 9, n even.For n � 4, n2 – 9 � 7, which is not composite.The statement is not true.

7. The expression to be examined is n2 � 9, n odd.For n � 5, n2 � 9 � 16, which is divisible by 8.For n � 7, n2 � 9 � 40, which is divisible by 8.For n � 9, n2 � 9 � 72.For n � 13, n2 � 9 � 160.For n � 15, n2 � 9 � 216.All are divisible by 8.

8. The expression to be tested is n2 � 3.For n � 5, n2 � 3 � 22.For n � 6, n2 � 3 � 33.For n � 7, n2 � 3 � 46.For n � 8, n2 � 3 � 61, which is prime.The expression is not a composite number for all values of n.

Exercise 1.2

2. Every odd integer n can be written as 2p � 1, where p � 1, 2, 3, …. Now, p itself can be either even or odd. If p is even, set p � 2k, k � 0, 1, 2, 3, …. If p is odd, set p � 2k � 1, k � 0, 1, 2, 3, ….Then every odd number can be written as n � 4k � 1or n � 4k � 3, k � 0, 1, 2, 3, ….If n � 4k � 1, then n � 7 � 4k � 8 � 4(k � 2).If n � 4k � 3, then n � 5 � 4k � 8 � 4(k � 2).Hence one of n � 5, n � 7 is always divisible by 4.

3. n3 � 4n � n(n2 � 4)� n(n � 2)(n � 2).

If n is even, then set n � 2k, k � 1, 2, 3, ….Now n3 � 4n � 2k(2k � 2)(2k � 2)

� 8k(k � 1)(k � 1).Since k � 1, k, k � 1 are three consecutive integersfor all values of k, one of them is divisible by 2 andone (possibly the same one) is divisible by 3.Then k(k � 1)(k � 1) is divisible by 6.Therefore n3 � 4n is divisible by 8 � 6 � 48.

4. Every odd integer n can be written as n � 2p�1,k � 0, 1, 2, ….Then n2 � 4p2 � 4p � 1

� 4p (p � 1) � 1.Now p and p � 1 are consecutive integers, so one ofthem is even. Then p(p � 1) � 2k. Therefore n2 � 8k � 1, where k is an integer.

5. Since 71 ends in 7, 72 ends in 49, 73 ends in 43, 74

ends in 01, then 75 � 71 � 74 ends in 07, 76 � 72 � 74 ends in 49, 77 � 73 � 74 ends in 43, and 78 � 74 � 74 ends in 01.Then 79 ends in 07, 710 ends in 49, 711 ends in 43, and712 ends in 01.This cycle continues.Hence 74k+1 ends in 07, 74k+2 ends in 49, and so on.Since 201 � 4 � 50 � 1, 7201 ends in 07.

6. If x and y are integers, then 2x is even and 4y is even.Hence the left side of the equation is divisible by 2,which is impossible if the right side is 5. Then it isimpossible for x and y both to be integers.

7. n5 � 5n3 � 4n � n(n4 � 5n2 � 4)� n(n2 � 1)(n2 � 4)� (n � 2)(n � 1)n(n � 1)(n � 2).

This is the product of 5 consecutive integers, so one ofthem is divisible by 5, and at least one is divisible byeach of 4, 3, and 2, provided that n � 3.Then the expression is divisible by 5 � 4 � 3 � 2 �120, for n � 3 and an integer.

Chapter 1: Introduction to Proof 1

Chapter 1 • Introduction to Proof

8. Since p and q are odd primes, then p � q is even, so 2divides p � q.Now, if q � p, then �

p �2

q� lies between p and q. It is

larger than p and smaller than q.But every number between p and q is composite,

since p and q are consecutive odd primes, so �p �

2q

�has at least two divisors.Then p � q has at least three divisors.

9. Every integer can be written as 3k, 3k � 1, or 3k � 2.Hence there are two possibilities. The first is that thereare three of a

1, a

2, a

3, a

4, a

5, of the same form. That

is, three of them are of the form 3k1, 3k

2, 3k

3, or they

are 3k1

� 1, 3k2

� 1, 3k3

� 1, or they are 3k1

� 2,3k

2� 2, 3k

3� 2. In each case their sum is divisible

by 3.The second case is that there are not three of the sameform, so there must be at least one of each form, say3k

1, 3k

2� 1, 3k

3� 2. But now this sum is 3(k

1� k

2� k

3� 1), which is divisible by 3.

Hence there is always at least one subset of threenumbers whose sum is divisible by 3.

10. Consider n2 � 4 � (n � 2)(n � 2).If n � 3, then each of n � 2 and n � 2 is greater than 1, and n2 � 4 is a composite number.

11. Every odd integer n can be written as n � 2k � 1,k � 2, if we wish integers greater than 5.Now n2 � 25 � (2k � 1) 2 � 25

� 4k2 � 4k � 24� 4(k)(k � 1) � 24.

Now k and k � 1 are consecutive integers, so one ofthem is even. Then 4k(k � 1) is divisible by 8.Since 24 is also divisible by 8, then n2 � 25 is divisible by 8 for n � 5.

Exercise 1.3

3. Let the angles in any quadrilateral have measures a�,b�, c�, d�, and let the exterior angles at oppositevertices have measures x� and y�, as shown.Then a � b � c � d � 360 (angles in aquadrilateral).Also b � x � 180 and d � y � 180 (straight angles).Then b � x � d � y � 360.Therefore a � x � c � y � 0 by subtraction and a � c � x � y.The sum of the exterior angles at opposite vertices is equal to the sum of the interior angles at the othertwo vertices.

4. Let ∠ABC be 2x� and ∠ACB be 2y�.Then 2x � 2y � 90 � 180 (angles in a triangle).

Therefore x � y � 45.Now ∠BDC � x � y � 180 (angles in a triangle).

Therefore ∠BDC � 135.∠BDC � 135�.

5. In ∆ PQR, ∠PQR� ∠PRQ (isosceles triangle).In ∆ QRS, ∠QRS � ∠QSR (isosceles triangle).

Then ∠SQR � 2 ∠QSR � 180 (angles in a triangle).Also ∠SQR � ∠PQR � 180.

Then ∠PQR � 2 ∠QSR.Then ∠PQS � 3 ∠QSR.

P

Q R

S

B

A C

x x

Dy

y

y

a

d

c

b x

2 Chapter 1: Introduction to Proof

6. If the polygon has n sides and n angles each of size x�,180(n � 2) � nx180n � nx � 360

n � �18

306�0

x� .

Note that (180 � x) must divide evenly into 360 if thepolygon is constructable.

7. Mark angles as shown, using the given information.Since AB � AD, ∠ADB � b.Since AB � BE, ∠BEA � a.Now 2x � a � 180

so x � 90 – �12

� a.

Also 2y � b � 180

so y � 90 – �12

�b.

In ∆ ABD, b � b � a � 90 – �12

� a � 180

a � 4 b � 180.

In ∆ ABE, a � a � b � 90 – �12

� b � 180

4a � b � 180.Adding 5a � 5b � 360

a � b � 72.In ∆ ABC, a � b � c � 180

Then c � 180 � 72� 108.

∠BCA is 108�.

8. Mark angles as shown.Since b � 2x � 180

x � 90 – �b2

�.

In ∆ ABD, �a2

� � b � 90 – �b2

� � t � 180 (angle sum)

t � �12

� (a � b) � 90

2t � (a � b) � 180

In ∆ ABC, c � (a � b) � 180 (angle sum)

Subtracting 2t � c � 0

t � �12

� C

or ∠APB � �12

� ∠ACB.

9. a. For the polygon ABCDEFG the sum of the exteriorangles is 360�.But each exterior angle occurs twice.Hence ∠P � ∠Q � ∠R�∠S�∠T � ∠U�∠V�720� 7 �180.Then ∠P � ∠Q � ∠R � ∠S � ∠T � ∠U � ∠V� 540.

VA

P

B Q

C

R

D

SE

T

FU

G

B

A

C

P

a2

bxx

t

c

a2

A

a xx

cC

byy

B

E

D


b. In general, if there are n sides then there are nsmall triangles surrounding the inner polygon. Letthe sum of the angles at the tips of the star be S.Then S � 720 � n �180

S � 180(n � 4).

Exercise 1.4

6. Name the vertices as in the diagram, using propertiesof the rectangle.

Then AC � �a�2�� b�2�and BD � �a�2�� b�2�Therefore AC � BD.Note that positioning the rectangle makes the proofsimple.

7. Name the vertices as in the diagram. Since we usemidpoints, we name vertices appropriately.Since W is the midpoint of A(0, 2b)and B(2a, 2b), its coordinates are (a, 2b).Similarly X has coordinates (2a, b), Y has coordinates (a, 0), and Z has coordinates (0, b).

Now WX � �(2�a� �� a�)2� �� (�b� �� 2�b�)2� � �a�2�� b�2�XY � �(2�a� �� a�)2� �� (�b� �� 0�)2� � �a�2�� b�2�YZ � �a�2�� b�2�ZW� �(0� �� a�)2� �� (�b� �� 2�b�)2� � �a�2�� b�2�.

These are all equal. Then WXYZ is a rhombus.

8. Let ∆ ABC be any triangle and let its coordinates be (2b, 2c), (0, 0), and (2a, 0), as shown.If D is the midpoint of AB, its coordinates are (b, c).If E is the midpoint of AC, its coordinates are (a � b, c).The slope of BC is 0 and the slope of DE is 0, so theyare parallel.The length of BC is 2a and the length of DE is a,

so DE � �12

� BC.

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal toone-half of it.

9. Let the parallelogram be positioned as shown, and letthe coordinates of A be (b, c), D be (0, 0), and C be(a, 0). Then since AB is equal and parallel to DC, thecoordinates of B are (a � b, c).Now AB2 � BC2 � CD2 � DA2 � (b2 � c2) � [(a � b � b) 2 � 02] � [(a � b � a)2 � c2] � a2

� b2 � c2 � a2 � b2 � c2 � a2

� 2 (a2 � b2 � c2).Also AC2 � BD2 � (a � b)2 � c2 � (a � b)2 � c2

� 2(a2 � b2 � c2).The sum of the squares of the sides is equal to thesum of the squares of the diagonals.

A(b, c) B(a + b, c)

D(0, 0) C(a, 0)

y

x

y

A(2b, 2c)

D(b, c)

B(0, 0) C(2a, 0)x

E(a + b, c)

y

A(0, 2b) W(a, 2b) B(2a, 2b)

X(2a, b)

x

Z(0, b)

D(0, 0) Y(a, 0) C(2a, 0)

y

A(0, b) B(a, b)

D(0, 0) C(a, 0)x


10. Let the triangle be positioned as shown, and let thecoordinates of B be (0, 0), C be (2a, 0), and A be (b, c).Since D is the midpoint of BC its coordinates are (a, 0).

Then AB2 � AC2 � b2 � c2 � (2a � b)2 � c2

� 4a2 � 2b2 � 2c2 � 4abAlso 2BD2 � 2AD2 � 2(a2) � 2 ((b � a)2 � c2)

� 4a2 � 2b2 � 2c2 � 4ab.The statement is true.

11. a. Let the rectangle be positioned as shown. Let P(x, y) be any point in the interior. Then 0 x aand 0 � y b.Now PA2 � PC2 � x2 � y2 � (a � x) 2 � (b � y2)

� 2x2 � 2y2 � a2 � b2 � 2ax � 2by.

Also PB2 � PD2 � x2 � (y � b)2 � (a � x)2 � y2

� 2x2 � 2y2 � a2 � b2 � 2ax � 2by.

Then PA2 � PC2 � PB2 � PD2.

b. Through P draw XY ⊥ AD and BC, as shown.Then AY � BX and YD � XC.Now PA2 � PY2 � AY2

and PC2 � PX2 � XC2

so PA2 � PC2 � PY2 � PX2 � AY2 � XC2.Also PB2 � PD2 � PX2 � BX2 � PY2 � YD2

� PX2 � AY2 � PY2 � XC2.Then PA2 � PC2 � PB2 � PD2.

12. Position the triangle as shown and let the coordinatesof the vertices be A (0, a), B (–b, 0), and C (c, 0).Then the perpendicular from A meets BC at O (0, 0),and the equation of AO is x � 0.

The slope of AB is �ab

�, so the slope of CE is – �ba

�, and

the equation of CE is y � 0 � � �ba

� (x � c).

The intersection of AO and CE is the point P and forthe coordinates of P we set x � 0 in the equation of CE.Then y � �

bac�.

P is the point �0, �bac��.

�bac� � 0

Then the slope of the line BF is –––––––– � �ac

�.0 � b

The slope of AC is ��ac

�.

Now ��ac

�� ac

�� 1.

Then BF ⊥ AC, and the altitudes are concurrent.

y

A(0, a)

F

xC(c, 0)OB(–b, 0)

E

P

B X C

p

A Y D

B(0, b) C(a, b)

P(x, y)

A(0, 0) D(a, 0)

y A(b, c)

B(0, 0) D(a, 0) C(2a, 0)x


13. Position the triangle as shown and let the coordinatesbe A(2a, 2b), B(–2c, 0) and C(2c, 0).Then the median from A meets BC at D (0, 0).The median from C meets AB at E (a � c, b).

The equation of AD is y � �ba

� x.

The equation of CE is y � 0 � �a �

b3c

� �x � 2c�.

These lines intersect at P, and for intersection

�ba

� x � �a �

b3c

� �x � 2c�(a � 3c) x � a (x � 2c)

–3cx � –2ac

x � �23a�

Then y � �ba

� x � �23b�.

P has coordinates ��23a�, �

23b��.

The median from B meets AC at F (a � c, b).

The equation of BF is y � �a �

b3c

� �x � 2c�.

We check that P lies on BF.

For y � �23b�, the right side is �

a �b

3c� ��

23a� � 2c�

� �a �

b3c

� ��2a �3

6c��

� �23b�.

Point P is on BF, and the medians are concurrent.

Chapter 1 Test

1. c(x, y) must lie on the right bisector of AB. Theequation of the right bisector is x � 2.

2. In a convex hexagon there are six vertices at each ofwhich there is an interior and an exterior angle. Thesum of the interior and exterior angles at a vertex is180�. Then the sum of the six interior and six exteriorangles is 6 � 180� � 1080�.The sum of the six interior angles is 180(6 � 2)� 720�.Therefore the sum of the exterior angles is 1080 � 720 � 360�.

3. Let the trapezium be positioned as shown and let thecoordinates of the vertices be A(2a, 2b), B(2c, 2b),C(2e, 0), and D(0, 0).Then the coordinates of P, Q, R, S, are P(a, b),Q(c � e, b), R(c, b), S(a � e, b).All of these points are b units above the x-axis, so allof them lie on the line whose equation is y � b.

4. Let the equal parts of ∠ABC be x� and let the equalparts of ∠ACD be y�. Now let ∠ACB be z�.Because ∠BCD � 180�, 2y � z � 180,

In ∆ ABC, 2x � z � 58� 1802x � z� 122

Then 2x � 2y � 2z� 302x � y � z � 151.

In ∆ EBC, ∠E � x � y � z � 180Therefore ∠E � 29.

58º

xx

BZ

yy

CD

EA

y

A(2a, 2b) B(2c, 2b)

C(2e, 0)x

D(0, 0)

PR S

• •• •Q

E

y

A(2a, 2b)

F

C(2c, 0)DB(–2c, 0)x


5. Solution 1.Let the three consecutive even numbers be 2k�2, 2k,2k�2.Then (2k � 2)2 � (2k)2 � (2k � 2)2

� 4k2 � 8k � 4 � 4k2 � 4k2 � 8k � 4� 12k2 � 8� 4(3k2 � 2)� 22(3k2 � 2).The expression is always divisible by 2, 2, and at leastone other number.

6. Let D be the midpoint of AB. D has coordinates

�0, �a2

��.

The right bisector of AB has equation y � �2a

�.Let E be the midpoint of BC.

E has coordinates ��b2

�, �2c

��.

The slope of BC is �bc

�.

Then the right bisector of BC has slope – �bc

�.

The equation of the right bisector of BC is

y � �2c

� � � �bc

� �x � �b2

��or y � � �

bc

� x � �2bc

2

� � �2c

�

� � �bc

� x � �b2

2�c

c2

�.

These right bisectors intersect at P. For thecoordinates of P solve the equations:

y � �a2

� and y � � �bc

� x � �b2

2�c

c2

�

�a2

� � � �bc

� x � �b2

2�c

c2

�

�bc

� x� �b2

2�c

c2

� � �a2

�

� �b2 �

2c2

c� ac�

x � �b2 �

2c2

b� ac�

P has coordinates ��b2 �2c2

b� ac�, �

a2

��.

We now show that if F ��b2

�, �a �

2c

�� is the midpoint of

AC, then PF is the right bisector of AC. This requiresonly that PF ⊥ AC, or that the product of the slopesof PF and AC be –1.

The slope of AC is �c �

ba

�.

The slope of DF is

�2

c�

�–-–––––––––––––––––––––––––––––––––––––––––––––––

�b2 � b2 �

b

c2 � ac�

c�–––––––––

�ac �

b

c2

�

� �a �

b

c�.

The product of these slopes is �c �

b

a� � �

a �

b

c� � �1.

Then PF is the right bisector of AC.Therefore the right bisectors are concurrent.

7. a. D � 92 � 4(8)(10)� �239

Since D 0 the equation has imaginary roots.

b. For the equation (n � 1) x2 � nx � (n � 1) � 0,D � n2 � 4(n � 1)(n � 1)

� n2 � 4(n2 � 1)��3n2 � 4.

Since the coefficients are positive integers, n � 2.Then D 0 for all n.Therefore the equation has imaginary roots for all n.

y

A(0, a)F

C(b, c)

P

E

xB(0, 0)

D

••

•

•

�a �

2c

� � �a2

� ��

�b2

� � �b2 �

2c2

b� ac�



Exercise 2.1

4. a. In ∆ADC and ∆CBAAD � CB � parallelogramDC � BAAC is commontherefore ∆ADC � ∆CBA (side-side-side).

b. In ∆PST, PS � PTtherefore ∠S � ∠T (isosceles triangle)and ∠PSQ � ∠PTR (supp. to ∠S and ∠T).Now in ∆PSQ and ∆PTRPS � PT (given)∠PSQ � ∠PTR (proven)SQ � TR (given)therefore ∆PSQ � ∆PTR (side-angle-side).

c. In ∆ABC and ∆DCB,AB � DC (given)∠ABC � ∠DCB (given)BC is commontherefore ∆ABC � ∆DCB (side-angle-side).

d. Diameters of a circle bisect each other. HenceOA � OB � OC � OD.In ∆AOB and ∆CODOA � OC � radiiOB � OD∠AOB � ∠COD (vert. opp.)therefore ∆AOB � ∆COD (side-angle-side).

6. In ∆PQR, PQ � PRtherefore ∠PQR � ∠PRQ (isosceles)and ∠SQR � ∠TRQ (supp. angles)Now PS � PT (given)

PQ � PR (given)therefore PS � PQ � PT � PR and QS � RT.In ∆QRS and ∆RQT : QS � RT∠SQR � ∠TRQ, QR is common.Therefore ∆QRS � RQT (side-angle-side)and ∠QRS � ∠RQT.

7. In ∆ABC and ∆ADCAB � AD (given)∠BAC � ∠DAC (given)AC � ACtherefore ∆ABC � ∆ADC (side-angle-side)therefore BC � DC and ∠BCA � ∠DCAtherefore AC bisects ∠BCD.

8. In ∆ABC and ∆DBEAB � BE (given)∠ABC � ∠DBE (opp. angles)∠CAB � ∠DEB � 90� (given)therefore ∆ABC ∆EBD (angle-side-angle)and AC � ED.

9. In ∆PQS and ∆PRSPQ � PR∠QPS � ∠RPSPS is common.Therefore ∆PQS � ∆QRS (side-angle-side)and ∠PQS � ∠PRS � x.In ∆PQR, PQ � PRtherefore ∠PQT � ∠PRThence ∠PQT � x � ∠PRT � xand ∠SQT � ∠SRT.In ∆AEB and ∆CEDAB � CD (given)∠BAE � ∠DCE (proven)∠AEB � ∠CED (opp. angles)therefore ∆AEB � ∆CEDand AE � CE, BE � DEand diagonally bisect each other.

10.

In ∆PQR and ∆SRQPQ � SRPR � SQ � (given)QR is commontherefore ∆PQR � ∆SRQ (side-side-side)and ∠PQR � ∠SRQ.

Q

P

RS

8 Chapter 2: Plane Figures and Proof

Chapter 2 • Plane Figures and Proof

11.

Given: Quadrilateral ABCD with AB � DCand AD � BC. Prove AC and DB bisect each other;i.e., AE � EC and DE � BE.Proof: In ∆ABC and ∆CDA

AB � CDBC � DA

and AC � CAtherefore ∆ABC � ∆CDA (side-side-side)and ∠BAC � ∠DCA.

12.

In ∆QTS and ∆RWS∠QST � ∠RSW (opp. angles)QS � RS∠QTS � ∠RWS � 90�therefore ∆QST � ∆RSW (angle-side-angle) and QT � RW.

13. Place vertex D of ∆DEF on vertex A of ∆ABC, and letDE coincide with AB. Since DE � AB, E will fall onB. Since EDF � ∠BAC, DF will fall along AC andsince DF � AC, F falls on C. Therefore ∆DEFcoincides with ∆ABC and the triangles are congruent.

14. a. Two angles and three corresponding sides.

b. Three angles and two corresponding sides.

c. No similarity for sides only.

15. Since A and B are the mid-points of equal sides SRand UV.SA � AR � UB � BVin ∆ASP and ∆BVW

SA � VB∠ASP � ∠BVW � 90�

PS � WVtherefore ∆ASP � ∆BVW (side-angle-side)

and AP � BW.In ∆ASX and ∆BUP

AS � BUSX � UP

∠ASX � ∠BUP � 90�therefore ∆ASX � ∆BUP (side-angle-side)

therefore AX � BP.

16. Since the faces are equilateral triangles,∠ABX � ∠ACY � 60�

and AB � AC.Now in ∆ABX and ∆ACY

AB � AC∠ABX � ∠ACY � 60�

BX � CYtherefore ∆ABX � ∆ACY (side-angle-side)

and AX � AY.

Exercise 2.2

5. Since A and B are the midpoints of PQ and SR,PA � AQ � SB � BR � a.Let the distance between PQ and SR be h.

Now �gm PQRS � 2ah

�gm ASBQ � ah

therefore �gm PQRS � 2�gm ASBQ

or �gm ASBQ � �12

��gm PQRS.

x

o

y

o x

Q

T

RS

W

P

A B

D C

E

Chapter 2: Plane Figures and Proof 9

6.

PS is a median therefore QS � SR � a.Let height of ∆PQR be h.

Therefore ∆RQS � �12

� ah

∆PSR � �12

� ah

and ∆PQS � ∆PSR.Therefore a median bisects the area of the triangle.

7. Since AD is a median ∆ADC � �12

� ∆ABC.

Since BE is a median ∆BEC � �12

� ∆ABC.

Therefore ∆ADC � ∆BEC.But quad DFEC is common to both triangles therefore∆ADC � quad DFEC � ∆BEC � quad DFECTherefore ∆AEF � ∆BFD.

8. AD is a median of ∆ABCtherefore ∆ABD � ∆ADCsimilarly ∆BED � ∆DECtherefore ∆ABD � ∆BED � ∆ADC � ∆DEC and ∆ABE � ∆ACE.

9. Since the diagonals of a parallelogram bisect eachother, PT � TR.Therefore ∆PTQ � ∆RTQ � aand ∆PST � ∆TSR � b.Also ST � TQ therefore ∆STP � ∆TQPhence a � b and ∆PQT � ∆PTS � ∆TSR � ∆TRQ.

10.

Since ∆PTS � ∆STR and both have the same heightthen the bases are equal. Then PT � RT. Similarly∆PTS � ∆PTQ and both have the same height.Therefore ST � QT.

Now in ∆PTS and ∆QTRTS � TQPT� TR

∠PTS � ∠QTRtherefore ∆PTS �∆RTQ (side-angle-side)therefore PS � RQ.Also ∠PST � ∠TQR therefore PS�QR (all ∠s).Since PS � QR and PS�QR.Therefore PSQR is a parallelogram.

11.

Construct a line through T that is parallel to PQ andSR meeting PS at X and RQ at Y.Now PXYQ is a parallelogram

and ∆PTQ � �12

��gm PXYQ.

Similarly ∆TSR � �12

��gm XSRY.

But �gm PSRQ � �gm PXYQ � �gm XSRY.

Therefore

∆PTQ � ∆TSR � �12

��gm PXYQ � �12

��gm XSRY

� �12

��gm PSRQ.

12.

Since ∆ADC � ∆ABC and AC is common, thealtitudes of ∆ADC and ∆ABC are equal, i.e., DX � BY.But DX and BY are both perpendicular to AC thereforeDX�BY. Hence BYDX is a parallelogram and thediagonals of a parallelogram bisect each other.Therefore BT � DT, i.e., AC bisects BD.

A

B C

D

Y

TX

P Q

X YT

S R

P

S

T

Q R

P

Q RS

h


13.

CF is a median of ∆ABC therefore

∆CFB � �12

�∆ABC

�gm ∆BCE � �12

�∆ABC.

Therefore ∆CFB � ∆BCE. But ∆COB is commontherefore ∆CFB � ∆COB � ∆BEC � ∆COBand ∆EOC � ∆FOB � a.OE is a median of ∆AOCtherefore ∆EOC � ∆OEA � aOF is a median of ∆AOBtherefore ∆AOF � ∆OFB � aand ∆AOE � ∆AOF.

14.

AC bisects parallelogram ABCD.Therefore ∆ABC � ∆ACD. The base AC is common.Therefore the height of each triangle is the same, say h.

Therefore ∆ADX � �12

� Ax · h

∆ABX � �12

� Ax · h.

Therefore ∆ADX � ∆ABX.

15.

∆AOD � �12

��gm ABCD

�gm ABCD � ∆AOD � �12

��gm ABCD

�gm ABCD � ∆AOD � ∆AOB � ∆DOC

therefore ∆AOB � ∆DOC � �12

��gm ABCD.

16.

Given quad ABCD construct through A a line parallelto BD meeting CD extended at E. Join EB.Then ∆EBC � quad ABCD.Proof: ∆BDC is common to both ∆BEC and quad ABCD.∆BDE � ∆BDA (same base, same altitude).Therefore ∆BDE � ∆ADC � ∆ABD � ∆ADC or∆BEC � quad ABCD.

17.

Since AB � AC � x

∆ABC � �12

� CD · x � �12

� BE · x

Therefore CD � BE.In ∆DBC and ∆BEC, BC is common, CD � BE∠BDC � 5BEC � 90�therefore ∆BDC � ∆CEB (hyp. side)therefore ∆BDC � ∆CEBnow BC is common; therefore altitudes from D and Eare equal. Therefore DE�BC.

18.

Extend BC to F so that EF ⊥ BF.Draw DG ⊥ BC.

A

D

BG K

E

C F

A

D E

B C

A

E

D

B C

A D

B O C

A

B

D

C

X

C

E

A BF

O


In ∆DGB and ∆EFC,BD � CD (given)

∠DBG � ∠ECF (∠ABC � ∠ACB � ∠ECF)∠DGB � ∠EFC (90�)∆DGB � ∆EFC

then DG � EF.Now in ∆DGK, ∆EFK,

DG � EF∠DGK � ∠EFK (90�)∠DKG � ∠EKF (vert. opp.)

then ∆DGK � ∆EFK (right-angled triangles)therefore DK � EK.

Exercise 2.3

6. Join S1

and S2. Construct the right bisector of S

1S

2meeting the circle at two points, A and B. Gates arelocated at A and B. Since they lie on the right bisectorof S

1S

2they will be equidistant from S

1and S

2.

7. Approximate the second road as a straight line andextend the roads to meet. The pumping station shouldlie on the river at the point determined by the bisectorof the angle formed by the two roads.

8. Bisect the angle formed by the intersecting lines. All points on this line are equidistant from the twointersecting lines. Construct the right bisector of AB.All points on this line are equidistant from A and B.Then the intersection of these bisectors gives therequired point.

9. Since a circle can always be drawn through three non-collinear points, the fourth point is restricted to be onthe circle if the four points are to be concyclic.

10. Construct the right bisector of YZ meeting the circle atX

1and X

2. X

1YZ and X

2YZ are isosceles triangles.

These two are always possible. If YZ is not a diameterthere are two others, X

3found by YZ � YX

3and X

4by

ZY � ZX4.

11.

Let the right bisectors of AB and AC intersect at O.Then OB � OA and OC � OA therefore OB � OC.Since O is equidistant from B and C, it lies on the rightbisector of BC. Therefore the right bisectors of thesides of a triangle pass through a common point.

13.

Let the bisectors of ∠B and ∠C meet at T. Draw aline through T parallel to BC meeting AB and AC at Dand E respectively. Now DT�BC. Therefore ∠DTB � ∠TBC � a, hence DT � DB.Similarly ET � CE.Therefore DT � TE � DB � CE

DE� DB � CE.

14.

Let the midpoint of BC be X. The right bisector of AXmeets AB at D. D is the required point.

15. If the angles form an arithmetic sequence, then let theangles be a � d, a, a � d.Now a � d � a � a � d � 180

3a� 180a � 60

therefore one of the angles is 60�.If one of the angles is 60� then one must be smallerthan 60� and the other larger than 60� (or else all are 60�). Let the angles be (60 � x)�, 60�, and (60 � y)�.Now 60 � x � 60 � 60 � y � 180

�x� y � 0x � y

Therefore the angles are in arithmetic sequence.

A

D

B X C

A

D T E

B C

A

BO

C

YX


16. Let the roots be p1

� p and q.Now (x � p)(x � p)(x � q) � 0

(x2 � p2)(x � q) � 0x3 � qx2 � p2x � p2q � 0

x3 � ax2 � bx � c � 0c � p2q, but p2 � �b

and q � �a.Therefore c � ab.Now if c � abx3 � ax2 � bx � c � 0becomes

x3 � ax2 � bx � ab � 0x2(x � a) � b(x � a) � 0

(x1 � a)(x2 � b) � 0x � a � 0 or x2 � b � 0

x2 � �bx � � ��b�

Therefore one of the roots is the negative of the other.

Exercise 2.4

1. Assume that ∠DBE � ∠DEB.Now ∠ACE � ∠DEB (DE�AC)and AC � AB (isosceles triangle)but AB > AC, therefore there is a contradiction, andhence ∠DBE ≠ ∠DEB.

3. Either the line intersects the curve or it doesn’t.Assume that it does. Thenx4 � 3x2 � 2x � 2x � 1x4 � 3x2 � 1 � 0

Since the left side is always greater than 1, there areno values of x that satisfy the equation. We have acontradiction. Therefore there is no intersection.

4. Either a, b, c are consecutive terms of a geometricsequence or they are not. Assume they are. Then wecan represent a, b, and c as a, ar, and ar2. Theequation becomesax2 � arx � ar2 � 0

x2 � rx � r2 � 0, since a ≠ 0

then x � ��r � �

2r2 � 4�r2��

�� r � �

2� 3r2�

�.

This equation has no real roots. Then a, b, and ccannot form a geometric sequence.

5.

Either CD is parallel to EF or CD is not parallel toEF. Assume that CD is not parallel to EF and let PTintersect CD, EF, and AB as in the diagram at Q, R,and S. If CD is not parallel to EF then ∠DQR ≠∠FRS. Since CD parallels AB, then ∠DQR � ∠BST.Since EF parallels AB, then ∠FRS � ∠BST. But then∠DQR � ∠FRS and it is a contradiction that they arenot equal. Then since ∠DQR � ∠FRS, CD�EF.

6.

Suppose that ∆ABC does not have two equal angles.Then ∠ABC ≠ ∠BAC. Since CE�AB, ∠BAC � ∠ACE(alt. angles) and ∠ABC � ∠ECD (corr. angles).Therefore ∠ACE ≠ ∠ECD.But this is a contradiction, because CE bisects ∠ACD.Then ∠ABC � ∠BAC and the triangle has two equalangles.

7.

In ∆ADX and ∆CDBXD � DB (given)

∠ADX � ∠CDB (vert. opp.)AD � DC (median)

then ∆ADX � ∆CDB (side-angle-side)therefore ∠DAX � ∠DCBtherefore AX�BC (alt. angles).Similarly AY �BC.Since AX�BC and AY�BC and the line AX and AY havethe common point A, then X, A, and Y are collinear.

AY X

E D

B C

A

BC

E

D

a

b ca

b

P

Q

R

S

F

B

T

C

E

A

D


8.

Since PX�QY∠PXM � ∠QYN � x (alt. angles).In ∆PMX, PX � PM (radii)therefore ∠PMX � ∠PXM � x (isosceles triangle).In ∆QNY, QY � QN (radii)therefore ∠QNY � ∠QYN � x (isosceles triangle)hence ∠PMX � ∠YNQ � x.Extend QN to T.∠TNM � ∠QNY � x (vert. opp.)therefore ∠TNM � ∠PMXand TQ�PM (corr. angles)i.e., PM�QN.

9.

Let l1

be the right bisector of BC and l2

be the rightbisector of CA. Either l

1and l

2meet or they do not

meet. Assume that they do not meet. Then l1�l ⊥

2.

Since l1⊥BC and l

2⊥CA if l

1�l

2then BC�CA.

But this is impossible since BC and CA are sides of atriangle. It is impossible for l

1to be parallel to l

2, and

the lines must meet.

10. Originally there are 32 white and 32 coloured squareson the checkerboard. Each domino tile will cover onewhite and one coloured square. By removing twowhite squares we are left with 30 white and 32 blacksquares. To cover the board we must have the samenumber of each colour, so it is not possible.

11. Either x < 1.1, or x � 1.1.If x � 1.1, x9 � 2.357 and 7x � 7.7. Then x9 � 7x >10.057. This is a contradiction, since x9 � 7x < 10.Then x < 1.1.

Exercise 2.5

5.

In ∆BFC, FD is a median.Therefore ∆BFD � ∆DFC � x, say.Similarly in ∆ABD and ∆ADC, BF and CF aremedians, hence ∆ABF � ∆BDF � x and ∆AFC �∆DFC � x.

a. Therefore �∆∆

A

A

B

B

C

F� � �

4

x

x� � �

1

4�

b. �∆∆

A

A

F

B

C

C� � �

4

x

x� � �

1

4�

c. and ∆ABF � ∆AFC � x.

6. a. If �a

b� � �

d

c�

then �a

b� � 1 � �

d

c� � 1

�a �

b

b� � �

c �

d

d�.

b. If �a

b� � �

d

c�

then �m

nb

a� � �

m

nd

c�

and �m

nb

a� � 1 � �

m

nd

c� � 1

Therefore �ma

n�b

nb� � �

mcn�d

nd�

multiplying by n gives �ma �

bnb

� � �mc �

dnd

�.

7.

∆ADE and ∆EDB have the same height ED and basesAD and DB.

Therefore �∆∆

A

E

D

D

E

B� � �

2

1�.

A

E D

C B

2

1

3

3k

4k

2k

A

B CD

F

x x

l1

l2

A

BC

Y

TN

x

xx

x

x

X

PM

Q


Since ED�CB, ED⊥AB, and ∆ADE is isosceles,so ED � 2. ∆EDB and ∆ECB have the same height DB and basesED � 2 and CB � 3.

Therefore �∆∆

E

E

D

CB

B� � �

2

3�.

Let ∆EDB � 2k (to avoid fractions).Then ∆ADE � 4k and ∆ECB � 3k.Therefore trapezoid DECB:∆ADE � 5:4.

8. In ∆ABD, EG�DB

therefore �A

E

E

B� � �

G

AG

D�.

In ∆ADC, GF�DC

therefore �G

AG

D� � �

F

A

C

F�.

Then in ∆ABC, �A

E

E

B� � �

F

A

C

F� and EF�BC.

10.

Join RX meeting YQ at A.

In ∆XZR, YA�ZR

therefore �XYZ

Y� � �

XA

AR�.

In ∆XPR, XP�AQ

Therefore �X

A

A

R� � �

Q

PQ

R�

hence �XYZ

Y� � �

PQ

QR�

11.

Let BX and DY meet AC at M and N. X and Y are themidpoints of AD and BC henceAX � XD � BY � CYalso XD�BYtherefore XDYB is a parallelogram so BX�YD.In ∆AND, MX�ND, AX � XDtherefore AM � MN.In ∆BMC, BM�YN, BY � YCtherefore MN � NC.Therefore AM � MN � NC and AC is trisected by BX and DY.

12.

Join BR. Since �A

P

P

B� � �

3

4�, then �

∆∆

A

B

P

P

R

R� � �

3

4�.

Let ∆BPR � 12kthen ∆APR � 9k.

Similarly �∆∆

B

A

R

B

C

R� � �

3

2�, so ∆ABR � 21k and ∆BRC

� 14k.

Join AZ.

�∆∆

C

AZ

Z

R

R� � �

3

2�.

Let ∆CZR � 2xthen ∆AZR � 3x.

Now �∆∆

A

B

P

P

Z

Z� � �

3

4�

therefore �9

2

k

6k

�

�

3x

2x� � �

3

4�

36k � 12x � 78k � 6x6x � 42kx � 7k

therefore ∆RCZ � 14k � ∆BCR.Since both triangles have the same height their baseswill be equal hence BC � CZ and C is the midpoint of BZ.

A

P

B C Z

R 3x

2x

9k

12k

14k

A X D

M

N

B Y C

X

Y

Z

A

P

Q

R


13.

Extend DA and CB to meet at P.Let AX � XD � x

BY � YC � y.Let AP � a and BP � b.

In ∆PDC, AB�DC

therefore �A

A

D

P� � �

B

P

C

B�

or �2

a

x� � �

2

b

y� from which �

a

x� � �

b

y�.

In ∆PXY, �a

x� � �

b

y� or �

P

A

A

X� � �

B

P

Y

B�

hence AB�XYtherefore AB�XY �DC.

14.

Join AD.In ∆ABC, KE�AB

therefore �C

K

K

B� � �

C

EA

E�

In ∆BDC, KF�BD

therefore �C

K

K

B� � �

F

C

D

F�

now in ∆CAD, �C

EA

E� � �

F

C

D

F�

therefore EF�AD.

Exercise 2.6

6. a. ∆ADE ~ ∆ABC (angles equal)

therefore �∆∆

AA

DBC

E� � �

AA

DB2

2

� � �499�

therefore ∆ABC � 81 � �499�

∆ABC � 441.

b. Quad DBCE � 441 � 81� 360.

8.

a. Given ∆ABC and ∆DEF with ∠A � ∠D, ∠B � ∠Eand ∠C � ∠F.Therefore ∆ABC ~ ∆DEF.Now ∠C and ∠F are bisected.Therefore ∠ACM � ∠DFN and ∆AMC ~ ∆DNF(angles equal)

therefore �CF

MN� � �

FC

DA�.

9. Given ∆ABC ~ ∆DEF.

Therefore �DAB

E� � �

DAC

F� � �

BE

CF� � k, say.

Now AB � k · DEAC � k · DF

and BC � k · EF.Adding: AB � AC � BC � k (DE � DF � EF)

�DAB

E�

�

AD

CF

�

�

BECF

�� k

i.e., the perimeter of ∆ABC (AB � AC � BC)to the perimeter of ∆DEF (DE � DF � EF) is equalto k which in turn is equal to

�DAB

C� � �

DAC

F� � �

BE

CF�.

10. In ∆PQO, AB�PQ so �OAP

A� � �

OBQ

B�.

In ∆OQR, BC�QR so �OBQ

B� � �

OC

CR� and

in ∆ORS, CD�RS therefore �OC

CR� � �

OO

DS�

therefore �OAP

A� � �

OO

DS� hence in ∆OPS, AD�PS.

A

M

B C

xx

D

N

FEx

x

A

BK

C

F

E D

P

A B

X Y

D C


11. a. ∆ABD ~ ∆ACD ~ ∆BCA.

b. (i) ∆ABD ~ ∆ADC

Therefore �BA

DD� � �

DAD

C� and (AD)2 � BD · DC.

(ii) ∆ABC ~ ∆ADC

Therefore �BA

CC� � �

DAC

C� and (AC)2 � BC · DC.

(iii) ∆ABC ~ ∆ABD

Therefore �BA

DB� � �

BA

CB� and (AB)2 � BC · BD.

c. From parts (ii) and (iii) addingAC2 � AB2 � BC · DC � BC · BD

� BC [DC � BD]� BC · BC

AC2 � AB2 � BC2.

12.

Join ED.Since E and D are midpoints of AB and AC, ED�BC.Now ∆AED ~ ∆ABC

therefore �AA

EB� � �

EB

DC� � �

12

�.

Now ∆EDF ~ ∆BCD (angles equal)

therefore �FE

CF� � �

FB

DF� � �

EB

DC� � �

12

�.

Therefore FC � 2EF and BF � 2FD.

13.

∆PAB ~ ∆PQS ∆EDF ~ ∆BCD

Therefore �PP

QA� � �

QAB

S� Therefore �

PP

CR� � �

BSR

C�

and ∆PAC ~ ∆PQR therefore �PP

QA� � �

PP

CR� � �

QAC

R�

therefore �QAB

S� � �

BSR

C� and �

BA

CB� � �

QSR

S�.

14.

Since quad ABCD ~ quad EFGH,

�EAD

H� � �

AE

BF� � �

FB

GC� � �

GCD

H�

and ∠A � ∠E, ∠B � ∠F, ∠C � ∠G, ∠D � ∠H.

In ∆ABD, ∆EFH, �EAB

F� � �

AE

DH� and ∠A � ∠E

therefore ∆ABD ~ ∆EFH.Similarly ∆BCD ~ ∆FGH.

Then �∆∆

EA

FB

HC

� � �AE

BF

2

2� � k

and �∆∆

FB

GCD

H� � �

BE

CF2

2

� � k

∆ABC � k ∆EFH and ∆BCD � k ∆FGH∆ABC � ∆BCD � k (∆EFH � ∆FGH)quad ABCD � k quad EFGH

or �qquuaadd

EA

FB

GCD

H� � k � �

GD

HC2

2�.

A

B

D C

EF

GH

P

A CB

QS R

A

E D

B C

y x

F

x y

A

B DC

y x

x y


15. The proof is similar to question 14. Divide thepentagons into one triangle and a quadrilateral and usethe result from question 14.

16.

Trap DBCE � �2245� ∆ABC

therefore if ∆ABC � 25 then ∆ADE � 1.Now ∆ADE ~ ∆ABC (angles equal)

therefore �∆∆

AA

DBC

E� � ��

AA

DB��

2� �

215�.

Therefore �AA

DB� � �

15

� and �AD

DB� � �

14

�.

Review Exercise

7. a. ∆BED � �15

� ∆ABD

∆ABD � �25

� ∆ABC

Therefore ∆BED � �15

� ��25

� ∆ABC� � �225� ∆ABC.

b. Using the fact that areas of triangles areproportional to bases with constant heights, weassign areas to various triangles as shown.

Therefore �∆∆

AB

DED

C� � �

7.x5x�, x ≠ 0.

8. a. Since XD � BY and XD�BY, XDYB is aparallelogram therefore BX�YD and BX � YD.

b. Since AX � XD and XH�DM (from a.), therefore

�XA

DX� � �

HAH

M�.

c. Since AX � XD and XH�DM, therefore AH � HM.d. Exactly the same reasoning as in previous parts.e. Using c. and d., AH � HM � CM, BX and DY

trisect AC.

9.

�28..28� � �

13a.2� and �

34

� � �3b.6�

Therefore 13.2 � 4a 4b � 10.8a � 3.3 b � 2.7

thus a � b � 3.3 � 2.7 � 6.0therefore y � 6.0.

10. a. 1:2

b. All areas are equal.

c. �rec

∆tAADBECD

� � �18

�

�rec

∆tAABBFCD

� � �16

�

12.

Since ∆ABE ~ ∆CDE (angles equal)

therefore �CA

DB� � �

CAE

E� � �

DBE

E� � �

12

�

thus, CE � 2AE and DE � 2BE.Therefore the diagonals of the trapezoid trisect eachother.

A B

D C

E

x

x

3.6

2.2

6.6

a b

13.2

k

3k

A

BD

E

C

4x7.5x

x

A

D E

B C


13. In ∆ABD, �AA

DP� � �

BE

DP�.

In ∆ADC, �AA

DP� � �

DPF

C�.

Therefore �BE

DP� � �

DPF

C�

or (EP)(DC) � (BD)(PF) but DC � BD, therefore EP � PF.

15. (i) If the three altitudes are equal then the triangle isequilateral. Let the three altitudes be h, and let thethree sides be a, b, and c as shown.

∆ABC � �12

� ha � �12

� hb � �12

� hc.

Thus ah � hb � hc or a � b � c (h ≠ 0).

(ii) If the triangle is equilateral then the three altitudesare equal.

Calculating the altitudes, if the sides are 2a inlength then the three altitudes must each be a�3�.

16.

Extend AD to meet the perpendicular from C at F.Draw the perpendicular from A to meet BC at E. AE isthe height of ∆ABC. CF is the height of ∆ACD. SinceAD�BC, AE � CF � h.If AD � a and BC � b,

then area trap ABCD � �12

� ah � �12

� bh

� �12

� (a � b)h.

17.

∆GBD � ∆GDC � x (equal bases, same height)and ∆GAF � ∆GBF � y (equal bases, same height)and ∆GAE � ∆GCE � z (equal bases, same height).

Since ∆ADB � ∆ADC and ∆GBD � ∆GCDtherefore ∆ADB � ∆GBD � ∆ADC � ∆GCD

or ∆AGB � ∆AGCtherefore 2y � 2z

y � z.Similarly, x � y and x � y � z. Each of the smaller triangles have equal areas.

18. Let AE � k and ED � 2ktherefore BC � 3k.Since ∆AEF ~ ∆CBF

therefore �∆∆

CAE

BFF

� � ��3kk��2

� �19

�.

Let ∆AEF � x and ∆CBF � 9x.

Since �EF

FB� � �

13

�, ∆AFB � 3x.

Since diagonal AC divides parallelogram into twotriangles of equal area, quad EFCD � 11x.

Therefore �qua

∆dAEBFFCD

� � �131xx

� � �131�.

Thus ∆ABF:quad EFCD � 3:11.

20. Let the area of parallelogram AHIE be P, the area ofparallelogram EIFB be Q, the area of parallelogramHIDG be R, and the area of parallelogram IGCF be S.Let AI:IC � x:y.Since HI�DC, then AH:HD � x:y and since IG�AD,then DG:GC � x:y. Now parallelogram AEGD andparallelogram EGCB have the same height so

P � R � �xy

� (Q � S).

Similarly using parallelogram AHFB and

parallelogram HDCF, P � Q � �xy

� (R � S).

Then R � Q � �xy

� (Q � R)

or R(x � y) � Q(x � y)

R � Q (since (x � y ≠ 0) or parallelogram HIDG � parallelogram EIFB.

A

F

B CD

G

Ey

y

x x

z

z

A

BE C

D F

A

c b

CBa

h


21.

Since �∆∆

CBD

DEE

� � �34

� and ∆BDE � 6, ∆CDE � 8.

Since �∆∆

CC

AB

DD

� � �12

� and ∆CBD � 14, ∆CAD � 7.

The total area is 14 � 7 � 21.

22.

Using parallelogram properties, mark equal angles as shown∆ADE ~ ∆FBE (equal angles)

then �AE

EF� � �

AB

DF�.

Also ∆ABF ~ ∆GDA (equal angles)

then �AA

GF� � �

AB

DF�

therefore �AE

EF� � �

AA

GF�.

Chapter 2 Test

1. a. Converse of Theorem If AD is a median then∆ADC � ∆ADB.

b. If AD is a median then ∆ADC and ∆ADB have thesame bases and equal heights. Thus, ∆ADC � ∆ADB.

c. ∆ABC � ∆ADB if and only if AD is a median.

2. a. A � �12 �

220

� � 120.

b. ∆ABC � 24∆ADB � 12

Therefore ∆DEB � 6.

c. ∆APB � ∆CPD � �12

�(10)(8 � x)� �12

� 10x

� 5 [(8 � x) � x]� 40.

d. 120 � d ��8 �

212

��d � 12.

3. AM � �12

� AB therefore AM � 5

∆AMY ~ ∆ACB

�AA

MC� � �

AA

BY� � �

MCB

Y�

�256� � �

A10

Y� � �

M24

Y�.

Therefore 26AY � 50 and 26MY � 120.

AY � 1�1123� MY � 4 �

183�

4.

∆AEC� 36therefore ∆CED� 108

then ∆ADC� 144therefore ∆ABC� 288.

A

Ea

3a

BD C

A B

D C10

xp

(8–x)

A

E

C D B

A B

D C

E F

G

z

yx

x y

C

AD

E

Bx 2x

3m

4m


5.

Draw PQ parallel to AB.Then ABPQ is a parallelogram and AP is a diagonal.Therefore ∆APQ � ∆ABP.Also QPCD is a parallelogram with DP as a diagonal.Therefore ∆QPD � ∆PCD.Then ∆APQ � ∆QPD � ∆APD � ∆ABP � ∆PCD.

6.

Mark angles as shown.Then ∆APD ~ ∆CBA

therefore �∆∆

PP

QBA

A� � �

PC

QB2

2

�

�∆P

3Q6

A� � �

49

�

therefore ∆PQA � 16.

Also, �AA

QB� � �

23

� (from above)

therefore �∆∆

PP

QBA

A� � �

23

�

therefore �∆P

16BA� � �

23

�

therefore ∆PBA � 24.Therefore trap BCQ has an area of

16 � 36 � 24 � 24 � 100.

7. Either the line from A to D is parallel to BC or it isnot. Assume that AD is not parallel to BC. There mustbe a point G on BD or BD extended so that AG�BC.

If AG�BC then ∆ABC � ∆GBC (same base, equalheight).Since we are given ∆ABC � ∆DBC, therefore ∆GBC� ∆DBC. This is clearly not possible so theassumption that AD is not parallel to BC must befalse. Then AD�BC.

A

B C

D

G

P Q

B C

A

y x

x y

A

Q

D

B

P

C



Exercise 3.1

5. Draw XY to cut AB at P. Since B is on the right bisector of XY (BX � BY) and A is on the right bisector of XY, AB is the right bisector of XY and ∠XPB � 90°. In right-angled triangle XAP let XP � 4 and AP � x. In ∆XPB, XP � 4 and PB � y. Then x2 � 16 � 25, so x � 3. Also y2 � 16 � 64, so y � 4�3�. Then AB � 3 � 4�3�.

6. Let the centre of the circle be O. Draw OA ⊥ PQ and OB ⊥ RS as in the diagram. Then PA � 4 (perpendicular bisects chord), and similarly RB � 6. Let OA be h units; then OB is 10 � h. In ∆OAP, OP is the radius r, and r2� 42 � h2

similarly, in ∆ORB, r2 � 62 � (10 � h)2

subtracting these equations, O � �20 � 100 � 20hh � 6

then r2 � 16 � 36 � 52r � 2�31�.

The radius is 2�31�.

P QA

R SB

O

X

A BP

Y

5 8

22 Chapter 3: Properties of Circles

Chapter 3 • Properties of Circles

8. Draw a line from O, the centre of the circle, to P. A � P. Draw a line perpendicular to OP, meetingthe circle at A and B. Then OP bisects AB.Proof: Since OP ⊥ AB, AP � PB (perpendicular to chord). It is always possible to draw a line through P that isbisected at P.

9. Let O be the centre of the circle and AB be a chord.OC is perpendicular to AB and is extended to meet thecircumference at D. Since OC ⊥ AB, AC � CB (chord bisector). In ∆OAC and ∆OBC, OA � OB (radii)

AC � BC (proven) ∠OCA � ∠OCB (right angles).

Then ∆OAC � ∆OBC (right-angled triangles).Therefore ∠AOC � ∠BOC.

Now arc AD � �∠3A6O0°

C� (2�r)

and arc DB � �∠3B6O0°

C� (2�r)

since ∠AOC� ∠BOC, arc AD � arc DB.

10. Let O be the common centre and let OA ⊥ XY. Since PQ is a chord of the inner circle, PA � AQ(perpendicular to chord). Similarly, XY is a chord of the outer circle,so XA � AY. Then XA � PA � AY � AQ

or XP � QY.

O

A BC

D

OB

P

A

11. Draw CA and CB from the centre C to meet PQ at Aso that CA ⊥ PQ and XY at B so that CB ⊥ XY. Join OC. In ∆OAC, ∆OBC, AC � BC (perpendiculars to equal chords)

OC � OC (common side) ∠OAC � ∠OBC (right angles)

then ∆OAC � ∆OBC (hypotenuse and side of right-angled triangles)therefore OA � OB. Now QA � YB (half of equal chords)then OA � QA � OB � YBor OQ � OY.

12. Since XP � XQ (radii)X is on the right bisector of PQ.Similarly YP � YQ (radii)Y is on the right bisector of PQ. Then XY is the rightbisector of PQ.

13. From O draw OM ⊥ AB and ON ⊥ CD as in thediagram.Since AB � CD; then OM � ON (equal chords).In ∆OMX, ∆ONX,

OX � OX (common)OM � ON (proven)

∠OMX � ∠ONX (right angles)then ∆OMX � ∆ONX (hypotenuse, side)Therefore ∠OXC � ∠OXB.

14. As in question 12, PQ is the right bisector of AB. Let AT have length h and PT have length x. Then TQ has length 21 � x.

in ∆PAT, x2 � h2 � 132

in ∆QAT, (21 � x)2 � h2 � 202

subtracting x2 � (21 � x)2 � 132 � 202

x 2 � 441 � 42x � x2 � 169 � 40042x � 210

x � 5.Then h2 � 169 � 24

h � 12444(h > 0).Therefore AB � 24.

A

P QT

B

AD

NX

MC

O

B

P

X Y

Q

O

QA

P

YB

X

C

X

Y

O

AP

Q

Chapter 3: Properties of Circles 23


Exercise 3.2

4. a.

In ∆OAP, OP � OAtherefore ∠OPA � ∠OAP � x (isosceles triangle)

∠AOB � 2x (exterior angle).

b.

In ∆OAP, OP � OA so ∠OPA � ∠OAP � x (isosceles triangle)

then ∠AOC � 2xin ∆OPB, OP � OB

so ∠OPB � ∠OBP � y (isosceles triangle) then ∠COB � 2y

therefore ∠AOB � 2x � 2y� 2(x � y)

and ∠APB � x � ytherefore ∠AOB � 2 ∠APB.

5. Join OA and OB. Draw AM ⊥ AB.

Then BM � �2r

�.

In ∆OMB, OM2 � OB2 � BM2

� r2 � �r4

2

�

� �34

� r2

OM � ��

23�

� r.

Then ∠MOB � 30° and ∠OBM � 60°therefore ∠AOB � 60° and reflex ∠AOB � 300°.

Then ∠ACB � �12

� ∠AOB

� 30°

and ∠ADB � �12

� reflex ∠AOB

� 150°therefore ∠ADB � 5 ∠ACB.

6. In �gm ABCD, ∠A � ∠C (opp. angles in �gm).Let ∠A � ∠C � a°.Join OB and OD.Then on arc BAD, ∠BOD � 2 ∠C � 2a°.Also, on arc BCD, ∠BOD � 2 ∠A � 2a°.Then 4a � 360

a � 90.Then ∠A � 90° and ABCD is a rectangle.

7. Since CD < AB, arc CED < arc CAD.

Then ∠CED � �12

� reflex ∠COD.

But reflex ∠COD > 180°therefore ∠CED > 90°.

A OB

DC

E

A D

B C

a

aO

O

A Bm

r

C

D

r2

P

O

AB

x

x

y

yC

P

O

AB

x

x

8. Join BD. Since ∠BOD � 2 ∠A,∠BOD � 180°.

Then BD is a diameter of the circle.Then BD2 � AB2 � AD2

� 8Now ∠BED � 90° (angle in semicircle)Therefore BE2 � DE2 � BD2

� 8By joining AC, in like fashion, AE2 � CE2 � 8.Then AE2 � BE2 � CE2 � DE2 � 16.

9. This is a generalization of question 8. The prooffollows the same steps.

10. Let ∠BAP be x°. Join PL.Since ∠BAL � 90°, BL is a diameter.Then ∠BPL � 90° (angle in semicircle)

also ∠BLP � ∠BAP � x° (subtended by chord BP)therefore ∠PBL � (90 � x)° (angles in ∆BPL).Then ∠QBK � (90 � x)° (opposite ∠PBL)therefore ∠KAQ � (90 � x)°since ∠KAQ � ∠QBK (subtended by chord QK).But ∠KAB � 90°then ∠QAB � 90° � (90 � x)°

� x°therefore ∠QAB � ∠BAP and AB bisects ∠QAP.

11. Join CD and CE. Let the intersection of DE and BCbe F, and let ∠BAC be 2x°, so ∠DAC is x°.Then ∠DEC is x° (angles subtended by chord DC).Now ∠EFC is 90°then ∠EFC � (90 � x) ° (angles in ∆EFC).Now ∠BAD � x° (half of ∠BAC)

so ∠BCD � x° (angles subtended by chord BD)then ∠ECD � (90 � x) ° � x°

� 90°

Then DE is a diameter of the circle.

12. Let ∠BOC be �°.Arc BC subtends ∠BOC at the centre and ∠BAC atthe circumference.

Then ∠BAC � ��

2°� (angle at the circumference).

In ∆ACX, ∠XAC � ��

2°� and ∠AXC � 90°.

Then ∠ACD � (90 � ��

2�)° (angles in a triangle).

Now ∠AOD � 2 ∠ACD

� (180 � �)° (subtended by arc AD).

Then ∠AOD � ∠BOC � (180 � �)° � �°� 180°.

13. Let D be the midpoint of BC.Since AB is a diameter, ∠ADB � 90° (angle in asemicircle).Similarly ∠ADC � 90°.In ∆ADB, ∆ADC, AD � AD (common)

BD � CD (given)∠ADB � ∠ADC (right angles)

then ∆ADB � ∆ADC (two sides, contained angle).Therefore AB � AC and the triangle is isosceles.

A

C

B

D

O

x�

AE

B F

D

C

x xx

AK L

B

QP

AE

B

D C

O


Exercise 3.3

6. Extend BC to E. ∠BAD � ∠BCD � 180° (cyclic quad)∠DCE � ∠BCD � 180° (straight angle).Therefore ∠BAD � ∠DCE.An exterior angle of a cyclic quadrilateral is equal tothe interior opposite angle.

7. Using the diagram given,∠ADB � ∠ABD � x° (isosceles triangle).Then ∠BAD � (180 � 2x)°.In ∆BDC, ∠BCD � ∠DBC � 2x° (isosceles triangle).Then ∠BAD � ∠BCD � (180 � 2x) ° � 2x°

� 180°Therefore ABCD is cyclic (opposite anglessupplementary).

8. Consider the trapezoid ABCD in a circle.Since AB � CD, ∠ABC � ∠BCD � 180°,so ∠BCD � 180° � ∠ABC.Since ABCD is cyclic, ∠ABC � ∠ADC � 180°,so ∠ADC � 180° � ∠ABC.Therefore ∠ADC � ∠BCD. The base angles are equal.Join AC and BD. In ∆ADC, ∆BCD, ∠ADC � ∠BCD (proven)

∠DAC � DBC (angles subtendedby chord DC)

Then ∠ACD � ∠BDC (third angles intriangles)

Since DC is commonthen ∆ADC � ∆BCD (two angles and one side).Therefore AC � BD.The diagonals are equal.

9. Using the diagram given, ∠RSQ is an exterior angle of ∆QST. Then ∠RSQ � ∠SQT � ∠STQ (exterior angle).Since ∠RSQ � 2 ∠SQT, ∠SQT � ∠STQ.Now since PR � PT, ∠STQ � ∠PRT (isoscelestriangle).Then ∠PRS � ∠SQP, angles subtended by SP.Therefore PQRS is a cyclic quadrilateral.

10. Join SR and QT. In ∆PQR, PQ � PRThen ∠PQR � ∠PRQ (isosceles triangle)

Let ∠PQR � �°. Then ∠PRQ � �.Since ST �QR, ∠PST � �°.Then ∠TSQ � (180 � �)°.Therefore ∠TSQ � ∠TRQ � 180° and STRQ is cyclic.In cyclic quadrilateral STRQ, chord SQ subtends equalangles at T and R. Then ∠QTS � ∠SRQ.

11. Join ST and QR. Since OT � OS, PQ � PR (equalchords).Then ∠PQR � ∠PRQ (isosceles triangle).

Now PT � �12

� PQ and PS � �12

� PR, so ST �RQ.

Then ∠PST � ∠PRQ � ∠PQR. But ∠PST � ∠TSR � 180°,so ∠PQR � ∠TSR � 180°.

Therefore STQR is a cyclic quadrilateral (oppositeangles supplementary).

P

S T

Q R

A B

D C

A B

D C

E

A

B CD


12. Join OA and OB and let X1

be on arc AB.

Since the octagon is regular, ∠AOB � �18

� (360°)

� 45°. Then reflex ∠AOB � 315°.

Chord AB gives ∠AX1B

where ∠AX1B � �

12

� reflex ∠AOB

� �31

25°�.

But this is true for all given angles.

Hence ∠AX1B � ∠BX

2C �…� ∠HX

8A � 8��

3125

��°� 1260°.

13. Join OS and OT. Since ST is fixed and OS � OT areradii, the triangle OST is fixed and ∠SOT is constant.

Since M is the midpoint of ST, ∠SOM � �12

� ∠SOT,and ∠SOM is constant as ST moves.Now ∠SMO � 90° (property of midpoint of chord),

and ∠SPO � 90°.Then SPOM is cyclic (opposite angles supplementary).Therefore ∠SPM � ∠SOM (subtended from chord SM).Since ∠SOM is constant, ∠SPM is constant.Note: This is a surprising result. Investigation withGeometer’s Sketchpad is advised.

Exercise 3.4

7. Since DF is tangent to the circle, DE and FE aretangents.Now DE � DB (tangents from D)also FE � FC (tangents from F)also AB � AC (tangents from A).

Then AD � DE � AB and AF � FE � AC.

The perimeter of the triangle is AD � DE � FE � AF� AB � AC� 2AB� 20.

8. Let the tangent contact points be W, X, Y, Z as in the diagram.Since tangents from an external point are equal,AW � AZ � a, say,BW � BX � b,CX � CY � c,DY � DZ � d.Then AB � CD � a � b � c � d

and AD � BC � a � b � c � dTherefore AB � CD � AD � BC.

A

B

C

D

W

X

Y

Z

BD

A EO

F C

TM

S

A BP O

A

H

G

F

E

D

C

B

O

TP Q

SO

R


9. Since BD is a tangent and PB is a radius,∠DBP � 90°.Join PC. Since AD is a tangent and PC a radius,∠ACP � 90°.In ∆ACP, AC2 � AP2 � PC2

� 36 � 4 � 32

AC � 4�2�Now AB � 8, AP � 6, PC � 2.

In ∆ACP, ∆ABD, ∠CAP � ∠BAD (same angle)∠ACP � ∠ABD (right angles)

Then ∆ACP ~ ∆ABD (equal angles).

Therefore �AA

CB� � �

BP

DC�.

Substituting, �4�

82�

� � �B2D�

BD � �41�62�

�

� 2�2�.

10. Using the given diagram,MA � MX (tangents from external point)MX � MB (same)

then MA � MBtherefore M is the midpoint of AB.

11. Join XO, PQ, POXP � XQ (tangents from external point)OP � OQ (radii).Then XO is the right bisector of PQ.Therefore ∠XRP � 90°.In ∆XRP, ∠XPR � 90° � ∠PXR � 180°.In ∆XOP, ∠XOP � 90° � ∠PXO � 180°.Therefore ∠XPQ � ∠XOP.

12. PR2 � PQ2 � QR2

� 162 � 302

� 1156PR � 34.Let I be the centre of the incircle. From I drawperpendiculars to the three sides of the triangle, andjoin IP, IQ, IR.

Now ∆PQR � �12

�(30)(16)

� 240.Also ∆PQR � ∆IPQ � ∆IPR � ∆IQR

� �12

�(16r) � �12

�(30r) � �12

�(34r)

� 40rthen 40r � 240

r � 6.The radius is 6.

Exercise 3.5

6. ∠CBE � ∠BAC (tangent-chord property).Since AC is a diameter, ∠ABC � 90° (subtended by diameter)then ∠ABD � ∠CBE � 90°.In ∆ADB, ∠ABD � ∠BAD � 90° (∠ADB � 90°)then ∠CBE � ∠BADtherefore ∠BAC � ∠BAD and AB bisects ∠CAD.

A

O

C

EBD

P

R30Q

16 rr

rI

P

XR

O

Q

D

BPO

A

EC


7. Join PQ, AQ, and BQ. Using the circle with AB as chord, ∠CAP � ∠AQP(tangent-chord property).Similarly, ∠CBP � ∠BQP.In ∆CAB, ∠CAB � ∠CBA � ∠ACB � 180°.Then ∠AQB � ∠ACB � ∠AQP � ∠BQP � ∠ACB

� ∠CAB � ∠BAC � ∠ACB� 180°.

Therefore AQBC is a cyclic quadrilateral.

8. Join PQ. ∠YXP � ∠XQP (tangent-chord property)∠XYP � ∠YQP (same)∠YXP � ∠XYP � ∠XPY � 180° (angle sum)then ∠XQP � ∠YQP � ∠XPY � 180°or ∠XQY � ∠XPY � 180°.

9. a. Let AB and AC be secants intersecting the circle atD and E. Join CD and BE.In ∆ABE, ∆ACD, ∠EAB � ∠CAD (same angle)

∠ABE � ∠ACD (subtended byarc DE)

Then ∆ABE ~ ∆ACD (equal angles)

Therefore �AA

CB� � �

AA

DE� or

AB · AD � AC · AE.

b. Let AB be a secant cutting the circle at E, and letAC be a tangent meeting the circle at D.We wish to prove that AD2 � AB · AE. Join DE.

In ∆AED, ∆ADB, ∠EAD � ∠DAB (same angle)∠ADE � ∠ABD (tangent-

chord property).Then ∆AED ~ ∆ADB (equal angles).

Therefore �AA

DE� � �

AA

DB�

or AD2 � AB · AE.

10. Join AT and BT. PT 2 � PA · PB (intersecting secantsproperty).Now PA � PO � OT (OA � OT) and PB � PO � OT (OB � OT).

Then PT2 � (PO � OT)(PO � OT) � PO2 � OT2

or PO2 � PT2 � OT2.Therefore ∠OTP � 90° (Pythagorus).Then PT ⊥ OT.

Review Exercise

11. Since AB �CD, ∠CAB � ∠ACD � 180°.Since ABDC is cyclic, ∠CAB � ∠BDC � 180°.Then ∠ACD � ∠BCD. Join AD and BC.In ∆ACD, ∆BDC, CD � CD (common)

∠ACD � ∠BDC (proven)∠CAD � ∠DBC (subtended by

chord CD)then ∆ACD � ∆BCD (two angles and side)therefore AC � BD.

P A O B

T

B

E

AD

C

AD B

E

C

X

Y

P

Q

B

C

A

Q

Py

x

xy


12. Since ∠ABC � 60° and ∠ACB � 45°,∠BAC � 75° (angle sum)

AH � AE (tangents from external point)then ∠AHE � ∠AEH

therefore ∠AHE � �12

�(180 � 75)°

� 52 �12

�°.

Similarly, ∠BHD � �12

�(180 � 60)°

� 60°.

Then ∠EHD � 180° � 60° � 52 �12

�°

� 67 �12

�°.

Similarly, ∠CED � �12

�(180 � 45)°

� 67 �12

�°.

Then ∠HED � 180° � 67 �12

�°

� 52 �12

�°

� 60°

and ∠HDE� 180° � 67 �12

�°

� 60°

� 52 �12

�°.

13. Let the isosceles triangle have AB as diameter of thecircle and let BC intersect the circle at D.We wish to prove that D is the midpoint of BC.Since AB is a diameter, ∠ADB � 90°.Then ∠ADC � 90°.In ∆ABD, ∆ACD, AB � AC (given)

AD � AD (common)∠ADB � ∠ADC (right angles).

Then ∆ABD� ∆ACD (hypotenuse, side, right triangles)Therefore BD � CD and D is the midpoint

of BC.

15. Assume that x is in degrees.Since TA is a tangent, ∠OAT � 90°.Then ∠AOT � (90 � x)°.Now ∆OAB is isosceles since OA � OB, and ∠AOT isan exterior angle of the triangle.Then ∠AOT � ∠OAB � ∠OBA

� 2∠OAB.

Therefore ∠OAB � �12

�(90 � x)°.

16. Using the given diagram, in ∆ADP we have ∠ADP � 180 � 4x.In ∆AQB, ∠ABQ � 180 � 5x.Since DBCA is a cyclic quadrilateral,

∠ADC � ∠CBA � 180 or 180 � 4x � 180 � 5x � 180

x � 20.

17. Since ∠RPT is an angle in a semicircle,∠RPT � ∠TPX � 90°.

Similarly, ∠TQX � 90°.We must now show that ∠PTQ � 90°.Draw TQ ⊥ to AB, meeting PQ at Y.Then TY is tangent to both circles.Therefore YP � YT � YQ (tangents from external point).Then ∆YPT is isosceles, so ∠YPT � ∠YTPSimilarly ∆YQT is isosceles, so ∠YQT � ∠YTQ.Then ∠PTQ � ∠PTY � ∠QTY � ∠YPT � ∠YQT.But ∠PTQ � ∠QPT � ∠PQT � 180° in ∆PQT.Therefore ∠PTQ � 90°.Then the angles in PXQT are all 90° and PXQT is a rectangle.

A

BO

Tx

A

BD

C

A

HE

B DC

x xA B

C D


18. Draw PR ⊥ OX and QS ⊥ OX and let QS have length r.Since P is equidistant from the arms of the angle,PO bisects ∠O and ∠POR � 30°.Then ∆POR has angles of 30°, 60°, and 90°.Therefore PO � 2PR � 20.Since the circles are tangent, PQ � 10 � r and QPOis a straight line.Then in ∆OQS, OQ � 30 � r.Now ∆OPR ~ ∆OQS (angles equal)

Then �OPR

P� � �

OQ

QS�

or �2100� � �

30r� r�

r � 30.The radius is 30.

19. Let the tangent at A meet the second circle at C. Join OB.Since OA and OB are radii, ∠OAB � ∠OBA.Since OA is a chord in the circle with centre Q and CAis tangent, ∠CAO � ∠OBA (tangent-chord property).

Then ∠CAO � ∠OAB.Therefore OA bisects ∠CAB.

20. Join BR, QR, and RP.Since BAQR is cyclic, ∠BRQ � ∠BAQ � 180°.Since ∠QAD � 180°, ∠BAD � ∠BAQ � 180°.Then ∠BAD � ∠BRQ.Since RBCP is cyclic, ∠BRP � ∠BCP � 180°.Since ∠DCP � 180°, ∠BCP � ∠BCD � 180°.Then ∠BCD � ∠BRP.But since ABCD is cyclic, ∠BAD � ∠BCD � 180°Therefore ∠BRQ � ∠BRP � 180°.Therefore QRP is a straight line.

21. Bisecting ∠A, ∠B, and ∠C gives ∠BAX � ∠XAC � 30°, ∠ABY � ∠YBC � 25°,and ∠BCZ � ∠ZCA � 35°.Now, using arc BX, ∠BAX � ∠BYX � 30° and using arc BZ, ∠BCZ � ∠BYZ � 35°.Then ∠XYZ � 65°.Similarly ∠YXZ � 60° and ∠XZY � 55°.

22. Using the given diagram, join AB. Then chord ABsubtends ∠APB in one circle and ∠AQB in the othercircle and the circles are identical. Therefore theangles are equal. Then ∆BPQ is isosceles and BP � BQ.

A

Y

C

X

B

Z

A

QR

PCD

B

A

C

O

B

Q

O

P

R

Q

S X

X

P

Y

Q

RA T B

S


23. Let AD and CB intersect at E. Since C and D arepoints on a semicircle, ∠ACB � ∠ADB � 90°.Then ∠FCB � ∠FDA � 90°.Since ∠FCB � ∠FDA � 180°, FCED is a cyclicquadrilateral.Join CD. Then ∠EFD � ∠ECD (subtended by chord ED).Now since chord BD subtends equal angles in thesemicircle, ∠BAD � ∠BCD.Therefore ∠BAD � ∠EFD.In ∆BAD, ∆FJB, ∠BAD � ∠JFB (proven)

∠DBA � ∠FBJ (same angle)then ∆BAD ~ ∆FJBtherefore ∠ADB � ∠FJBbut ∠ADB � 90° (angle in semicircle)therefore ∠FJB � 90°then FJ ⊥ AB.

Chapter 3 Test

1. a. x � 50 (tangent-chord property).b. a2 � 4.12 (secant property)

a � 4�3�.

c. 3b � 2 � 8 (secant property)

b � �136�.

d. x � 120 (exterior angle of cyclic quadrilateral).

y � 100 � 180 (cyclic quadrilateral)y � 80.

2. Using the given diagram, join AD.Then ∠CAD � 90° because it is an angle in a semicircle.Then ∠BAD � 140°.Since BADC is cyclic,∠BCD � ∠BAD � 180°Therefore ∠BCD � 40°.

3. Using the given diagram, M is the midpoint of ABbecause OM ⊥ AB, so AM � MB.In right-angled triangle OMB, OM � 8 and OB � 17.Then MB2 � 172 � 82

�225MB � 15.

In right-angled triangle OMC, OM � 8 and OC � 10.Then MC2 � 102 � 82

� 36MC � 6.

Then AC � MA � MC� 15 � 6 � 9.

4. Using the given diagram, and joining OD,∠ODC � 90° (angle in a semicircle).Then OD ⊥ CE.Therefore CD � DE (perpendicular to chord)

5. RP is tangent and RB is a secant in one circle.Then RP2 � RA. RB (tangent-chord property) and RQ2 � RA. RB.

Then RP2 � RQ2

or RP � RQand BR bisects PQ.

6. a. Let AB and CD be any two of the equal chords in acircle with centre O. Let X and Y be the respectivemidpoints. Then AX � CY. In ∆AXO, ∆CYO, AX � CY (proved)

OA � OC (radius)∠AXO � ∠CYO � 90°

(midpoint of chords).Then ∆AXO � ∆CYO (side, hypotenuse in

right triangles).Therefore OX � OY.

By a similar argument the lines joining everymidpoint of the equal chords to the centre areequal. Hence a circle with centre O and radius OXpasses through the midpoint of every one of theequal chords.

b. Since every one of these chords meets the circle ata point such that the line connecting this point isperpendicular to the chord, each chord is a tangent.

P

R

Q

A

B

F

C D

A BJ

E


7. Let AP extended meet tangent XTY at X. Join TP and TB.Then ∠XTP � ∠PAT (tangent-chord property)

and ∠YTB � ∠BAT (same)but ∠PAT � ∠BAT (bisected)

therefore ∠XTP � ∠YTB.Let ∠YTB be x°, so ∠XTP � x°.In ∆ATB, ∠ATB � 90° since it is subtended bydiameter AB.Then ∠ABT � (90 � x)° using the angle sum property.Now PTBA is a cyclic quadrilateral, and ∠XPT is anexternal angle, so ∠XPT � ∠ABT � (90 � x)°.In ∆XPT, ∠XPT � ∠XTP � (90 � x)° � 90°.Therefore ∠PXT � 90°, and XTY ⊥ APX.

Cumulative Review Chapters 1–3

13. Let AC and BT intersect at X. Join AT and BC.In ∆SBC, ∆NTA, SC � AN (given)

SB � NT (given)∠BSC � ∠TNA (opposite angles

in parallelogram)Then ∆SBC � ∠NTA (side-angle-side).Therefore BC � AT and ∠SCB � ∠NAT.Now, since MN � SR, ∠NAC � ∠SCA.

Then ∠NAC � ∠NAT � ∠SCA � ∠SCBor ∠TAX � ∠BCX.

In ∆BXC, ∆TAX, BC � AT (proved)∠BCX � ∠TAX (proved)∠BXC � ∠TXA (opposite)

then ∆BXC � ∆TAX (side-angle-angle)therefore BX � XTtherefore AC bisects BT.

14. Extend BA and DC to meet at X.By the secant property, XB · XA � XD · XCor (XA � AB)XA � (XC � CD)XCor XA2 � XA · AB � XC2 � XC · CDor XA2 � XC2 � �AB(XA�XC) (since AB � CD) or (XA �XC)(XA � XC) � �AB(XA �XC).Now either XA � XC � 0 or XA � CX � �AB.It is not possible that XA � XC � �AB, sinceXA � XC > 0.Therefore XA � XC � 0, and XA � XC.Then ∠ABC � ∠BDC since ∆XBC is isosceles.Since ABDC is a cyclic quadrilateral,∠ABD � ∠ACD � 180°.Since ∠BDC ∠ABD, ∠BDC � ∠ACD � 180°.Then AC � BD.

15. Let AB have length 2. Then BC � 1.Let BF have length x. Then FC � 1 � x.Now ∆ABC ~ ∆EFC since ∠C is in both triangles and∠EFC � ∠ABC � 90°.

Then �FB

CC� � �

EA

FB�,

or �1 �

1x

� � �2x

�

2 � 2x � x

x � �23

�.

Then square DEFB : ∆ABC � x2 : �12

� (2)(1)

� �49

� : 1

� 4 : 9.

AD

B

E

C

F

AX

C

D

B

x

M A N

T

RCS

B

X

P

A B Y

T

X

A

B

O

C

YD


16. In a regular pentagon each angle is 3 � �18

50°� � 108°.

In ∆ABC, ∠B � ∠C � ∠A � 180°.

Also ∠A � �12

� ∠B and ∠C � ∠B.

Then �52

� ∠B � 180°

∠B � 72°.Then ∠XBC � 36° and ∠C � 72°,

so ∠BXA � 108° (exterior angle of triangle).Now we must prove that XA � XB.Since ∠B � 72°, ∠A � 36°.Also ∠XBA � 36°.Therefore ∆XAB is isosceles and XA � XB.Then XA and XB are consecutive sides of a regularpentagon.

17. Using the given diagram and the property of tangentsfrom an external point, AX � AZ, BX � BY, and CY � CZ. Then AX · BY · CZ � BX · YC · ZA.

18. Let the polygon have n sides. Then the sum of allangles is 180(n � 2)°.There are n angles in the polygon, so the sum of the angles is 100 � 140 � 120(n � 2)° � 120n°.

Then 180(n � 2) � 120n60n � 360

n � 6.There are six sides to the polygon.

19. Using the given diagram, ∠ABC � ∠ACB since thetriangle is isosceles.Also ∠ABC � ∠BDE � 180° since DE � AC.Then ∠ACB � ∠BDE � 180°, and DBCE is a cyclicquadrilateral.

20. Since CD is a diameter, ∠CBA � 90° � 30° � 120°.Since BCDA is cyclic, ∠CBA � ∠ADC � 180°.Therefore ∠ADC � 60°.

21. Part 1. We prove that AE � BF if AE ⊥ BF. Since AE ⊥ BF, ∠FTE � 90°.Then ∠FCE � ∠FTE � 180° (∠FCE is an angle in a square).Then FCET is a cyclic quadrilateral.Therefore ∠TFC � ∠TEC � 180°.

Now ∠TEB � ∠TEC � 180° (straight angle).Then ∠TEB � ∠TFC.In ∆ABE, ∆BCF, AB � BC (sides of square)

∠ABE � ∠BCF (right angles)∠AEB � ∠BFC (proved).

Then ∆ABE � ∆BCF (side-angle-angle).Therefore AE � BF.

Part 2. We prove that if AE � BF then AE � BFIn ∆ABE, ∆BCF, AB � BC (sides of square)

AE � BF (given)∠ABE � ∠BCF (right angles).

Therefore ∆ABE � ∆BCF (side, hypotenuse in right-angled triangle).Then ∠AEB � ∠BFC.Now ∠AEB � ∠TEC � 180° (straight angle).Therefore ∠BFC � ∠TEC � 180°.Then FTEC is a cyclic quadrilateral.Therefore ∠FCE � 90° (angle in a square).Therefore ∠FTE � 90° and AE ⊥ BF.

D F C

ET

A B

A

B C

X


22. Since M and N are midpoints of AB and AC, MN � BC.Then ∆BNC and ∆BMC have common base BC andequal height.Therefore ∆BNC � ∆BMC.Since ∆BYC is common,∆BNC � ∆BYC � ∆BMC � ∆BYC

or ∆BYM � ∆CYN.Since N is the midpoint of AC,∆ABN � ∆NBC.Then ∆ABN � ∆BMY � ∆NBC � ∆CYN

Therefore quad AMYN � ∆BYC.

23. Join QT and RT. Since RP � RS, ∆RPS is isosceles.Then ∠RPS � ∠RSP.Then ∠PRQ � 2 ∠RPS (external angle).Since PQ � PR, ∆PQR is isosceles.Then ∠PQR � ∠PRQ � 2 ∠RPS.Now chord RT subtends ∠RPT � ∠RQT.

Therefore ∠RQT � �12

� ∠PQR, and QT bisects ∠PQR.

24. Using the given diagram and recalling the CosineLaw, x2 � 22 � 32 � 2(2)(3) cos � where � is ∠ABC.Since ABCD is cyclic,∠ADC � 180° � ∠ABC � 180° � �.Then x2 � 42 � 62 � 2(4)(6) cos (180° � �)

� 42 � 62 � 2(4)(6) cos �.Then 13 � 12 cos � � 52 � 48 cos �

60 cos � � �39

cos � � ��1230�.

Therefore x2 � 13 � 12 cos �

� 13 � �359�

� 20.8x � 4.6.

25. Note that ∠AHE is incorrectly labelled as a rightangle. In fact ∠BHC is 90°. Using tangent TAN andchord AC, ∠NAC � ∠CBA.Since ∠BHC � ∠BEC � 90°, a circle with BC asdiameter will pass through both H and E.Then HBCE is a cyclic quadrilateral, and ∠HEA is an external angle.Therefore ∠HEA � ∠HBC � ∠NAC.Since ∠HEA � ∠NAC, then TAN � HE.

T

A

N

HE

BC

P

T

Q R S

A

M N

B C

Y



Exercise 4.1

12. a.

The first leg of the trip is OA��. 240 km/h for 1�14

�

hours gives �OA�� 240 � �54

� � 300 km.

The second leg of the trip is AB�� and �AB��

240 � 2 � 480 km.

In ∆OAB, ∠OAB � 70° � 10° � 80°.

From the cosine law �OB��2 � 3002 �

4802 � 2 · 300 · 480 cos 80° �OB�� 519.99.

Length of the third leg is 520 km.

Let ∠ABO � � then from the sine law

�s3in00

��

si

�n

OB�8�0

�°

� � � 34.62°.

The displacement vector for the third leg, BO��,has a magnitude of 520 km at S 45° W.

b. The total distance the aircraft travelled is 300 � 480 � 520 � 1300 km. The time taken is

�1234000

� � 5.42 hours or 5 hours 25 minutes.

13. Since �(k � 2)v�� < �4v��k � 2��v�� < �4��v��.

Since v�� ≠ 0, �v�� ≠ 0 and �k � 2� < �4��4 < k � 2 < 4

and �2 < k < 6.

14. If u�� kv��, u�� and v�� have the same direction for k > 0

and opposite directions for k < 0. In either case u�� will

be parallel to v��.

If u�� is parallel to v��, u�� and v�� will have the same

direction or opposite directions hence u�� is some

multiple of v�� and u�� kv��.

Exercise 4.2

7. a.

In the diagram u�� v�� AC�� and

u�� v�� BD��.

AC�� and BD�� are the diagonals of �gm ABCD.

Now if �u�� v�� u�� v�� , the diagonals of the

parallelogram are equal, hence the parallelogram is

a rectangle and u�� ⊥ v��.

b. If �u�� v�� > �u�� v��, the angle between u�� and v��

will be acute.

c. If �u�� v�� < �u�� v��, the angle between u�� and v�� willbe obtuse.

9.

Let R�� 5a�� 2b��. Since �a�� 2 and �b�� 3,

�5a�� 10 and �2b�� 6. From the cosine law

�R��2 � 102 � 62 � 2 · 10 · 6 cos 50°

�R�� 7.672

∠AOC � � then from the sine law

�sin

6�

� – �sin

�R��5

�0°

�, � � 36.80°

5a�� 2b�� has a magnitude of 7.7 and makes an angleof 37° with 5a�� and 93° with �2b��.

b

a–2b

B C

R

A

b

50º5a

50º 10O

→

→

→ →

→

B C

A D

v

u

→

→

B

A

O

�10º

20º300 km

480 km

36 Chapter 4: Vectors

Chapter 4 • Vectors

12. a�� 3x�� 2y�� ➀

b�� 5x�� 4y�� ➁

Solving the two equations for x�� and y��

➀ � 2 � ➁

11x�� 2a�� b��

x�� 121�a��

111�b��.

Substitute into ➀ a�� 161�a��

131�b�� 2y��

11a�� 6a�� 3b�� 22y��

y�� 252�a��

232�b��.

16. Since a�� and b�� form the sides of a parallelogram and

since �a�� b��, the parallelogram will be a rhombus.

a�� b�� and a�� b�� will be the diagonals of this

rhombus and since the diagonals of a rhombus are

perpendicular to each other so will (a�� b��) be

perpendicular to (a�� b��).

17.

AB�� ED��. Then ABDE is a parallelogram

and AB�� AE�� AD��.

AF�� CD��. Hence ACDE is a parallelogram

and AC�� AF�� AD��.

Now AB�� AC�� AD�� AE�� AF��

� AB�� AE�� AC�� AF�� AD��

� AD�� AD�� AD��

� 3AD��.

But O is he midpoint of AD and ∆ABO is equilateral

therefore �AO�� 1 and �AD�� 2.

Now �AB�� AC�� AD�� AE�� AF�� 3�AD�� 6.

Since the sum is equal to 3AD��, the direction of the

sum is along AD�� which makes an angle of 60° with

AF��and 60° with AB��.

18.

In ∆ABC, ∠ABC � 180° � � and from the cosine

law �x � y�2 � �x��2 � �y��2 �2�x��y�� cos (180° � �).

But cos (180° � �) � �cos �.

Hence �x�� y��2� �x��2 � �y��2 � 2�x��y�� cos �. ➀

From ∆ABD, ∠ABD � � and from the cosine law

�x�� y��2 � �x��2 � �y��2 �2�x��y�� cos �. ➁

Adding ➀ and ➁

�x�� y��2 � �x�� y��2 � 2�x��2 � 2�y��2.

Now �x�� 11, �y�� 23, �x�� y�� 30.

Hence �x�� y��2 � 302 � 2 · 112 � 2 · 232 �x�� y�� 20.

19. Vectors u�� v�� and u�� v�� are the diagonals of a

parallelogram where u�� and v�� are adjacent sides. Since

�u�� v�� < �u�� v�� the angle between u�� and v�� will

be obtuse. Since the diagonals of a parallelogram

bisect each other, draw u�� v�� and u�� v�� bisecting

each other.

AC�� u�� v��

DB�� u�� v��

with AB�� u��, AD�� v��.

A

B

C

D

u

v

→

→

A �

C

D

B

y

– yx – y

x

y

– y

x + y→

→

→ →

→

→ →

→

→

A F

E

DC

B O

Chapter 4: Vectors 37


20.

a. FG�� FH�� HG��

� AC�� DA��

� j � k.

b. BH�� BG�� GH��

� j � k

DH�� DE�� EH��

� j � i

CH�� CG�� GH��

� i � k

FE�� FH�� HE��

� j � i

EG�� EH�� HG��

� i � k.

c. BD�� BG�� GC�� CE��

� j � i � k

BE�� i � j � k.

d. AH�� i � j � k

CF�� i � j � k

GD�� i � j � k.

e. Face diagonal is FG��.

�FG�� 1 � 1�� 2�.

Body diagonal is AH��.

�AH�� 1 � 1�� 1�� 3�.

21.

In ∆OAB, ∠OAB � 180° � �

and cos (180° � �) � �cos �.

From the cosine law

�u�� v��2 � �u��2 � �v��2 � 2�u��v�� cos (180° � �)

�u�� v��2 � �u��2 � �v��2 � 2�u��v�� cos �. ➀

In ∆OAC, AC�� v��, �AC�� v�� v��, ∠OAC � �.

From the cosine law

�u�� v��2 � �u��2 � ��v��2 � 2�u��v�� cos �

�u�� v��2 � �u��2 � �v��2 � 2�u��v�� cos �. ➁

Adding ➀ and ➁

�u�� v��2 � �u�� v��2 � 2�u��2 � 2�v��2.

�u�� v��2 � �u�� v��2 � 2(�u��2 � �v��2).

Exercise 4.3

9.

Represent the tensions in the cords by T��1

and T��2

asshown in the diagram.

�98 N

→–v

v u + v

u

u – v

C

– v

A

v

B

180º – ��

→

→ → →

→

→

→→

�O

D E

FH

B

A

G

C

ij

k

ˆˆ

ˆ


From the force diagram and the sine law,

�sin

�T��61�0°

� � �sin

�T��42�5°

� � �sin

9875°�

�T��1� � �

98si

snin75

6°0°

�

�T��2� � �

98si

snin75

4°5°

�

�T��1� � 87.9

�T��2� � 71.7.

The tension in the cord making an angle of 45° withthe ceiling is approximately 87.9 N and the tension inthe other cord is approximately 71.7 N.

10.

Represent the magnitude of the forces by a.From the cosine law we have

�R��2 � a2 � a2 � 2 · a · a cos 120°

2a2 � 2a2 · �12

� � 302

3a2 � 900a2 � 300a � 10�3�.

The magnitude of each force is 10�3� N.

11. An object will be in a state of equilibrium when theresultant of all the forces acting on it is zero. This meansthat the sum of any two magnitudes must be greater thanor equal to the magnitude of the third force.

a. Since 5 � 2 � 7 < 13, equilibrium cannot beachieved.

b. 7 N, 5 N, and 5 N can be arranged to produceequilibrium.

c. 13 � 14 � 17, hence equilibrium can be achieved.In this case the three forces would be collinear.

d. Since 12 � 13 � 24 < 26, equilibrium cannot be achieved.

12. a.

b. From the triangle of forces and the cosine law

82 � 52 � 72 � 2 · 5 · 7 cos �

cos � � �52

2�

·75

2

·�

782

�

� � 82°.The angle between the 5 N and 7 N forces will be 180° � 82° � 98°.

13.

Represent the tensions in the cords by T��1

and T��2

as

shown in the diagrams. From the triangle of forces

and the sine law,

�sin

�T��71�0°

� � �sin

�T��62�5°

� � �sin

68465°

�

�T��1� � �

68s6insi

4n57°0°

�

�T��1� � 911.6

�T��2� � �

68s6insi

4n56°5°

�

�T��2� � 879.3.

→

→

70º

686 N

65º25º

20º25º

20º 25º

70 kg

686 N

→ →

8

7

�

5

7

82º 98º

8

5

→aR

60º 120º

a

The tension in the rope making a 25° angle with thehorizontal is approximately 911.6 N and in the otherrope is approximately 879.3 N.

14.

The 20-, 15-, and 25- metre lengths form a right-angled triangle as shown in the diagram. Since the 375 N force is collinear with the 15 m steel wire, itwill have a tension of 375 N and the tension along the25 m steel wire will be 0 N.

15.

Let T��1

represent the tension in the wire and T��2

thecompression in the steel brace as in the diagrams.

Now sin 65° � �8

�T�5�1

0

��

�T��1� � �

sin85

605°

�

�T��1� � 937.9

tan 65° � �8

�T�5�2

0

��

�T��2� � �

ta8n5605°

�

�T��2� � 396.4.

The tension in the wire is approximately 937.9 N andthe compression in the steel brace is approximately396.4 N.

16.

Let �p��x� and �p��

y� represent the components of p�� along

the x-axis and y-axis respectively.

Now �u��y� � 5, �u��

x� � 0

�v��x� � 9 cos 40° � 6.9

�v��y� � �9 sin 40° � �5.8

�w��x� � �12 sin 65° � 10.9

�w��y� � �12 cos 65° � �5.1.

If p�� w�� v�� u�� then �p��x� � 12 sin 65° � 9 cos 40°

� 3.98

and �p��y� � �9 sin 40° � 12 cos 65° � 5

� �10.86.

17.

The horizontal component moving the log is 1470 cos 15° N � 1420 N.

18. a.

Let �F��p� and �F��

n� represent the parallel and

perpendicular components respectively.

�F��p� � 196 sin 28° �F��

n� � 196 cos 28°

� 92 � 173

The component parallel to the plane is 92 N andperpendicular to the plan 173 N.

→

→

28º

28º

20 kg

196 N

15º1470 cos 15º

1470 N

y

x

|u| = 5→

|v| = 9→|w| = 12→

65º40º

65º

850 N

65º

25º

850 N→

→

→

→

25º

375 N

15 m

20 m

25 m


b. The component normal to the ramp pushes downagainst the ramp and it in turn pushes back with anequal but opposite force. The component parallel to and down the ramp contributes to the luggage

sliding down the ramp. If �F��p� is greater than the

force of friction opposing F��p

, then the luggage willslide down the ramp.

19.

Let F�� represent the horizontal force and T�� the tensionin the rope. � is the angle the rope makes with thehorizontal.

Now cos � � �15.5�

� � 72.54°

sin � � �3

�T4��

3

��

�T�� sin

37423.54°�

�T�� 359.6

tan � � �3

�F�4

�

3

��

�F�� tan

37423.54°�

�F�� 107.9.

A force of 107.9 N will hold the girl in this positionand the tension in the rope is 359.6 N.

20. a.

Let �F��v� represent the vertical component and �F��

H�

be the horizontal component.

Now �F��v� � 66 cos 8°

� 54.5

�F��H� � 55 sin 8°

� 7.7.

b. The vertical component is approximately 54.5 kNand is the component that gives the helicopter lift.The horizontal component is approximately 7.7 kNand is the component that moves the helicopter in ahorizontal plane.

21.

The component that is parallel to the ramp is 2450 sin 25° � 1035.4. The force of friction,to oppose this, must have a magnitude of at least1035.4 N.

22.

If �F��x� is the horizontal component then

�F��x� � 320 cos 42°

� 237.8. The horizontal component causing the roller to moveis approximately 238 N.

320 N

42º

50 kg

490 N

→

25º

250 kg

2450 N

8º

55 kN

5

1.5

→

T

→F

35 kg

343 N

→

T

→F

343 N

�

�

�


23.

Since the forces are perpendicular to each other,consider them acting along the edges of a rectangularsolid with dimension 15 by 10 by 6. Now the

magnitudes of the forces �AD�� 6 N, �AE�� 10 N,and �AB�� 15 N. AG�� will be the sum of these forces

where �AG��2 � 62 � 102 � 152

�AG�� 19.

In ∆AGB, ∠GAB � �

and cos � � �1159�

� � 38°.

In ∆AEG, ∠EAG �

and cos � �1109�

� 58°.

The magnitude of the resultant is 19 N and it makesangles of approximately 58° and 38° with the 10 Nand 15 N forces respectively.

24.

Let R�� represent the vector along which the shipmoves. From the parallelogram and cosine law, we

have �R��2 � �F��2 � 4�F��2 � 2 · �F�� · 2�F�� cos 150°

� 5�F��2 � 2�3��F��2

�R�� 5 � 2��3�� F��

From the sine law

�s

�i

F

n��

��

sin

�R�1�5

�0°

�

sin � � ��

s

5

in

�

15

2

0

��°

3��

� � 9.89°.

The ship will move approximately 20° � 9.89° � 10°off the starboard bow.

25. a.

Let T��1

and T��2

be the tension in each length of

string. Since the mass is suspended from the

midpoint of the cord, �T��1� � �T��

2�. From diagram 1

the vertical components of T��1

and T��2

are �T��1� sin �

and �T��2� sin �. For equilibrium the sum of these

vertical components will be 400.

Therefore �T��1� sin � � �T��

2� sin � � 400.

But �T��1� � �T��

2� therefore 2�T��

1� sin � � 400.

From the 100, 80, 60 triangle, sin � � �45

�

hence �85

��T��1� � 400

�T��1� � 250.

The tension in each length would be 250 N hencethe string will support the weight.

OR

From diagram 2 and the sine law

�sin (9

�T�0

�

°1�� )

� � �si4n020�

�.

But sin (90° � �) � cos �, sin 2� � 2 sin � cos �

and sin � � �45

�.

Hence �T��1� � �

24s0in0

�

cocsos

�

��

s2in00

�� 250.

Conclusion as above.

→

→

�

→

→

100 cm

� �

400 NDiagram 1 Diagram 2

60 cm

80 cm

400 N

�

�

�

�

→

→

|F|

2 |F|

10º20º

R

150º|F|

→

→

→

→

�

H

E

G

F

C

B

D

A

10

15

6


b.

Represent the tensions as T��1

and T��2

and the angles

in ∆ABC and � and � as shown in the diagram.

Since AC � AB � 120, ∠ACB � ∠ABC � �.From the cosine law

802 � 1202 � 1202 � 2 · 120 · 120 cos �

cos � � �22··112200

2 �

· 128002

�

� � 38.94°.

Also � � 2� � 180°

therefore � � 70.53°.

From the sine law

�sin (9

�T0

��

°1�� )� � �

sin (9

�T�0

�

°2�� )

� � �sin (

4�

00� �)�

90° � � � 51.06°90° � � � 19.47°

� � � � 109.47°

�T��1� � �

40s0in

s1in09

5.14.70°6°

� � 330.0

�T��2� � �

40s0in

s1in09

1.94.74°7°

� � 141.4.

Since the tension �T��1� � 330 N > 300 N, the string

will not support the 400 N weight.

Exercise 4.4

2. a.

Let the angle to the bank be �. The componentperpendicular to the bank will be 2 sin �, the speedthat takes him across the river, and the componentparallel to the bank is 2 cos �. For the man to swimdirectly across the river then 2 cos � � 1

cos � � �12

�

and � � 60°.The man must swim at an angle of 60° to the bankif he is to reach a point directly across from hisstarting point.

b. If the speed of the current if 4 km/h, 2 cos � � 4,cos � � 2 which is not possible since �cos �� 1.He will not be able to swim to a point directlyacross the river in this case. As long as the currentis less than 2 km/h, he will be able to swim to apoint directly across the river.

3.

Let v��s, v��

b, and v��

trepresent the velocities of the

streetcar, bus, and taxi respectively and v��s

� 35,v��

b� �42, v��

t� 50 where north is positive and

south is negative.

N

W E

S

v tv s

v b

→

→

→

2 cos � � 2 km/h

2 sin �

1 km/h

400 N

�

90º – α

90º – �

�

α

→

→

�

400 N

120 cm

120 cm

A B�

�

C

80 cm

α

α

→

→


a. The velocity of the streetcar relative to the taxi,v��

s� v��

t� 35 � (�50) � �15, is 15 km/h south.

b. The velocity of the streetcar relative to the bus,v��

s� v��

b� 35 � (�42) � 77, is 77 km/h north.

c. The velocity of the taxi relative to the bus,v��

t� v��

b� 50 � (�42) � 92, is 92 km/h north.

d. The velocity of the bus relative to the streetcar,v��

b� v��

s� �42 � (35) � �77, is 77 km/h south.

4. a.

The distance downstream will be the distance

travelled in 6 min at 6 km/h, �110� � 6 � 0.6, 0.6 km.

He will touch the bank 0.6 km downstream fromthe marina and will be there in 6 minutes.

b. The boat will proceed across the river at a speed of20 km/h regardless of the speed of the current.Hence the time it takes to cross the river will be thetime it takes to travel 2 km at 20 km/h,

�220� � �

110�, 6 minutes.

5.

Let the resultant velocity be v��.

Now �v��2 � 4502 � 1002

�v�� 460.9772

and tan � � �140500

�

� � 12.53°.

a. The plane will travel a distance of

3 � �v�� 1383 km in 3 hours.

b. The direction of the plane is approximately N 13° E.

6.

Adding the vectors creates ∆OAB where OA�� is the

velocity of the aircraft, �OA�� 175, AB�� is the velocity

of the wind, �AB�� 40, and ∠BAO� 90° � 10° � 8°

� 72°.

OB�� v�� is the resultant velocity and ∠BOA � �.From the cosine law

�v��2 � 402 � 1752 � 2 · 40 · 175 cos 72°

�v�� 167.03.

From the sine law

�si4n0�

� � �sin

�v��7

�2°

�

sin � � �40 s

�i

v�n��

72°�

� � 13.17°.

The ground velocity is approximately 167 km/h in adirection N 5° W (13° � 8° � 5°).

W E

S

8º

B10º

A

N8º

�

O

N

W E

S

100

450 � v→

2 km

6 km/h

20 km/h


7.

Adding the vectors creates ∆OAB where OA�� is the

boat’s velocity, �OA�� 3; AB�� is the current’s velocity,

�AB�� 2, and ∠BAO � 55°, ∠BOA � �.

From the cosine law

�v��2 � 32 � 22 � 2 · 3 · 2 cos 55°

�v�� 2.473.

From the sine law

�sin

2�

� � �sin

�v��5

�5°

�

sin � � �2 si

�n

v��55°�

� � 41.48°.

The velocity is approximately 2.5 m/s in a direction ofN 56° W (41.48° � 15° � 56.48°).

8.

Adding the vectors forms ∆OAB where the plane

steering east at 240 km/h is represented by OA��,

the wind from the northwest is represented by AB��,

�AB�� 65, and the plane’s actual velocity is v�� where

∠AOB � �.From the cosine law

�v��2 � 2402 � 652 � 2 · 240 · 65 cos 135°

�v�� 289.63.

A

� 45º

B→v

O

A

B

40º15º

40º

→v

O


From the sine law

�si6n5�

� � �sin

�v1��35°�

sin � � �65 si

�n

v��135°�

� � 9.1°.The plane’s actual direction is approximately S 81° E.

9.

a. The horizontal component is 215 cos 18° � 204 km/h. The vertical component is 215 sin 18° � 66 km/h.

b. The horizontal component is the speed that the jetadvances. The vertical component is the speed atwhich the jet gains vertical altitude.

10.

OA�� represents the vector along which the plane steers.

�OA�� 520 km/h, ∠BOA � � hence the plane steers

at S (20 � �)° E. AB�� represents the wind velocity,

�AB�� 46 km/h and ∠ABO � 80° � 20° � 100°.

OB�� v�� represents the velocity with respect to the

ground.

From the sine law

�si4n6�

� � �sin

521000°� � �

sin ∠�v��

O

�AB

�

sin � � �46 s

5in20

100°�

� � 4.997°.

N

W E

SB

A46

520

10º80º

20º

�

→v

O

18º

215 km/h


∠OAB � 180° � � � 100°� 75°

∴ �v�� 520

sisnin

10∠0O°AB

�

�v�� 510.04.

The pilot should steer in a direction S 25° E and theplane’s ground speed will be approximately 510 km/h.

11.

The destroyer travels in a direction � as in the diagramand will intercept the sub in t hours. Hence thedistance DB � 30t nautical miles and SB � 20tnautical miles. ∠OSB � 135° and from the sine law

�s2in0t

��

sin3103t5°

�

sin � � �20 si

3n0135°�

� � 28°, 0 � 90°The destroyer should travel in a direction of N 62° Eto intercept the submarine.

12. a.

Represent the velocity of the aircraft as v�� and thewind velocity as w��, v > w. Let the distance betweenToronto and Vancouver be x km. The speed ingoing from Vancouver to Toronto with the wind is(v � w) km/h and from Toronto to Vancouver willbe (v � w) km/h. The time to go from Vancouver

to Toronto will be �v �

xw

� h and from Toronto to

Vancouver �v �

xw

� h.

Total time is Ta

� �v �

xw

� � �v �

xw

�

� x ��v(v�

�

ww

�

)(vv�

�

ww)

��

v22�

xvw2�.

b. When there is no wind, the time required to travel

from Vancouver to Toronto is �xv

� h and from

Toronto to Vancouver is �xv

� h.

Total time is Tb

� �2vx�.

Now Ta

� Tb

� �v2

2�

xvw2� � �

2vx�

� 2x��v2

v�

(v2(v�

2 �

w2)w2

��

v(v22x�

w2

w2)� > 0

Therefore Ta

� Tb

> 0

Since Ta

� Tb

> 0, Ta

> Tb, it takes longer to travel

from Vancouver to Toronto and back when there is

a wind.

13.

The speed relative to the ocean floor is represented byOA, a diagonal of a rectangular solid with sides oflength 0.5, 3, and 12, as shown in the diagram.

OA2 � (0.5)2 � 122 � 32

OA � 12.379The speed of the sailor relative to the ocean floor isapproximately 12.4 m/s.

14. Let v��c

represent the velocity of the car and v��tthe

velocity of the truck. Vector v��R, the velocity of the

truck relative to the car, is such that v��R

� v��t� v��

c.

80 km/h

50 km/h

→vc

→vt→vt

→vc →vR

�

A

N

E

0.5 m/s

3 m/s

12 m/s

O

V T

w

D S

B

30t20t

45º8

�

�v��R�2 � 802 � 502

�v��R� � 94.34

� is the angle between v��R

and v��t.

tan � � �8500�

� � 58°.The velocity of the truck relative to the car isapproximately 94.3 km/h in a direction N 32° E.

Review Exercise

7.

Represent the tension in the string by T�� and the anglethe string makes with the vertical by � as shown in the diagrams. Since the system is in equilibrium the sum of the threeforces will be O�� as shown in the triangle diagram.

Now �T��2 � 29.42 � 122

�T�� 31.75

tan � � �2192.4�, � � 22.2°.

The tension in the string has a magnitude of 32 N andthe string makes an angle of 22° with the vertical.

8. a.

Represent the resultant by R��. From the cosine law

�R��2 � �F��1�2 � �F��

2�2 � 2�F��

1��F��

2� cos 125°

� 542 � 342 � 2 · 54 · 34 cos 125°

�R�� 78.601.

The magnitude of the resultant is approximately 79 N.

b. �F��1� � 21, �F��

2� � 45, � � 140°

�R��2 � �F��1�2 � �F��

2�2 � 2�F��

1��F��

2� cos 40°

� 212 � 452 � 2 · 21 · 45 cos 40°

�R�� 31.909

The magnitude of the resultant is approximately 32 N.

9.

From the sine law

�sin

�R��5

�0°

� � �si

�

n

F��

71�

5°� � �

sin

�F��

52�

5°�

�R�� 480 N

�F��1� � �

�R��s

�insi

5n07°5°

�

�F��1� � 605.2

�F��2� � �

�R��s

�insi

5n05°5°

�

�F��2� � 513.3.

The magnitudes of the two forces are approximately 605 N and 513 N.

→R

75º

75º55º

→

F2

→

F1

180º – 130º= 50º

→F 2

F 1

→

R→

55º180º – 55º= 125º

→� �

12 N

3 kg

29.4 N

29.4

12

TT→


10.

Resolve the 2 N and 5 N forces into rectangularcomponents along and perpendicular to the 12 N force.

Let R�� be the resultant of F��1

and F��2

where

�F��1� � 12 � 2 cos 20° � 5 cos 40° � 17.7096

and �F��2� � 5 sin 40° � 2 sin 20° � 2.5299.

Now �R��2 � �F��1�2 � �F��

2�2

�R�� 17.8894

tan � � ��

�

F

F�

�

�

�2

1

�

��

� � 8.13°.The resultant has a magnitude of approximately 17.9 N and makes an angle of 8° with the 12 N forceand 32° with the 5 N force.

11.

Let T��1

and T��2

represent the tensions in each string and

� and � be the angles that the strings make with the

ceiling as shown in the diagram. In ∆OAB, OA � 7,

OB � 5, AB � 10. From the cosine law

72 � 102 � 52 � 2 · 5 · 10 cos �

cos � � �102

2�

· 55·

2

1�

072

�

cos � � �1295�, � � 40.5°.

Also 52 � 72 � 102 � 2 · 7 · 10 cos �

cos � � �3315�, � � 27.7°.

From the sine law

�sin (�

98� �)� � �

sin (9

�T0

��

°

1�

� �)� � �

sin (9

�

0

T��

°

2�

� �)�.

But sin (90° � �) � cos �, sin (90° � �) � cos �

and � � � � 68.2°

�T��1� � �

98si

cno6s82.72.°7°

�

� 93.5

�T��2� � �

98si

cno6s84.02.°5°

�

� 80.3.

The tension in the 5 m string is 93.5 N and in the 7 mstring is 80.3 N.

�α

α

90º – �

90º – α

98 N

→

T1

→

T2

A B10 m

�

→

T2

→

T1

�

α

α

10 kg

98 N

O

2 sin 20º

5 sin 40º

20º2 N

40º12 N

5 N

�

→

F2

→

F1

→

R


12.

To fly due east let the bearing of the plane be � northof east, and v�� represent the velocity due east. From thesine law

�si8n0�

� � �sin

801035°� � �

si

�n

v��

�

sin � � �sin

11035°�

� � 4.1°� � 180° � 135° � 4.1°

� 40.9°.

Now �sin

�4

v��0

�.9°

� � �sin

801035°�

�v�� 800

sinsin

13450°.9°

�

�v�� 740.8.

a. The plane’s heading should be N 85.9° E.

b. The time required to go 800 km at 740.8 km/h is

�784000.8

� � 1 hour 5 min.

13.

The wind will be from the north-west to push theplane on a flight of S 15° E. If w�� is the wind velocity,then from the cosine law

�w��2 � 4802 � 5282 � 2 · 480 · 528 cos 15°

�w�� 139.9

The wind speed is approximately 140 km/h.

14.

a. Resolve the velocity of 3 m/s into rectangularcomponents, 3 cos 30° with the current and 3 sin 30° perpendicular to the current. Her speeddownstream will be (4 � 3 cos 30°) m/s. Herdistance downstream in 10 s will be 10(4 � 3 cos 30°) � 65.98 m.

b. Her speed going across the river is 3 sin 30° m/s.The time required to go 150 m is

�3 s

1in50

30°� � 100 s.

15. a.

Let v�� represent the ground velocity. From thetriangle of vectors let � be the angle between v�� and450 km/h as shown. The angle opposite 450 is 60° � 25° � 85°. From the sine law

�si7n3�

� � �sin

45805°

�

sin � � �73 s

4i5n085°

�

� � 9.3°The pilot should steer on a heading of N 69° E.

73 km/h

25º

60º �450 km/h

→v

150 m

4 m/s

3 sin 30º

3 m/s30º

3 cos 30º

480 km/h 528 km/h

15º

→w

45º

80 km/h

800 km/h

45º

α

�→v


b. �sin

�v��8

�6°

� � �sin

45805°

�

�v�� 45

s0insi

8n58°6°

�

�v�� 450.62

The ground speed is approximately 451 km/h.

c. The time to fly 350 km is �345501

� � 0.776 hours or 47 min.

16.

Let v��R

be the relative velocity of the tanker to the cutter.

V��tthe velocity of the tanker

V��c

the velocity of the cutter.

Now v��R

� v��t� v��

c

�v��R� � 19 knots, �v��

c� � 12 knots.

From the cosine law

�v��t�2 � 122 � 192 � 2 · 12 · 19 cos 14°

�v��t� � 7.9.

From the sine law

�si1n2�

� � �sin

�v��1

t�4°

�

� � 21.5°.The actual velocity of the tanker is 7.9 knots on abearing of N 54° E.

17.

Let the displacement vector be s� and ∠FMN be �.∠MFN � 110°.

From the cosine law

�s��2 � 1732 � 2172 � 2 · 173 · 217 cos 110°

�s�� 320.46.

Now sin � � �217 s

�i

s�n

�110°�

� � 39.52°.The displacement vector has a magnitude of 320 kmwith a bearing of S 70° E.

18. au�� bv�� O��

au�� bv��

If u�� and v�� are not collinear then a � b � 0.

If u�� and v�� are collinear and have opposite directions

then �au�� bv��.

�a��u�� b��v��

Let ��a

v��

� � ��u

b��

��

� � k, k � R.

Now a � k�v�� and b � k�u��.

19. Case I.

�u�� > �v��. In the diagram, let OA�� u��, AB�� v��. Therefore

OB�� u�� v��. Locate C in OA so that CA � �v�� hence

OC � �u�� v��.In ∆CAB, CA � AB therefore ∠ACB � ∠ABC andeach of these angles is obtuse. In ∆OCB, ∠OCB is thelargest angle, therefore OB is the longest side, hence

OC < OB, i.e., �u�� v�� < �u�� v��.

OC A

B

→v

→u

→|v|

u + v→→

→b v

→a u

20º 50º173

70º

217M

F

N

�

→s

14º

14º� →

vR

→vc

→vt


Case II.

�u�� < �v��. Similar proof to the above (see diagram).

AD � �u��

DB � ��u�� v��In either case, ��u�� v�� < �u�� v��.

Equality holds if u�� is parallel to v�� but in the

opposite direction or if �u�� v�� 0 hence

��u�� v�� u�� v��.

Chapter 4 Test

1. �u�� v�� u�� v�� when u�� and v�� are collinear and

have the same direction.

2.

a. u�� a�� 3c��

b. v�� b�� a��.

c. w�� 23

�b�� 5c�� a��.

3. 3(4u�� v��) � 2u�� 3(u�� v��)

� 12u�� 3v�� 2u�� 3u�� 3v��

� 7u�� 6v��.

4. 4(a�� b��) � 4a�� 4b�� is called the distributiveproperty.

5. From the parallelogram law the resultant R�� OC�� and∠OBC � 55°.

Now �R��2 � 152 � 112 � 2 · 15 · 11 cos 55°

�R�� 12.5.The magnitude of the resultant is approximately 12.5 N.

→

AC

B

R

O

11 N

15 N

55º125º

4b

4aa

a + b

4(a + b)→

→→

→ →

→ →

b→

→– a

→v →

b

→u

→a →

a

→w

→c3

– 5→c

23– →

b

→c

→

b

→a

→|u|

O

B

A

D

→u

u + v→

→

→v

→ →

→u

u + v→v

→ →


6.

From the position diagram, the 6-m, 8-m, and 10-mlengths form a right triangle.

Therefore sin � � �35

�, sin � � �45

�, and � � � � 90°.

From the force diagram, � � � � 90°

and sin (90° � �) � �4

�T�

9

�2

0

��

therefore �T��2� � 490 sin �

� 490 � �45

�

�T��2� � 392

sin (90° � �)� �4

�T�

9

�1

0

��

therefore �T��1� � 490 sin �

� 490 � �35

�

�T��1� � 294.

The tension in each part of the cable is 294 N(8-m length) and 392 N (6-m length).

7.

Let the boat steer at an angle of � to the bank asshown in the diagram. The component of the boat’sspeed against the current is 4 cos � and the componentperpendicular to the current is 4 sin �. To go directlyacross the river, the component against the currentmust equal the current.

Therefore 4 cos � � 1.5

cos � � �14.5�

� � 67.976The boat must steer at an angle of 68° to go directlyacross. The speed going across the river is 4 sin 68°

m/s therefore the time to cross is �4 s

6in50

68°� � 175.29 s

or 2.9 min.

8.

Let v��a

represent the velocity of the aircraft, v��frepresent

the velocity of the fighter jet, and u�� represent therelative velocity of the fighter jet with respect to theaircraft. Hence u�� v��

f� v��

a.

Now �u��2 � 7352 � 3002 � 2 · 735 · 300 cos 60°

�u�� 640.0976.

Also �s3in00

��

sin

�u��6

�0°

�

sin � � �300

�s

u�i�n

�60°

�

� � 23.95°.

The relative velocity of the fighter jet with respect tothe aircraft is 640 knots with a direction of S 44° E.

50º50º

60º

70º

� →u

→va–

→vf

→va

1.5 m/s

4 cos �

4 sin �

�

4 m/s

α

α

�

�

→

T1

→

T2

10 m

8 m6 m

50 kg490 N

�

�

α

90º – α

90º – �

490 N

→

T1

→

T2

Position Diagram Force Diagram



Exercise 5.1

13. c. OG�� (�9, 12)

�OG��2 � (�9)2 � (12)2

� 81 � 144

� 225

�OG�� 5

tan � � ��192�.

Since G is in the second quadrant � ≅ 180� � 53�� ≅ 127�.

e. OJ��

2

5��, � �

�6

5��

�OJ��2 � �45

� � �65

�

� 2

�OJ�� 2�

tan � � � ��

2

6��.

Since J is in the fourth quadrant,� ≅ 360� � 51�� ≅ 309�.

14. a. If a�� (�12, �4, 6)

then �a��2 � 144 � 16 � 36 � 196

�a�� 14.

c. If c�� 1247�, �

�

2272

�, ��

277��

then �c��2 ��196 �

742894 � 49�

� 1�c�� 1.

15. If two unit vectors are the sides of an equilateraltriangle then their sum as well as their differencecould be a unit vector.

If a�� b�� c�� where �a�� b�� 1 and the angle

between a�� and b�� is 120� then �c�� 1. Also c�� b�� a��

where as above �a�� b�� c�� 1.

16. a. a�� (2, 3, �2)

�a��2 � 4 � 9 � 4

�a�� 17�.

b. ��a�1�� a��

2

17��, �

�3

17��, � �

�2

17��

Now ��2

17��

2� ��

3

17��

2� ��

2

17��

2� 1

therefore ��a�1�� a�� is a unit vector.

17. a. v�� 2i �3j �6k� (2, �3, �6)

�v��2 � 4 � 9 � 36 � 49

therefore �v�� 7.

b. A unit vector in the direction of v�� is

v�� v1�� v��

v�� 27

�i � �37

�j � �67

�k

v�� 27

�, ��37

�, ��67

��.

18. v�� (3, 4, 12)�v��2 � 9 � 16 � 144

� 169�v�� 13.

A unit vector in a direction opposite to

v�� is ��133�, ��

143�, ��

1123��.

→c

→a

→

b

60º

Chapter 5: Algebraic Vectors and Applications 53

Chapter 5 • Algebraic Vectors and Applications

19. Let a�� be a vector in two dimensions making an angle� with the x-axis. Now a�� in component form is

a�� a�� cos �, �a�� sin ��. A unit vector in the

direction of a�� is ��a�1�� a�� (cos �, sin �) therefore any

unit vector in two dimensions can be written as (cos �, sin �).

21. a. u�� (a, b, c)

cos α � ��ua��, since α is acute, a is positive.

b. u�� (a, b, c)

cos β � ��ub�� since β is obtuse, b is negative.

22. Since the direction angles α, β, and γ are all equal, say� then cos2 � � cos2 � � cos2 � � 1

cos2 � � �1

3�

cos � � ��11

3�� or cos � � �

�1

3��

� ≅ 55� � ≅ 125�The direction angles are 55� or 125�.

23.

In ∆OAB, ∠OAB � 90� therefore OB2 � OA2 � AB2.In ∆OBP, ∠OBP � 90� therefore OP2 � OB2 � BP2.Hence OP2 � OA2 � OB2 � BP2 but OA � a,OB � b, BP � OC � c.

Therefore �OP�� a2 � b�2 � c2�.

24. A vector in R4, u�� (4, 2, �5, 2) might have a

magnitude of �u��2 � 42 � 22 � (�5)2 � 22

�u��2 � 16 � 4 � 25 � 4

�u�� 49�, �u�� 7.

It is very tempting to think that a geometricinterpretation can be given. Mathematically we wishto consider vectors with n elements, n any integer, anda geometric interpretation is not possible for n ≥ 4.

Exercise 5.2

8. a. Given the points P(15, 10), Q(6, 4), R(�12, �8)PQ�� (�9, �6), PR�� (�27, �18)

� 3(� 9, �6).

Since PR�� 3PQ��, P, Q, and R are collinear.

b. D(33, �5, 20), E(6, 4, �16), F(9, 3, �12)

DE�� (�27, 9, �36), EF�� (e, �1, 4)� �9(3, �1, 4)

� �9EF��.

Since DE�� 9EF��, D, E, and F are collinear.

aA B

b

C c

P (a,b,c)

O

z

y

x

bß

→u

z

y

x

α→u

a

54 Chapter 5: Algebraic Vectors and Applications

9. b. A(0, 1, 0), B(4, 0, 1), C(5, 1, 2), D(2, 3, 5)

AB�� (4, �1, 1), CD�� (�3, 2, 3)

Since AB�� k � CD��, k � R, AB is not parallel

to CD.

�AB�� 16 � 1� � 1�� 18�� 3�2�

�CD�� 9 � 4�� 9�� 22�

�AB�� CD��.

10. PQRS is a parallelogram. The coordinates are in cyclic

order therefore PQ�� SR��. If R has coordinates (a, b)

PQ�� (�10, �1) and SR�� (a � 3, b � 4).

Since PQ�� SR��, a � 3� �10, a � �13b � 4 � �1, b � �5

The coordinates of S are (�13, �5).

11. Let the three vertices be A(�5, 3), B(5, 2), C(7, �8).There will be 3 possible positions for the fourthvertex. Let one position be P

1(a

1, b

1) for

parallelogram ACBP1.

CA�� BP��1

and (�12, �11) � (a1

� 5, b1

�2).Therefore a

1� �7, b

1� �9.

Let P2

(a2, b

2) be a vertex of parallelogram AP

2CB.

Now AP��2

� BC�� and (a2

� 5, b2

�3) � (2, �10)

therefore a2

� �3, b2

� �7.

P3(a

3, b

3) is a vertex of parallelogram ABP

3C.

Now BP��3

� AC�� and (a3

�5, b3

�2) � (12, �11)

therefore a3

� 17, b3

� �9.

The possible coordinates of the fourth vertex are P

1(�7, �9), P

2(�3, �7) and P

3(17, �9).

12.

Opposite faces of a parallelepiped are congruentparallelograms.

Now OP�� OC�� OB��

� (4, 0, �1) � (3, 6, 1)

OP�� (7, 6, 0).

OQ�� OA�� AR�� RQ��

� OA�� OB�� OC��

� (2, 4, �2) � (3, 6, 1) � (4, 0, �1)

� (9, 10, �2).

OR�� OA�� OB��

� (2, 4, �2) � (3, 6, 1)

OR�� (5, 10, �1)

OS�� OA�� OC��

� (2, 4, �2) � (4, 0, �1)

OS�� (6, 4, �3)The other 4 coordinates are (7, 6, 0), (9, 10, �2),(5, 10, �1), and (6, 4, �3).

P Q

SR

C(4, 0, –1)

O(0, 0, 0) A(2, 4, –2)

B(3, 6, 1)

y

x

B(5, 2)A(–5, 3)

C(7, –8)

Q(–6, 1)

y

x

R(a, b) S(–3, –4)

P(4, 2)


13. Let the midpoint in each case be M and the positionvector OM��.

a. A(�5, 2), B(13, 4)

OM�� OA��

2OB��

�

��(�5, 2) �

2(13, 4)�

OM�� (4, 3).

b. C(3, 0), D(0, �7)

OM�� OC��

2OD��

�

� �12

�(3, �7)

OM�� 32

�, �72

��.

c. E(6, 4, 2), F(�2, 8, �2)

OM�� OE��

2OF��

�

� �12

�(4, 12, 0)

OM�� (2, 6, 0).

d. G(0, 16, �5), H(9, �7, �1)

OM�� OG��

2OH��

�

� �12

�(9, 9, �6)

OM�� 92

�, �92

�, �3�.

14. a. 3(x, 1) �2(2, y) � (2, 1)(3x, 3) � (4, 2y) � (2, 1)(3x �4, 3 �2y) � (2, 1)

Equating components3x � 4 � 2, 3 � 2y � 1

x � 2 y � 1.

b. 2(x, �1, 4) �3(4, y, 6) ��12

�(4, �2, z) � (0, 0, 0)

Expanding and equating components2x � 12 �2 � 0x � �5,

�2 �3y � 1 � 0

y � ��13

�,

8 �18 ��12

�z � 0

z � �20.

15. Given points X(7, 4, �2) and Y(1, 2, 1)

XY�� (�6, �2, 3).

The magnitude of XY�� is �XY�� 36 � 4� � 9�� 7.

A unit vector in a direction opposite to XY has

components �67

�, �27

�, and ��37

�, or YX � ��67

�, �27

�, ��37

��.

16. a. A point on the y-axis has coordinates P(0, a, 0).Since it is equidistant from A(2, �1, 1) and B(0, 1, 3)

�AP�� BP�� or �AP��2 � �BP��2

therefore 4 � (a � 1)2 � 1� (a � 1)2 � 9a2 � 2a � 6� a2 � 2a � 10

4a� 4a� 1.

The point on the y-axis equidistant from A and B is(0, 1, 0).

b. The midpoint of AB is the point Q(1, 0, 2) which isnot on the y-axis and is equidistant from A and B.

17. a.

Since AM is a median of ∆ABC, M will be themidpoint of BC. Therefore the coordinates of M

are �2, ��12

�, ��52

��.

C(1, 3, –7) M

G

B(3, –4, 2)

A(2, 3, –4)2

–


Now AM�� 0, �2, �32

��and �AM�� 4 � �

94

�� 52

�.

The length of median AM is �52

�.

b. Let G be the centroid of ∆ABC.

Since AG:GM � 2:1 AG � �23

� AM � �53

�.

The distance from A to the centroid is �53

�.

18. In each case let the centroid be G with position

vector OG��.a. A(1, 2), B(4, �1), C(�2, �2)

OG�� OA�� O

3B�� OC��

� �13

�(1 � 4 � 2, 2 � 1 � 2)

OG�� 1, �13

��.

The centroid is �1, ��13

��.

b. I(1, 0, 0), J(0, 1, 0), K(0, 0, 1)

OG�� OI�� O

3J�� OK��

OG�� 13

�, �13

�, �13

��.

The centroid is ��13

�, �13

�, �13

��.

c. A1(3, �1), A

2(1, 1), A

3(7, 0), A

4(4, 4)

OG�� 14

� (OA��1

� OA��2

� OA��3

� OA��4)

� �14

� (3 � 1 � 7 � 4, �1 � 1 � 4)

� ��145�, 1�.

The centroid is ��145�, 1�.


d. O(0, 0, 0), I(1, 0, 0), J(0, 1, 0), K(0, 0, 1)

OG�� 14

� (1, 1, 1)

OG�� 14

�, �14

�, �14

��.

The centroid is ��14

�, �14

�, �14

��.

19. a. OG��

�

OG�� 1181�, ��

4111��.

The centre of mass is ��181�, ��

4111��.

b. OG��

� �111� (1 � 6 � 7, 4 � 21, �1 � 3 � 70)

OG�� 121�, ��

1171�, �

7121��.

The centre of mass is ��121�, ��

1171�, �

7121��.

Exercise 5.3

7. b. c�� (1, �2, 3), d�� (4, 2, �1)

c�� · d�� c��d�� cos �

c�� · d�� 4 � 4 � 3 � �3

�c�� 1 � 4�� 9�� 14�.

d�� 16 � 4� � 1�� 21�.

cos � � ��1

�

4� �3

21��

cos � ≅ 0.1750.

1(1, 4, �1) � 3(�2, 0, 1) � 7(1, �3, 10)��

1 � 3 � 7

(12 � 5 � 11, 3 � 35 � 9)��

11

2(0, 0) � 3(4, 1) � 5(�1, �7) � 1(11, �9)��

2 � 3 � 5 � 1

8. c. i � (1, 0, 0); �i � � 1

m�� (1, 1, 1); �m�� 3�i · m�� 1

i · m�� i ��m�� cos �

cos � � ��

1

3��

� ≅ 55�The angle between vectors i and m�� is 55�.

d. p�� (2, �4, 5); �p�� 4 � 16� � 25� � �45�q�� (0, 2, 3); �q�� 4 � 9� � �13�p�� · q�� 8 � 15 � 7

p�� · q�� p��q�� cos �

cos � � ��45�

7

�13��

� ≅ 73�

The angle between vectors p�� and q�� is 73�.

9. a�� (2, 3, 7); b�� (�4, y, �14)

a. a�� and b�� are collinear if a�� kb��, k � R

therefore (2, 3, 7) � k (�4, y, �14)

2 � �4k, 3 � ky, 7 � �14k

k � ��12

� k � ��12

�

Since k � ��12

�, y � �6, a�� and b�� will be collinear.

b. If the vectors are perpendicular a�� · b�� 0�8 � 3y � 98 � 0

3y � 106

y � �1036

�.

If y � �1036

�, a�� and b�� will be perpendicular.

10. u�� 3j � 4k � (0, 3, 4)v�� 2i � (2, 0, 0).

Let w�� (a, b, c).

Since w�� ⊥ u��, w�� · u�� 0, and 3b � 4c � 0

w�� ⊥ u��, w�� · v�� 0, and 2a � 0.Solving these equations, we have a � 0 and ifb � 4, c � �3.A possible vector w�� is (0, 4, �3).

11. Since a�� (2, 3, 4) and b�� (10, y, z) are

perpendicular a�� · b�� 0. Therefore 20 � 3y � 4z � 0

and y � ��43

�z � �230�.

12. u�� (1, 5, 8), v�� (�1, 3, �2).

a. LS � u�� · v��

� �1 � 15 � 16 � �2

RS � v�� · u��

� �1 � 15 � 16 � �2

therefore u�� · v�� v�� · u��.

b. LS � u�� · u��

� 1 � 25 � 64 � 90

RS � �u��2� 1 � 25 � 64 � 90

therefore u�� · u�� u��2.LS � v�� · v��

� 1 � 9 � 4 � 14

RS � �v��2� 1 � 9 � 4 � 14

therefore v�� · v�� v��2.

c. LS � (u�� v��) · (u�� v��) � (0, 8, 6) · (2, 2, 10) � 16 � 60 � 76

RS � �u��2 � �v��2� 90 � 14 � 76

therefore (u�� v��) · (u�� v��) � �u��2 ��v��2.


d. LS � �u�� v�� · �u�� v�� (0, 8, 6) · (0, 8, 6) � 64 � 36 � 100

RS � �u��2 � 2 u�� · v�� v�� 2

� 90 � 2 (�2) � 14 � 90 � 4 � 14 � 100

therefore (u�� v��) · (u�� v��) � �u��2 � 2 u�� · v�� v��2.

e. (2u��) · v��

� (2, 10, 16) · (�1, 3, �2) � �2 � 30 � 32 � �4

u�� · (2v��) � (1, 5, 8) · (�2, 6, �4) � �2 � 30 � 32 � �4

z(u�� · v��) � 2 (�2) � �4

therefore (u��) · v�� u�� · (2v��) � 2(u�� · v��).

13. u�� (2, 2, �1), v�� (3, �1, 0), w�� (1, 7, 8)

LS � u�� · (u�� w��) � (2, 2, �1) · (4, 6, 8) � 8 � 12 � 8 � 12

RS � u�� · v�� u�� · w��

� 6 � 2 � 2 � 14 � 8 � 12

therefore u�� · �v�� w�� u�� · v�� u�� · w��.

14. a. (4i � j ) · j � 4i · j � j · j

� 0 � � j �2

� �1.

b. k · (j � 3k) � k · j � 3k · k� 0 � 3(1) � �3.

c. (i � 4k) · (i � 4k) � i · i � 8i · b � 16k · k� 1 � 0 � 16 � 17.

15. a. (3a�� 4b��) · (5a�� 6b��)

� 15a�� · a�� 38a�� · b�� 24b�� · b��

� 15�a��2 � 38a�� · b�� 24�b��2.

b. (2a�� b��) · (2a�� b��)

� 4a�� · a�� b�� · b��

� 4�a��2 � �b��2.

16. a�� î � 3j � k, b�� 2î � 4j � 5k

� (�1, �3, 1) � (2, 4, �5)

3a�� b�� (�3, �9, 3) � (2, 4, �5)

� (�1, �5, �2)

2b�� 4a�� (4, 8, �10) � (�4, �12, 4)

� (8, 20, �14)

(3a�� b��) · (2b�� 4a��) � (�1, �5, �2) · (8, 20, �14)

� �8 � 100 � 28

(3a�� b��) · (2b�� 4a��) � �80.

17. Since 2a�� b�� is perpendicular to a�� 3b��,

(2a�� b��) · (a�� 3b��) � 0

therefore 2�a��2 � 5a�� · b�� 3�b��2 � 0

2�a��2 � 5�a��b�� cos � � 3�b��2 � 0.

But �a�� 2�b��.Substituting gives 8�b��2 � 10�b��2 cos � � 3�b��2 � 0

10 cos � � 5

cos � � �12

�

� � 60�.The angle between a and b is 60�.

18. Since a and b are unit vectors, a � �b� � 1.

a. (6a � b) · (a � 2b)� 6�a�2 � 11�a��b� cos � � 2�b�2

� 6 � 11 cos � � 2 � 4 � 11 cos �.

But � � 60� therefore 4 � 11 cos � � 4 � 11 cos 60�

� 4 � �121�

� ��32

�

(6a � b) · (a �2b) � ��32

�.


b.

Let � be the angle between the unit vectors a and b��.From the cosine law

�a � b�2 � �a�2 � �b�2 � 2�a��b� cos(180� � �).

Now �a � b� � �3� and cos(180� � �) � �cos �

therefore 3 � 1 � 1 � 2�a� �b� cos �1 � 2a · b

a · b � �12

�.

Now (2a � 5b) · (b � 3a) � �13a · b � 6�a�2 � 5�b�2

� ��123� � 6 � 5

(2a � 5b ) · (b � 3a) � ��121�.

19. a�� 3i � 4j � k � (3, �4, �1)

b�� 2i � 3j � 6k � (2, 3, �6)

a�� · b�� 6 � 12 � 6

� 0.

Since a�� · b�� 0, a�� ⊥ b�� and the parallelogram will be

a rhombus.

Let u�� and v�� be the sides of the rhombus.

u�� v�� a�� (3, �4, �1)

u�� v�� b�� (2, 3, �6)

Add: 2v�� (5, �1, �7)

Subtract: 2u�� (1, �7, 5)

The angle between u�� and v�� is the same as the angle

between 2u�� and 2v��.

Therefore (2u��) · (2v��) � �2u��2v�� cos �

5 � 7 � 35 � (1 � 49 � 25) cos �

cos � � ��2735�

� ≅ 108�.The angles between the sides of the rhombus are 108�and 72�.

Now �2u�� 75� � 5�3�, �u�� 5�

23�

�

the lengths of the sides of the rhombus are �5�

23�

�.

�

b

a

b

180º – �

ab+


20. a.

OA�� A��, OB�� AC�� b��, OC�� a�� b��.

Since a�� is perpendicular to b�� ∠OAC � 90�.

In ∆OAC, �OC��2 � �OA��2 � �AC��2

i.e., �a�� b��2 � �a��2 � �b��2.The usual name of this result is the PythagoreanTheorem.

b.

From ∆OAB, OA�� a��, OB�� b��

BA�� c�� a�� b�� and ∠BOA � �.

Now �BA��2 � �OA��2 � �OB��2 � 2�OA��OB�� cos �

�c��2 � �a��2 � �b��2 � 2 �a�� b�� cos �.

The result here is called the Cosine Law.

21.

In questions such as this a specific example canillustrate the desired result. Suppose a�� (3, �1)

and b�� (3, 4) then a�� · b�� 5

Now, for c�� (p, q), if a�� · c�� a�� · b��

3p � q � 5.There is an infinite number of possibilities for c��, oneof which is b��.

→a

→c

→

b

y

x

3x – y = 5a, b

�→a

B

A

→

b→c =

→a –

→

b

O

B C

A

→

b→

b

→a

→ a→

b+

O

However, others such as (1, �2) have c�� ≠ b��.

In fact c�� is any vector having its end point on the line 3x � y � 5.

22. Given vector a�� 4i � 3j � k� (4, �3, 1)

A vector parallel to the xy-plane has the formu�� (p, q, 0).

Since a�� ⊥ u��, a�� · u�� 0 and 4p � 3q � 0.

Choosing p � 3 and q � 4 gives vector u�� (3, 4, 0) which

is perpendicular to a��.

Now �u�� 9 � 16� � 5 therefore a unit vector in the

xy-plane perpendicular to a�� is u�� 35

�, �45

�, 0�.

23. Given that x�� y�� z� � 0 and �x�� 2, �y�� 3, �z�� 4.

Now (x�� y�� z�) · (x�� y�� z�) � o�� · o�� 0.

Therefore x�� · x�� y�� · y�� z� · z� � 2x�� · y�� 2x�� · z� � 2y�� · z� � 0

�x��2 � �y��2 � �z��2 � 2(x�� · y�� x�� · z� � y�� · z�) � 0

4 � 9 � 16 � 2(x�� · y�� x · z� � y�� · z�) � 0

and x�� · y�� x�� · z� � y�� · z� � ��229�.

24.

Two body diagonals of the cube are OP�� (1, 1, 1) and

AQ�� (�1, 1, 1).

Now OP�� · AQ�� OP��AQ�� cos � where � is an angle

between the body diagonals.

OP�� · AQ�� 1 � 1 � 1 � 1

�OP�� 3�, �AQ�� 3�.

Therefore cos � � �13

� and � ≅ 71�

The body diagonals of a cube make angles of 71� and 119�to each other.

z

y

xA(1, 0, 0)

B(0, 1, 0)

P(1, 1, 1)

C(0, 0, 1) Q(0, 1, 1)

O


25. a. Since (a�� b��) · (a�� b��) � 0, (a�� b��) ⊥ (a�� b��).

a�� b�� and a�� b�� represent the diagonals of a

parallelogram having sides a�� and b�� . Since the diagonals are perpendicular to each other, theparallelogram is a rhombus with �a�� b�� .

b. Since �u�� v�� u�� v��, �u�� v��2 � �u�� v��2.But (u�� v��) · (u�� v��) � �u�� v��2

and (u�� v��) · (u�� v��) � �u�� v��2

therefore (u�� v��) · (u�� v��) � (u�� v��) · (u�� v��)

u�� · u�� 2u�� · v�� v�� · v�� u�� · v�� 2u�� · v�� v�� · v��

4u�� · v�� 0

u�� · v�� 0.

Therefore u�� ⊥ v��.

u�� and v�� represent the sides of a rectangle whereas

a�� and b�� were the sides of a rhombus.

26. Since a�� · b�� a�� b�� cos � and �cos �� ≤ 1

�a�� · b�� ≤ �a�� b��.Equality holds when cos � � 1; i.e., a�� and b��

are collinear.

If a�� a1, a

2� and b�� b1, b

2�then �a

1b

1� a

2b

2� ≤ �a

12 ��a

22� �b

12 ��b

22�.

If a�� a1, a

2, a

3� and b�� b1, b

2, b

3�then �a

1b

1� a

2b

2� a

3b

3�

≤ �a12 ��a

22 ��a

32� �b

12 ��b

22 ��b

32�.

For a general solution to the Cauchy-Schwarzinequality refer to Exercise 12.2 question 18.

Exercise 5.4

7. i � (1, 0, 0), j � (0, 1, 0), k � (0, 0, 1)a. i j � (0, 0, 1) � k.b. k j � (�1, 0, 0) � �i.

8. a. Let u�� (u1, u

2, u

3)

v�� (v1, v

2, v

3)

�v�� (�v1, �v

2, �v

3)

u�� v�� (u2v

3� v

2u

3, v

1u

3�u

1v

3, u

1v

2� v

1u

2)

� v�� u�� (�v2u

3� u

2v

3, � u

1v

3� v

1u

3, � v

1u

2

� u1v

2)

� u�� v��.


b. If u�� and v�� are collinear, u�� kv��.

Let v�� (a, b, c) then u�� (ka, kb, kc)

u�� v�� (kbc � bkc, akc � kac, kbc � bkc) � (0, 0, 0)

u�� v�� 0.

9. (a�� · a��)(b�� · b��) � (a · b��)2

� �a��2 �b��2 � (�a��b�� cos �)2

� �a��2 �b��2 (1 � cos2 �)

� �a��2 �b��2 sin2 �

therefore RS � ��a��2�b��2�sin2 �� a��b��sin ��.

But 0� ≤ � ≤ 180� therefore sin � ≥ 0

and RS � �a��b�� sin �

� �a�� b��

therefore �a��b�� (a�� · a��)(�b · b) �� (a�� · b��)�2�.

10. a�� (2, 1, 0), b�� (�1, 0, 3), c�� (4, �1, 1)

a. a�� b�� · c�� (3, �6, 1) · (4, �1, 1)

� 12 � 6 � 1

� 19.

b. b�� c�� · a�� (3, 13, 1) · (2, 1, 0)

� 19.

c. c�� a�� · b�� (�1, 2, 6) · (�1, 0, 3)

� 19.

d. (a�� b��) c�� (3, �6, 1) (4, �1, 1)

� (�5, 1, 21).

e. (b�� c��) a�� (3, 13, 1) (2, 1, 0)

� (�1, 2, �23).

f. (c�� a��) b�� (�1, 2, 6) (�1, 0, 3)

� (6, �3, 2).

11. Let u�� v�� (a, b, c) and w�� (p, q, r)

v�� w�� (br � qc, pc � ar, aq � bp)

u�� (v�� w��) � [b(aq � bp) � c(pc � ar),

c(br � qc) � a(aq �bp), a(pc � ar) � b(br � qc)]

u�� v�� (0, 0, 0); (u�� v��) w�� (0, 0, 0)

hence u�� (v�� w��) ≠ (u�� v��) w��.

12.

a�� b�� is a vector that is perpendicular to both a�� and

b��. Let n�� a�� b��. Now n�� a�� is a vector that is

perpendicular to both n�� and a��. Therefore n�� a��, n�� and

a�� are perpendicular to each other; i.e., (a�� b��) a��,

a�� b�� and a�� are mutually perpendicular.

13. Let u�� (u1, u

2, u

3), v�� (v

1, v

2, v

3), and

w�� (w1, w

2, w

3).

Now v�� w�� (v2w

3� w

2v

3, w

1v

3� v

1w

3,

v1w

2�w

1v

2)

u�� · (v�� w��) � u1v

2w

3� u

1w

2v

3� u

2w

1v

3� u

2v

1w

3

� u3v

1w

2� u

3w

1v

2.

Also u�� v�� (u2v

3� v

2u

3, v

1u

3� u

1v

3, u

1v

2� v

1u

2)

and (u�� v��) · w�� w1u

2v

3� w

1v

2u

3� w

2v

1u

3

� w2u

1v

3� w

3u

1v

2� w

3v

1u

2

� u · (v�� w��).

14. a. We show this by choosing a�� (3, 1, 2), b � (�1,

1, 1), and c�� (p, q, r). Now a�� b�� (3, �5, �2)

and a�� c�� (r 2q, 2p � 3r, 3q � p).

If a�� b�� a�� c�� then r � 2q � 3 ➀

2p � 3r � �5 ➁

3q � p � �2 ➂

Choose q � k, from ➀ r � 2k � 3, from ➂

p � 3k � 2. These values for r and k satisfy ➁.

This shows that there are an infinite number of

possibilities for c��. For one such value choose

k � �2.

r � �1, p � �4, q � �2, and

c�� (�4, �2, �1)

a�� b�� a�� c�� (3, �5, �2) and b ≠ c.

n = a bb

a

(a b) a

→

→

→ → →

→ → →

b.

15. a. Given a�� (1, 3, �1), b�� (2, 1, 5), v�� (�3, y, z),

and a�� v�� b��.

a�� v�� (3z � y, 3 � z, y � a) � (2, 1, 5)Equating components gives y � �4, z � 2, and 3z � y � 6 � 4 � 2. Therefore v�� (�3, �4, 2).

b. Let v�� (a, b, c).

Now a�� v�� (3c � b, �a � c, b � 3a) � (2, 1, 5).

Equating components gives 3c � b � 2 ➀�a � c � 1 ➁b � 3a � 5. ➂

From ➀ and ➁ we have b � 2 � 3c and a � �1 � c.b � 3a � 2 � 3c � 3 (�1 � c) � 5 whichsatisfies ➂.

This shows that choosing any value for c in ➀,

substituting to find b and a from ➀ and ➁ will

satisfy ➂ hence giving another vector v��. Let

c ��2, from ➀ b � 8 and from ➁ a � 1.

Therefore v�� (1, 8, �2) is another vector so that

a�� v�� b��.

c. We see from part b that c is any real number, hencethere will be an infinite number of vectors v��.

Exercise 5.5

5. a.

A body diagonal is OA�� (1, 1, 1); an edge is OB��

� i � (1, 0, 0). The projection of i onto OA�� is

�i

�O· O

A��A�

��

� · ��O

O

A�

A�

�

��

(1, 0, 0

�

) ·

3�

(1, 1, 1)� · �

(1,

�

1,

3�

1)�

� �13

�(1, 1, 1)

� (�13

�, �13

�, �13

�).

b. The projection of a body diagonal onto an edge isthe projection of OA�� onto i which is i � (1, 0, 0).

6. a. a�� (1, 2, �2)

b�� (�1, 3, 0)

The area of the parallelogram is

�a�� b��.a�� b�� (6, 2, 5)

therefore �a�� b�� 36 � 4� � 25� � �65�b. c�� (�6, 4, �12) � �2(3, �2, 6)

d�� (9, �6, 18) � 3(3, �2, 6).

Since c�� 23

�d��, c�� and d�� are collinear therefore no

parallelogram is formed, hence its “area” is zero.

7. a. A triangle with vertices A(7, 3, 4), B(1, 0, 6) andC(4, 5, �2).

Two sides are defined by AB�� (�6, �3, 2)

and AC�� (�3, 2, �6)

AB�� AC�� (14, �42, �21)

� 7(2, �6, �3)

�AB�� AC�� 7�4 � 36� � 9�� 49

the area of ∆ABC � �12

��AB�� AC��

� �429�.

z

y

x

A(1, 1, 1)

B(1, 0, 0)

O

ac

b

a b

→

→

→

→ →


b. A triangle with vertices P(1, 0, 0), Q(0, 1, 0),R(0, 0, 1).

Now two sides are PQ�� (�1, 1, 0)

and PR�� (�1, 0, 1)

PQ�� PR�� (1, 1, 1), �PQ�� PR�� 3�.

The area of ∆PQR is ��23�

�.

8. Given a parallelepiped defined by a�� (2, �5, �1)

b�� (4, 0, 1)

c�� (3, �1, �1).

Now b�� c�� (1, 7, �4).

The volume of the parallelepiped is �a�� · (b�� c��)�

� �(2, �5, �1) · (1, 7, �4)�

� �2 � 35 � 4�

� 29.

9. Work W � F�� · d�� F��d�� cos �

a. W � 220 15 cos 49�

≅ 2165 J.

b. W � 4.3 2.6 cos 85�

≅ 1.0 J.

c. W � 14 6 cos 110�

≅ �29 J.

d. �F�� 4000 kN

� 4 106 N

�d�� 5 km

� 5 103 m

W � �F��d�� cos 90�

� 0 J.

10. To overcome friction, the applied force F�� must havemagnitude greater than 150 N. � � 0�, cos � � 1.

Therefore W > �F��d�� cos �

W > 150 1.5 1W > 225.

The work done is greater than 225 J.

11. �F�� 30 9.8 � 294 N.

W � �F��d�� cos �

� 294 40 cos 52°≅ 7240.

The work done is 7240 J.

12. Since �F�� 110 N, �d�� 300 m, � � 6�, W � 110 300 cos 6�≅ 32819.The work done is 32819 J.

13.

W � 78.4 3 cos 70�≅ 80.

The work done against gravity is approximately 80 J.

14.

Consider the “same” force as a force acting at 20� tothe direction of motion. The work done dragging the trunk up the ramp is 90(10) cos 20�. The workdone dragging the trunk horizontally is 90(15) cos 20�. Total work done is 900 cos 20� � 1350 cos 20�≅ 2114 J.

15. a. F�� 2i d�� 5i � 6j� (2, 0) � (5, 6)

W � F�� · d��

W � 10.

b. F�� 4i � j d�� 3i � 10j� (4, 1) � (3, 10)

W � F�� · d��

� 12 � 12W � 22.

c. F�� (800, 600) d�� (20, 50)

W � F�� · d��

� 16000 � 30000W � 46000.

d. F�� 12i � 5j � 6k d�� 2i � 8j � 4k� (12, �5, 6) � (�2, 8, �4)

W � F�� · d��

� �24 � 40 � 24W � �88.

12º20º 20º

F F

8 kg ≅ 78.4 N

20º70º

78.4 cos 70º

|d| = 3→


16. A 10 N force acts in the direction of a vector (1, 1).

A unit vector along F�� is ��

1

2��, �

�1

2�� therefore

F�� 10 ��

1

2��, �

�1

2�� (5�2�, 5�2�)

the displacement vector is d�� PQ�� (7, 5).

W � F�� · d�� 35�2� � 25�2�� 60�2�.

The work done is 60�2� N.

17.

The 30 N force acts along a�� (�2, 1, 5)

�a�� 4 � 1�� 25� � �30�

â � ��

3

2

0��, �

�1

30��, �

�5

30��.

The force vector F�� 30â

F�� (�2�30�, �30�, 5�30�)

The displacement d�� AB�� (1, �2 �3)

Therefore W � F�� · d��

� �2�30� 2�30� � 15�30�

� �19�30�.

The work done in moving the object from A to B is

�19�30� J.

18. a.

�F�� 50 N �r�� 20 cm � 0.2 m

� � 30�

�T�� r� F��

�T�� r� F��

� �r��F�� sin �

� 0.2 50 sin 30�

�T�� 5

The torque on the bolt is 5 N.

b. Since �T�� r��F�� sin �, maximum torque can be

achieved when sin � is a maximum. This maximum

is 1 when � � 90�. Therefore �T�� r��F�� 10

and the maximum torque that can be achieved is

10 J.

19. a. Proj(u�� onto v��) � Proj(v�� onto u��) is a true statement

when

i) u�� v�� or

ii) u�� ⊥ v��, in which case the projection vector is O��.

b. �Proj(u�� onto v��)� � �Proj(v�� onto u��)� is a true

statement when

i) u�� ⊥ v�� in which case the projection has

magnitude 0.

ii) when �u�� v�� and the angle between u�� and v��

is 45� or 135�

iii) when u�� v�� or u�� v��.

F

r

30º

Bolt→

→

A(2, 1, 5) B(3, –1, 2)

a→


Review Exercise

7. Given a�� 6i � 3j � 2k,

a�� (6, 3, �2)

b�� 2i � pj � 4k,

b�� (�2, p, �4)

and cos � � �241�, � is the angle between a�� and b��.

Now a�� · b�� a��b�� cos �.

Therefore

�12 � 3p � 8 �

�36 � 9� � 4� �4 � p2� � 16� �241�

3p � 4 � �241� · 7 �p2 � 2�0�

9p � 12 � 4 �p2 � 2�0� ➀

Squaring both sides:81p2 � 216p � 144 � 16p2 � 32065p2 � 216p � 176 � 0

(p � 4)(65p � 44) � 0

p � 4 or p � � �4645�.

We see that p � � �4645� does not satisfy ➀ and

p � 4 does; therefore the only value for p is 4.

8. Let a�� i � j � k � (1, 1, 1)

b�� λ2i � 2λj � k � (λ2, �2λ, 1)

Since a�� ⊥ b��, a�� · b�� 0

therefore λ2 � 2λ � 1 � 0

(λ � 1)2 � 0

λ � 1

If a�� ⊥ b�� then λ � 1.

9. If �x�� 3, �y�� 4 and the angle between x�� and y�� is

60� then

(4x�� y��) · (2x�� 3y��) � 8�x��2 � 10 x�� · y�� 3 �y��2

� 72 � 10�x��y�� cos 60� � 48

� 24 � 10(3)(4)��12

��(4x�� y��) · (2x�� 3y��) � 84.

10. uu�� has direction angles α1, β

1, γ

1. A unit vector along u��

is u� (cos α1, cos β

1, cos γ

1). Similarly a unit vector

along v�� is v � (cos α2, cos β

2, cos γ

2). Since u�� ⊥ v��,

u ⊥ v and u · v � 0 therefore cos α1

cos α2

� cos β1

cos β2

� cos γ1

γ2

� 0.

11. �x�� y��2 � (x�� y��) · (x�� y��)

� x�� · y�� 2 x�� · y�� y�� · y��

� �x��2 � 2 x�� · y�� y��2

2x�� · y�� x�� y��2 � �x��2 � �y��2

therefore x�� · y�� 12

� �x�� x��2 � �x��2 � �y��2.

12. Given ∆ABC with vertices A(�1, 3, 4), B(3, �1, 1),and C(5, 1, 1).

a. AB�� (4, �4, �3), AC�� (6, �2, �3), and

BC�� (2, 2, 0).

Since AB�� · BC�� 0, AB�� ⊥ BC��, and ∠ABC � 90�

therefore ∆ABC is a right-angled triangle.

b. Since ∆ABC has a right angle at B, the area of

∆ABC � �12

� �AB��BC��.

Now �AB�� 16 � 1�6 � 9��41�

�BC�� 4 � 4�� 2�2�

∆ABC � �12

� �41� 2�2�

� �82�the area of ∆ABC is �82�.

c. �AC�� 36 � 4� � 9�� 7

The perimeter of ∆ABC is AB � AC � BC �

�41� � 7 � 2�2� ≅ 16.2.

d. Let the fourth vertex to complete the rectangle beD(a, b, c)

CD�� BA�� and (a � 5, b � 1, c � 1) � (�4, 4, 3)equating components, a � 1, b � 5, c � 4, and thecoordinates of the fourth vertex are (1, 5, 4).


13. Given the vector u�� (17, �3, 8).

a. The projection of u�� onto each of the coordinatearea will be 17i, �3j and 8k.

b. The projection of u�� (17, �3, 8) onto the xyplane is (17, �3, 0), onto the xz plane is (17, 0, 8),and onto the yz plane is (0, �3, 8).

14. Since the vertices lie in the xy plane, the coordinatesin R3 will be A(�7, 3, 0), B(3, 1, 0), and C(2, �6, 0).

Now AB�� (10, �2, 0)

AC�� (9, �9, 0)

and AB�� AC�� (0, 0, �72).

Area of ∆ABC � �12

� �AB�� AC��

� �12

� 72

� 36.

The area of ∆ABC is 36.

15. a. Consider the base of the tetrahedron as a triangle inthe xy-plane with O(0, 0, 0), A(0, 1, 0).

Now ∆OBC is a 30�, 60�, 90� triangle with OB � 1

therefore OC � ��23�

� and CB � �12

�

hence the coordinates of B are ��

6

3��,�

12

�, 0�.

The centroid of ∆ABC is D��

6

3��,�

12

�, 0�.

D will be the foot of the perpendicular from thefourth vertex E to D. Let the coordinates of E be

��

6

3��, �

12

�, a�.

Now �OE�� 1

and �336� � �

14

� � a2 � 1

a2 � �2346� � �

23

�

a � ��

2�3�

�

the fourth vertex has coordinates ��

6

3��, �

12

�, ��

3

6��

or ��

6

3��, �

12

�, ��

3

6��.

The coordinates of the fourth vertices are

O(0, 0, 0), A(0, 1, 0), B��

6

3��

12

�, 0� and

E��

6

3��, �

12

�,� ��

3

6��.

b. The x-component of the centroid will be

�14

� �0 � 0 � ��

2

3��

�6

3��

�6

3��

the y-component is �14

� �1 � �12

� � �12

�� 12

�

the z-component is �14

� ��

3

6��

�12

6��.

The coordinates of the centroid are

G��

6

3��, �

12

�, � ��12

6��.

c. The distance from each vertex to the centroid will

be the same, say �OG��.

�OG�� 336� ��

14

� � �1�644��

� ��112� ��

14

� � �2�14��

� ��38

��

O(0, 0, 0) A(0, 1, 0)y

D

C

x

B( , ,0)212

3


� ��4

6��.

The centroid is ��46�

� units from each vertex.

16. a. a�� b�� is a vector that is perpendicular to all

vectors in the plane of a�� and b��. Let n�� a�� b��.

Now

n�� c�� is a vector perpendicular to both n�� and c��.

Since n�� c�� is perpendicular to n��, as are a�� and b��,

n�� c�� (a�� b��) c��, a�� and b�� will be coplanar;

i.e., (a�� b��) c�� lies in the plane of a�� and b��.

b. Let a�� (a1, a

2, a

3), b�� (b

1, b

2, b

3), and

c�� (c1, c

2, c

3).

Now

a�� b�� (a2b

3� a

3b

2, b

1a

3� a

1b

3, a

1b

2� a

2b

1)

and LS � (a�� b��) c��

� (c3b

1a

3� c

3a

1b

3� c

2a

1b

2� c

2a

2b

1,

c1a

1b

2� c

1b

1a

2� c

3a

2b

3� c

3b

2a

3,

c2a

2b

3� c

2a

3b

2� c

1b

1a

3� c

1a

1b

3)

a�� · c�� a1c

1� a

2c

2� a

3c

3

(a�� · c��)b � [(a1c

1� a

2c

2� a

3c

3)b

1,

(a1c

1� a

2c

2� a

3c

3)b

2,

(a1c

1� a

2c

2� a

3c

3)b

3]

and (b�� · c��)a�� [(b1c

1� b

2c

2� b

3c

3)a

1,

(b1c

1� b

2c

2� b

3c

3)a

2,

(b1c

1� b

2c

2� b

3c

3)a

3]

RS � (a�� · c��)b�� (b�� · c��)a��

� (a2c

2b

1� a

3c

3b

1� b

2c

2a

1� b

3c

3a

1, a

1c

1b

2

� a3c

3b

2� b

1c

1a

2� b

3c

3a

2, a

1c

1b

3� a

2c

2b

3

� b1c

1a

3� b

2c

2a

3).

Since LS � RS, (a�� b��) c�� (a�� · c��)b�� (b�� · c��)a��.

17. The volume of a tetrahedron is given by the formula

v � �13

�(area of the base)(height) � �13

� Ah.

Consider the base to be the triangle with vertices A(1, 1, 2), B(3, �4, 6), C(�7, 0, �1)

now AB�� (2, �5, 4) and AC�� (�8, �1, �3)

the area of the base will be A � �12

� �AB��AC��.


Since AB�� AC�� is a vector perpendicular to the plane of

∆ABC, the height of the tetrahedron will be the magnitude

of the projection of AD�� on (AB�� AC��)

therefore h ��AD��

�A·

B�(�A

B��

AC��A

�C��)�

�.

The volume V � �13

� �12

� �AB�� AC�� AD��

�A·

B�(�A

B��

AC��A

�C��)�

�

V � �16

� �AD�� · (AB�� AC��)�.

Now AB�� AC�� (19, �26, �42), AD�� (�2, 4, 6)

where D is the fourth vertex, D (�1, 5, 8)

AD�� · (AB�� AC��) � �38 � 104 � 252

� �394

V � �16

� 294

� �1937

�

The volume of the tetrahedron is �1937

�.

Chapter 5 Test

1. a. If u�� · v�� 0 then u�� is perpendicular to v��.

b. If u�� · v�� u��v�� then cos � � 1, � � 0� and u�� and v��

will have the same direction, i.e., u�� kv��, k > 0.

c. If u�� v�� 0�� then u�� and v�� are collinear,i.e., u�� kv��, k > R.

d. If �u�� v�� u��v�� then sin � � 1, � � 90� andu�� ⊥ v��.

e. If (u�� v��) · u�� 0, no conclusion can be made about u��

and v�� since u�� v�� is perpendicular to both u�� and v�� and

the dot product of perpendicular vectors is zero.

f. If (u�� v��) u�� 0�� then u�� v�� and u�� are collinear. But

u�� v�� is perpendicular to both u�� and v��. This

is true only if u�� v�� 0 in which case u�� and v��

are collinear.


2. Given u�� 6i � 3j � 2k � (6, 3, 2)

v�� 3i � 4j � k� (�3, 4, 1)

a. 4u�� 3 v�� (24, 12, 8) � (�9, 12, 3)

� (33, 0, 5)

� 33i � 5k .

b. u�� · v�� (6, 3, 2) · (�3, 4, 1)

� �18 � 12 � 2

� �4.

c. u�� v�� (�5, �12, 33)

� �5i � 12j � 33k.

d. �u�� v�� 25 � 1�44 � 1�089�

� �1258�.

A unit vector perpendicular to both u�� and v�� is

��

12

5

58��, �

��

1

1

2

2

58��, �

�3

1

3

258��.

3. a.

i) OP�� is the position vector of point P(3, �2, 5)

ii) the projection of OP�� onto the z-axis is

OA�� (0, 0, 5)

iii) the projection of OP�� onto the xy plane is

OB�� (3, �2, 0).

b. �OA�� 5

�OB�� 9 � 4� � �13�.

4. a.

ABCD is a parallelogram with coordinates A(�1, 2, �1), B(2, �1, 3), C(p, q, r),D(�3, 1, �3).

Now AB�� DC��

therefore (3, �3, 4) � (p � 3, q � 1, r � 3) and

p � 0, q � �2, r � 1 and the coordinates of C are

(0, �2, 1).

b. To determine the angle at A we use the dot product

AB�� · AD�� AB��AD�� cos A

AB�� (3, �3, 4)

�AB�� 9 � 9�� 16�

� �34�AD�� (�2, �1, �2)

�AD�� 4 � 1�� 4�� 3

AB�� · AD�� 6 � 3 � 8

� �11

cos A � ��3�

11

34��

A ≅ 129�

The angle at A is approximately 129�.

c. The area of parallelogram ABCD � �AB�� AD��

AB�� AD�� (10, �2, �9)

�AB�� AD�� 100 �� 4 � 8�1�

� �185�The area of parallelogram ABCD is �185.�

B

C

A

D

z

A

B

x

3

y

P(3, –2, 5)

O

5.

A force F�� acting at a direction of 35� to the horizontal

has magnitude �F�� 75 N moves a box a distance

�d�� 16 m.The work done is

W � �F��d�� cos �

� 75 16 cos 35�.

The same force acting at 35� to the horizontal has a

magnitude of �F�� 75 N and acts at an angle of 15�

to the line of motion where �d�� 8 m.

The work done is W � 75 8 cos 15�.

Total work done is

75 16 cos 35� � 75 8 cos 15�

≅ 1562.5 J.

6.

a. The force should act at right angles to the wrenchto produce maximum torque.

b. T�� r� F��

�T�� r� F�� r��F�� sin �.

If � � 90�, maximum torque is

�T�� 0.18 50 � 9 J.

The direction of T�� is perpendicular to the plane

of r� and F�� so that r�, F��, and r� F�� form a right-

handed system.

c.

Since sin 30� � �12

� a force applied at an angle of

30� will produce half the maximum torque

�T�� r��F�� sin 30�

� 0.18 50 �12

�

�T�� 4.5 J.

7.

Diagonal AC�� a�� b��

and BD�� a�� b��.

Let the angle between AC�� and BD�� be �.

Now AC��BD�� AC��BD�� cos �

AC�� · BD�� (a�� b��) · (a�� b��)

� �a��2 � �b��2

�AC�� a��2 � ��b��2�

�BD�� a��2 � ��b��2�

therefore �a��2 � �b��2

� ��a��2 � ��b��2� ��a��2 � ��b��2� cos �

�a��2 � �b��2 � (�a��2 � �b��2) cos �

and cos � � ��a

a

�

�

�

�

��

2

2

�

�

��b

b

�

�

�

�

��

2

2�.

For 0 < � < 90�, cos � � ��a

a

�

�

�

�

��

2

2

�

�

��b

b

�

�

�

�

��

2

2� for �a��2 > �b��2.

B C

A D

�

→a

→

b

30º

→

F

→r

→

T→

F

→r

20º

20º15º

→

F

→

d

→

F

→

d

35º


Chapter 6: Linear Combinations 71

Chapter 6 • Linear Combinations


Exercise 6.1

7. a. Let a�� (1, �1, 1), b�� (0, 1, 1), and c�� (1, 0, 2).

If a��, b��, and c�� are coplanar then one of these

vectors can be expressed as a linear combination

of the other two, i.e., a�� sb�� tc��.

By inspection of s � �1 and t � 1 we have

(1, �1, 1) � � (0, 1, 1) � (1, 0, 2) or

a�� b�� c, and a��, b��, and c�� are coplanar.

b. Let p�� (1, 0, 1), q�� (1, 1, 1), and r�� (1, 0, �1).

p�� q�� (�1, 0, 1) � �r��. Since p�� q�� is

perpendicular to the plane of p�� and q�� and collinear

with r��, r�� is perpendicular to the plane of p�� and q��

therefore p��, q��, and r�� are not coplanar.

8. a. Given u�� (2, 1, 1), v�� (�1, 1, 3).

i) (4, 2, 2) � su�� tv��. By inspection we have

s � 2, t � 0 and (4, 2, 2) � 2u�� 0v��.

ii) (1, 2, 4) � su�� tv��. By inspection we have

s � 1, t � 1 and (1, 2, 4) � u�� v��.

iii) (1, 5, 11) � su�� tv��

� s(2, 1, 1) � t(�1, 1, 3).

Equating components gives2s � t � 11 ➀

s � t � 51 ➁

s � 3t � 11 ➂

Add ➀ and ➁: 3s � 6, s � 2.From ➁ t � 3.

Check in ➂: 2 � 3(3) � 2 � 9 � 11

therefore (1, 5, 11) � 2u�� 3v��.

iv) (4, 5, 8) � su�� tv��

� s(2, 1, 1) � t(�1, 1, 3).

Equating components gives

2s � t � 4 ➀

s � t � 5 ➁

s � 3t � 8 ➂

Add ➀ and ➁: 3s � 9, s � 3.

From ➁ t � 2.

Verify in ➂: 3 � 3(2) � 3 � 6 � 9 ≠ 8.

Since s � 3 and t � 2 does not satisfy all three

equations, (4, 5, 8) cannot be expressed as a linear

combination of u�� and v��.

b. To determine another vector that can be written in

the form su�� tv��, choose values for s and t, i.e., let

s � 4 and t � � 3.

Now 4u�� 3v�� 4(2, 1, 1) � 3(�1, 1, 3)

� (8, 4, 4) � (3, �3, �9)

� (11, 1, �5).

c. To determine another vector that cannot be written

in the form su�� tv��, proceed as in b but change

one component, i.e., (11, 1, 7) cannot be written as

su�� tv�� since s � 4 and t � �3 will generate the

first two components but not the third.

9. u�� xa�� 2yb��, v�� 2ya�� 3yb��, and w�� 4a�� 2b��.

Since 2u�� v�� w��,

we have 2(xa�� 2yb��) � (�2ya�� 3yb��) � 4a�� 2b��

(2x � 2y) a�� (4y � 3y) b�� 4a�� 2b��

(2x � 2y� 4) a�� (y � 2) b�� o��.

Since a�� and b�� are not collinear, 2x � 2y � 4 � 0 and

y � 2 � 0 therefore y � �2 and x � 4.

10. a. a(2, 1, 0) � b(�3, 4, 5) � c(2, 0, 3) � (�4, 10, 7)Equating components gives

2a � 3b � 2c � �4 ➀a � 4b � 10 ➁

5b � 3c � 7 ➂2 � ➁ � ➀: 11b � 2c � 24 ➃2 � ➂ � 3 � ➃: 43b � 86

b � 2.Back substituting: a � 2 and c � �1.

b. a(3, �1, 2) � b(�1, 1, 3) � c(2, 1, 5) � (2, 5, 16)Equating components gives 3a � b � 2c � 26 ➀

�a � b � c � 56 ➁2a � 3b � 5c � 16 ➂

➀ � ➁: 2a � 3c � 7 ➃3 � ➀ � ➂: 11a � 11c � 22 � a � c � 2 ➄

➃ � 2 � ➄: c � 3.Back substituting: a � �1, b � 1.

11. a. u�� (1, 3, 2), v�� (1, �1, 1), w�� (5, 1, �4).

Since u�� · v�� 1 � 3 � 2 � 0, u�� ⊥ v��,

u�� · w�� 5 � 3 � 8 � 0, u�� ⊥ w��,

v�� · w�� 5 � 1 � 4 � 0, v�� ⊥ w��, and u��, v��, and

w�� are mutually perpendicular.

b. i � (1, 0, 0). Now (1, 0, 0) � a(1, 3, 2) � b(1, �1, 1) � c(5, 1, �4).Equating components a � b � 5c � 1 ➀

3a � b � c � 0 ➁2a � b � 4c � 0 ➂

➀ � ➁: 4a � 6c � 1 ➃➁ � ➂: 5a � 3c � 0 ➄➃ � 2 � ➄: 14a � 1

a � �1

1

4�, c � �

4

5

2�, b � �

1

3�

Therefore i � �1

1

4� u��

1

3� v��

4

5

2� w��.

j � (0, 1, 0), therefore a � b � 5c� 0 ➀3a � b � c � 1 ➁

2a � b � 4c � 0 ➂➀ � ➁: 4a � 6c � 1 ➃➁ � ➂: 5a � 3c � 1 ➄➃ � 2 � ➄: 14a � 3

a � �1

3

4�, c � �

4

1

2�, b � �

1

3�.

Therefore j � �1

3

4� u��

1

3� v��

4

1

2� w��.

k� (0, 0, 1), therefore a � b � 5c� 0 ➀3a � b � c � 0 ➁

2a � b � 4c � 1 ➂➀ � ➁: 4a � 6c � 0 ➃➁ � ➂: 5a � 3c � 1 ➄➃ � 2 � ➄: 7a � 1

a � �1

7�, c � ��

2

2

1�, b � �

1

3�

therefore k � �1

7� u��

1

3� v��

2

2

1� w��.

12. u�� (5, �5, 2), v�� (1, 8, �4), w�� (�2, �1, 2), and

x�� (�3, 6, 8).

a. Now x�� au�� bv�� cw�� gives rise to the equations 5a � b � 2c � �3 ➀

�6a � 8b � c � 6 ➁2a � 4b � 2c � 8 ➂

➀ � ➂: 7a � 3b � 50 ➃2 � ➁ � ➂: �8a � 12b � 20 ➄

➃ � ➄ � 4: 5a � 10

a � 2, b � 3, and c � 8

and x�� 2u�� 3v�� 8w��.

72 Chapter 6: Linear Combinations

b. û � ��u1�� u�� therefore u�� u��û

�u�� 25 � 2�5 � 4� � �54� � 3�6�

u�� 3�6� û.

Similarly v�� v�� v, �v�� 1 � 64� � 16� � 9,

v�� 9v

w�� w�� w � �w�� 4 � 1�� 4� � 3, w�� 3w

therefore x�� 6�6� û � 27v � 24w.

13. a.

Let x�� au�� bv��. Draw lines from the tip of x��

parallel to u�� and v�� as in the diagram. The sides of

the parallelogram are the vectors �au�� and bv��.

From ∆OAB

�s

�i

�

n

a

8

u�

0

��

� � �si

�n

bv��

7

�0�

� � �sin

�x��3

�0�

� � �sin

4

30�� 8

�si

a

n

�u�8

��0�

� � 8

a � �8 sin

280��

� 4 sin 80°

a ≅ 3.9392

�si

b

n

�v��7

�0�

� � 8

b � �8 si

1n070��

� �45

� sin 70�

b ≅ 0.75175

x�� 4 sin 80� u�� 45

� sin 70� v��

x�� 3.94 u�� 0.75 v��.

y

B

A

x

80º

30º

20º50º

40º 40º 30º20º

→v

→u

→vb

→u– a

→x

b. û � ��u1�� u�� therefore u�� 2u . Similarly v�� 10v.

Therefore x�� 8 sin 80� u � 8 sin 70� v

x�� 7.88 u � 7.52 v.

Exercise 6.2

6. Vectors u�� and au�� are scalar multiples, hence will be

parallel. Similarly v�� and bv�� are parallel. Since u�� and v��

are linearly independent they will not be parallel,

hence au�� is not parallel to bv��. Therefore au�� and bv��

are linearly independent.

7. Assume v�� ai � bj � ck and v�� pi � qj � rk

therefore ai � bj � ck� pi � qj � rk

(a � p) i � (b � q) j � (c � r)k� o��.

Since i, j, and kare linearly independent a � p � 0, a � p b � q � 0, b � qc � r � 0, c � rtherefore vectors expressed in terms of i, j, and kare unique.

8. If (�1, 1, 1), (1, �1, 1), and (1, 1, �1) can be used asbasis in 3-space then a(�1, 1, 1) � b(1, �1, 1) � c(1, 1, �1) � o�� where a,b, and c are not all zero.Equating components �a � b � c � 0

a � b � c � 0a � b � c � 0

Solving we have a � b � c � 0. The vectors arelinearly independent hence can be used as a basis in 3-space.

9. To show a��, a�� b�� and a�� (a�� b��) can be used as

a basis for vectors in space, we need to establish

that the three vectors are not coplanar. Since a�� is

perpendicular to a�� b��, a�� and a�� b�� will not be

collinear, hence will define a plane. Now a�� (a�� b��)

is a vector perpendicular to both a�� and a�� b��

(i.e., perpendicular to the plane of a�� and a�� b��),

hence does not lie in the plane of a�� and a�� b��.

Therefore a��, a�� b��, and a�� (a � b��) are not coplanar

and can be used as a basis for vectors in space.

10. a. v��1

� (1, 2), v��2� (3, 5) and v��

1≠ kv��

2therefore v��

1and

v��2

can form bases for two dimensional space.

Now v�� av��1

� bv��2

� (8, 7)

a � 3b � 8 ➀

2a � 5b � 7 ➁

2 � ➀ � ➁: b � 9, a � �19.

The coordinates of v�� with v��1

and v��2

bases is

(�19, 9).

b. v��1

� (3, 5), v��2

� (6, 10). Since v��2

� 2v��1, v��

1, and v��

2

are collinear hence cannot form a bases in R2.

11. a. v��1

� (�1, 0, 1), v��2

� (2, 1, 1), v��3

� (3, 1, 1).

Now v��1

� v��2

� (�1, 3, �1)

and v��3

· (v��1

_ v��2) � �3 � 3 � 1 � �1 ≠ 0.

Therefore v��1, v��

2, and v��

3are not coplanar and can

form a bases in R3.

av��1

� bv��2

� cv��3

� v�� (1, 2, 3).

Equating components �a � 2b � 3c � 1 ➀

b � c � 2 ➁a � b � c � 3 ➂

➂ � ➁: a � 1 ∴ 2b � 3c � 2b � c � 2

from which c � �2 and b � 4.

The coordinates of v�� will be (1, 4, �2).

b. v��1

� (1, 3, �1), v��2

� (2, 1, 1), v��3

� (�4, 3, �5).

Now v��1

� v��2

� (4, �3, �5)

and v��3

· (v��1

� v��2) � �16 � 9 � 25 � 0

therefore v��1, v��

2, and v��

3are coplanar and do not

form a bases in R3.

c. v��1

� (1, 0, 0), v��2

� (1, 1, 0), v��3

� (1, 1, 1). We

have v��1

� v��2

� (0, 0, 1) and v��3

· (v��1

_ v��2) � 1 ≠ 0

therefore v��1, v��

2, and v��

3are not coplanar and can

form a bases in R3.

av��1

� bv��2

� cv��3

� v�� (1, 2, 3).

Equating components a � b � c � 1b � c � 2

c � 3.Solving gives c � 3, b � �1 and a � �1. The

coordinates of v�� will be (�1, �1, 3).


12. Since (3u�� 4v��) and (6u�� 2v��) are linearlyindependent,a(3u�� 4v��) � b(6u�� 2v��) � 0�� and a � b � 0

3au�� 4av�� 6bu�� 2bv�� 0��

(3a � 6b)u�� (4a � 2b)v�� 0��.

Since a � b � 0, 3a � 6b � 0, and 4a � 2b � 0,

giving 0u�� 0v�� 0�� hence u�� and v�� are linearly

independent.

13. Suppose that d�� ka�� lb�� mc�� is not a unique

representation. There then exists another

representation, say d�� pa�� qb�� rc��. Subtracting,

we have (k � p)a�� (l � q)b�� (m � r)c�� 0��. Since

a��, b��, and c�� are basis vectors and hence not coplanar,

then k � p � 0, l � q � 0, and m � r � 0 from

which k � p, l � q, and m � r and a contraction is

evident. Hence d�� ka�� lb�� mc�� is a unique

representation.

14. u��, v��, and w�� are mutually perpendicular, linearly

independent vectors. If a(u�� v��) � b(v�� w��) �

c(w�� u��) � 0�� and a � b � c � 0, then u�� v��, v�� w��,

and w�� u�� will be linearly independent.

Expanding and collecting terms gives

(a � c)u�� (a � b)v�� (b � c)w�� 0��. Since u��, v��,

and w�� are linearly dependent,

a � c � 0 ➀

a � b � 0 ➁

b � c � 0 ➂

Solving: ➀ � ➁ � ➂ 2a � 0, a � 0, c � 0,

and b � 0. Therefore u�� v��, u�� w��, and v�� w�� are

linearly independent.

15. Since (1 � s)u�� 23

�v�� and (3u�� sv��) are parallel

(1 � s)u�� 23

�v�� k(3u�� sv��), k � R.

Expanding and collecting terms gives

(1 � s � 3k)u�� (��23

� � ks)v�� 0��.

Since u�� and v�� are linearly independent,

1 � s � 3k � 0 ➀

and ��23

� � ks � 0 or 2 � 3ks � 0 ➁

From ➁ k � ��32s�

Substitute into ➀ 1 � s � �2s

� � 0

s2 � s � 2 � 0(s � 2)(s � 1) � 0

s � 2 or s � �1.If the vectors are parallel, s � 2 or s � �1.

Exercise 6.3

5.

Since T divides AB in the ratio 2:�1, T will be

external to AB as shown in the diagram.

In ∆OAT,

OT�� OA�� AT��

but AT�� 2AB��

therefore OT�� OA�� 2AB��.

Now in ∆OAB, AB�� OA�� OB��

therefore OT�� OA�� 2(�OA�� OB��)

OT�� OA�� 2OB��.

6.

Given OB�� 23

� OC�� 13

� OD��.

From ∆OBC,

CB�� OC�� OB��

� �OC�� 23

� OC�� 13

� OD��

C

B

D

O

A

B

T2

O


� � �13

� OC�� 13

� OD��

� �13

� ��OC�� OD��.

But CD�� OC�� OD��

therefore CB�� 13

� CD�� and C, B, and D are collinear.

11.

Let P divide AB in the ratio 1:2

therefore OP�� 23

� OA�� 13

� OB��

� �23

� (3, 6, 8) � �13

� (6, 0, �1)

� (4, 4, 5). The coordinates of P are (4, 4, 5). The midpoint of PB is Q(5, 2, 2). The points of trisection are (4, 4, 5)and (5, 2, 2).

13.

Since OE�� 25

� OD�� 75

� OF��,

E: is external to DF,is beyond F,divides DF in the ratio 7:�2.

Since OG�� 15

� OD�� 45

� OF��,

G: is between D and Fdivides DF in the ratio 4:1.

In both cases DF is 5 units.

a. D divides GE in the ratio �4:7.

b. F divides GE in the ratio 1:2.

14.

P divides AB in the ratio a:b. Let AP � ak and PB � bk. Now the altitude of ∆OAP � h and thealtitude of ∆OBP � h.

∆OAP � �12

� akh

∆OBP � �12

� bkh

�∆∆

OO

BAP

P� � �

ab

�.

15.

Given OD�� rOA�� sOB�� tOC�� and r � s � t � 1.

Now r � 1 � s � t

therefore OD�� (1 � s � t)OA�� sOB�� tOC��

OD�� OA�� s(�OA�� OB��) � t(�OA�� OC��).

In ∆OAB, �OA�� OB�� AB��,

In ∆OAC, �OA�� OC�� AC��.

Therefore OD�� OA�� sAB�� tAC��

OD�� OA�� sAB�� tAC��.

In ∆OAD, OD�� OA�� AD��

therefore AD�� sAB�� tAC��.

AD�� is expressed as a linear combination of AB�� and

AC�� therefore AD��, AB��, and AC�� are coplanar, hence A,

B, C, and D are coplanar.

AB

C

DO

A PB

ak bk

h

O

D

G

F

E

4

1

7

2

O

P

A(3, 6, 8)

1

2

Q

B(6, 0, –1)


Exercise 6.4

1.

a. Let P and Q be midpoints of AB and AC respectively.

Therefore BP�� PA�� a�� and

CQ�� QA�� b��.

In ∆PAQ, PQ�� a�� b��.

In ∆BAC, BC�� 2a�� 2b��

� 2(a�� b��)

BC�� 2PQ��.

Therefore BC�PQ and BC � 2PQ

or PQ � �12

� BC.

b. From the midpoint theorem OP�� 12

� OA�� 12

� OB��

and OQ�� 12

� OA�� 12

� OC��

Now PQ�� OP�� OQ��

� ��12

� OA�� 12

� OB�� 12

� OA�� 12

� OC��

12

��OB�� OC��but BC�� OB�� OC�� therefore PQ��

12

� BC��

hence PQ�BC and PQ � �12

� BC.

A

Q

C

B

P

→a

→a

→

b

→

b

O


2.

a. Since P and Q trisect BC, let BP�� PQ�� QC�� a��.

In ∆ABP, AB�� AP�� a��.

In ∆ACQ, AC�� AQ�� a��.

Adding gives AB�� AC�� AP�� AQ��.

b. Since P and Q trisect BC

OP�� 23

� OB�� 13

� OC��

and OQ�� 13

� OB�� 23

� OC��.

Adding OP�� OQ�� OB�� OC��.

c. In ∆AOB, AB�� OA�� OB��.

In ∆AOC, AC�� OA�� OC��.

Adding: AB�� AC�� 2OA�� OB�� OC��.

Similarly from ∆AOP, AP�� OA�� OP��

∆AOQ, AQ�� OA�� OQ��

and AP�� AQ�� 2OA�� OP�� OQ��

but OP�� OQ�� OB�� OC��.

Therefore AB�� AC�� AP�� AQ.��

3. A

FD

B EC

B

A

P Q C→a

→a

→a

Since D, E, and F are the midpoints of the sides of∆ABC as shown,

OD�� 12

� OA�� 12

� OB��,

OE�� 12

� OB�� 12

� OC��,

and OF�� 12

� OC�� 12

� OA��.

Adding gives OD�� OE�� OF�� OA�� OB�� OC��.

4.

ABCD is a quadrilateral. Diagonals BD and AC bisecteach other at E

therefore AE�� EC�� a��

and BE�� ED�� b��

In ∆AED, AD�� a�� b��.

In ∆BEC, BC�� b�� a��.

Therefore AD�� BC�� and ABCD is a parallelogram.

5.

G is the centroid of ∆ABC.

a. D divides BC in the ratio 1:1.

b. G divides AD in the ratio 2:1.

c. OG�� 13

� OA�� 23

� OD��

D is a midpoint, therefore OD�� 12

� OB�� 12

� OC��.

Substituting: OG�� 13

� OA�� 23

��12

� OB�� 12

� OC��OG��

13

� OA�� 13

� OB�� 13

� OC��.

6.

D, E, and F are the midpoints of the sides of ∆ABC

with AE�� EC�� b��,

CD�� DB�� a��,

and BF�� FA�� c��.

G is the centroid and divides each median in the

ratio 2:1

therefore AG�� 23

� b�� 13

� (�2c��) � �23

� b�� 23

� c��.

Similarly BG�� 13

�(2c��) � �23

�(�a��) � �23

� c�� 23

� a��

CG�� 13

�(2a��) � �23

�(�b��) � �23

� a�� 23

� b��.

Adding gives the required result AG�� BG�� CG�� 0��.

→a

→a

→c

→c

→b

→b

A

F

B DC

E

G

A

BD

C

G

2

1

A

B

C

D

→a

→a

E

→b

→b


7.

ABCD is a parallelogram with

AD�� BC�� c��

BA�� CD�� d��

The diagonals intersect at E.

Let BE�� (1 � s)a��, ED�� sa��

AE�� mb��, EC�� (1 � m)b��.

In ∆ABE, d�� (1 � s)a�� mb��.

In ∆DEC, d�� (1 � m)b�� sa��.

Therefore (1 � s)a�� mb�� (1 � m)b�� sa��

(1 � s � s)a�� (� 1 � m � m)b��

(1 � 2s)a�� (2m � 1)b��.

a�� and b�� are linearly independent

therefore 1 � 2s � 0, s � �12

�

and 2m � 1 � 0, m � �12

�

BE�� (1 � s)a��

� �12

� a��

ED�� sa��

� �12

� a��

therefore BE�� ED��.

Similarly AE�� EC�� 12

� b�� and the diagonals bisecteach other.

8.

AB is a chord of circle with centre O. A line l through

O is perpendicular to AB meeting AB at E. We are to

show that E is the midpoint of AB.

Let AE�� a�� and EB�� ma��, m > 0

OE�� c��.

In ∆OAE, OA�� c�� a��.

In ∆OBE, OB�� c�� ma��.

But �OA�� OB�� therefore �OA��2 � �OB��2

and OA�� · OA�� OB�� · OB��.

OA�� · OA�� (c�� a��) · (c � a��)

� c�� · c�� 2c�� · a�� a�� · a��

OB�� · OB�� (c�� ma��) · (c�� ma��)

� c�� · c�� 2mc�� · a�� m2a�� · a��.

Since OA�� · OA�� OB�� · OB��

a�� · a�� 2c�� · a�� m2a�� · a�� 2mc�� · a��.

But c�� ⊥ a�� therefore c�� · a�� 0

and a�� · a�� m2a�� · a��

m2 � 1

m � but m > 0 therefore m � 1 and

AE�� EB�� a��, hence E is the midpoint of AB.

9.

∆ABC is right angled at B. D is the midpoint of AC.

Let AD�� DC�� b��, BA�� c�� and BC�� a��.

Now c�� BD�� b��

and a�� BD�� b��

since a�� ⊥ c��, a�� · c�� 0

therefore (BD�� b��) · (BD�� b��) � 0

�BD��2 � �b��2 � 0

and �BD�� b��hence the midpoint of the hypotenuse is equidistantfrom the three vertices of the triangle.

A

B

D

C

→c

→a

→b

→b

l

A BE

→c

O

A

B C

D→c

→c

→d

→dE

→bm

→b(1 � m)→a(1 � s)

sa→


10.


AB�� DC�� a��

AD�� BC�� b��.

Prove �DB��2 � �AC��2 � �AD��2 � �DC��2 � �CB�2 �

�BA��2 or �DB��2 � �AC��2 � 2�a��2 � 2�b��2.In ∆ADB: DB�� a�� b��.

In ∆ABC: AC�� a�� b��.

Now �DB��2 � DB�� · DB�� (a�� b��) · (a�� b��)

� �a��2 � 2a�� · b�� b��2

�AC��2 � AC�� · AC�� (a�� b��) · (a�� b��)

� �a��2 � 2a�� · b�� b��2.Adding �AC��2 � �DB��2 � 2�a��2 � 2�b��2, hence the sum of the squares of the diagonals of anyparallelogram is equal to the sum of the squares of the four sides.

11.

ABCD is a trapezoid with AB�� nDC��. DB and AC

intersect at K. Since AB�� nDC��,

∠BAK � ∠KCD

∠ABK � ∠KDC

then ∆ABK ~ ∆KDC.

Now ��D

AB�

C�

�

��

� � ��C

AK�

K�

�

��

� � ��D

BK�

K�

�

��

�.

But AB�� nDC��

therefore ��D

AB�

C�

�

��

� � �n1

� and ��K

BK�

D�

�

��

� � �n1

�

therefore ��D

AB�

C�

�

��

� � �n1

� and ��K

BK�

D�

�

��

� � �n1

�

hence K divides BD in the ratio n:1

and AK�� n �

n1

� AD�� n �

11

� AB��.

12.

∆ABC is inscribed in a circle with centre X. A point P

is such that XP�� XA�� XB�� XC��. Since X is the

centre, �XA�� XB�� XC��.

a. In ∆PXC, CP�� XC�� XP��

CP�� XC�� XA�� XB�� XC��

CP�� XA�� XB��.

b. From ∆AXB, AB�� XA�� XB��

CP�� · AB�� (XA�� XB��) · (�XA�� XB��)

� ��XA��2 � �XB��2

But X is the centre of the circle and �XA�� and �XB��are radii, therefore �XA�� XB�� and CP�� · AB�� 0

Since CP�� · AB�� 0, CP�� ⊥ AB��.

In ∆PXB, BP�� XB�� XP��

BP�� XA�� XC��.

Now AC�� XA�� XC��

BP�� · AC�� XA��2 � �XC��2

� 0.

therefore BP�� ⊥ AC��

In ∆PXA, AP�� XA�� XP��

AP�� XB�� XC��.

Now BC�� XB�� XC��

AP�� · BC�� (XB�� XC��) · (�XB�� XC��)

� 0.

Therefore AP�� ⊥ BC��.

c. Since AP��, BP��, and CP�� are perpendicular to BC,AC, and AB respectively, P is common to all threevectors, and these altitudes are concurrent.

A

C

BPX

A

D C

B

K

D

AB

C→a

→a

→b

→b


13. a.

ABCD is a rectangle with AB�� DC�� a��

AD�� BC�� b��.

Let O be any point.

In ∆OAD, OA�� OD�� b��.

In ∆OCD, OC�� OD�� a��.

OA�� · OC�� (OD�� b��) · (OD�� a��)

� OD�� · OD�� OD�� · (�b�� a��) � a�� · b��.

But a�� ⊥ b��, therefore a�� · b�� 0 and a�� b�� DB��.

Therefore OA�� · OC�� OD�� · OD�� OD�� · DB��

� OD�� · (OD�� DB��)

OA�� · OC�� OD�� · OB��

(in ∆DOB, OD�� DB�� OB��).

b. In ∆OAB, OA�� OB�� a��

OA�� · OA�� (OB�� a��) · (OB�� a��)

�OA��2 � �OB��2 � 2a�� · OB�� a�� · a��. ➀

In ∆DOC, OC�� OD�� a��

OC�� · OC�� (OD�� a��) · (OD�� a��)

�OC��2 � �OD��2 � 2a�� · OD�� a�� · a��. ➁

Adding ➀ and ➁:

�OA��2 � �OC��2 � �OB��2 � �OD��2 � 2a�� · (OD�� OB��)

� 2a�� · a��.

Now 2a�� · (OD�� OB��) � 2a�� · a�� 2a�� · (OD�� OB�� a��)

� 2a�� · b��

(from quadrilateral ABOD).

But a�� ⊥ b��, therefore 2a�� · b�� 0

and �OA��2 � �OC��2 � �OB��2 � �OD��2.

A B

D C

→a

→a

→b

→bO


14.

ABCDEF is a regular hexagon with centre O. DiagonalsAC and BE intersect at K. Join AO and OC. ∆AOB and∆BOC are congruent equilateral triangles. AOCB is aparallelogram. Diagonals AC and OB bisect each other.Therefore K divides AC in the ratio 1:1.

Now BK � KO � aEO � OB � 2a

and BK:KE � 1:3 therefore K divides BE in the ratio 1:3.

15. a.

Let AF�� 2b��, FC�� 3b��,

BE�� a��, EC�� 2a��.

BD�� sc��, DF�� (1 � s)c��

and AD�� td��, DE�� (1 � t)d��.

From ∆BDE

sc�� a�� (1 � t)d��. ➀

From ∆BFC

3a�� c�� 3b��

a�� 13

� c�� b��.

A

F

B CE→a 2

→a

→sc

D

(1 – t) d→

3b→

2b→

td→

(1 –

s)c→

A B

C

DE

F

K

O

Substituting for a�� in ➀

sc�� 13

� c�� b�� (1 � t)d��. ➁

In ∆ADF, 2b�� td�� (1 � s)c��.

Substitute for b�� in ➁

sc�� 13

� c�� 12

� td�� 12

� (1 � s)c�� (1 � t)d��.

Collecting terms

(s � �13

� � �12

� � �12

� s)c�� (�12

�k � 1 � t)d��

(�32

� s � �56

�)c�� (�32

� t � 1)d��.

Since c�� and d�� are linearly independent

�32

� s � �56

� � 0 and �32

� t � 1 � 0

s � �59

� t � �23

�

Now BD�� 59

� c�� AD�� 23

� d��

DF�� 49

� c�� DE�� 13

� d��

�AD��:�DE�� 2:1

��D

BD�

F�

�

��

� � �54

�

�BD��:�DF�� 5:4

D divides AE in the ratio 2:1 and BF in the

ratio 5:4.

b.

We place the ratios given in the problem andcalculated in part a on the diagram. We use theproposition that areas of triangles having the samealtitude are proportional to their bases. Join CD

∆BED:∆DEC � 1:2.Let ∆BED � 5k

therefore ∆DEC � 10k.Also ∆ABD:∆BDE � 2:1

and ∆BDA � 10k.Now ∆ABD:∆ADF � 5:4

therefore ∆ADF � 8kand ∆ADF:∆FDC � 2:3

and ∆FDC � 12k.Quad CEDF � 22k∆ABC � 45k.Quad CEDF:∆ABC � 22:45.

16. Given parallelogram ABCD, DC is extended to E sothat DE:EC � 3:�2 therefore C divides DE in the

ratio 1:2, i.e., �DCD

C� � �

12

�.

AB�DE therefore ∠BAF � ∠FEC, ∠ABF � ∠EFC,and ∆ABF ~ ∆ECF

hence �AE

FF� � �

CBF

F� � �

CAB

E� � �

DC

CE� � �

12

� (AB � DC).

Therefore F divides BC in the ratio 1:2 and AE in theratio 1:2.

A

F

CBE

D

2

3

21

2

4

5

10k

12k

10k

8k

5k1


The following is a vector solution to question 16.

Since DE:EC � 3:�2, DC:CE � 1:2

ABCD is a parallelogram.

Let AB�� DC�� a��.

CE�� 2a��

BF�� tb��, FC�� (1 � t)b��

AF�� sc��, FE�� (1 � s)c��

In ∆ABF, tb�� a�� sc��. ➀

In ∆FCE, 2a�� (1 � t)b�� (1 � s)c��

therefore a�� 12

� (t � 1)b�� 12

� (1 � s)c��.

Substitute for a�� in ➀

tb�� 12

� (1 � t)b�� 12

� (1 � s)c�� sc��

2tb�� b�� tb�� c�� sc�� 2sc��

b��(3t � 1) � c��(3s � 1).

Since b�� and c�� are linearly independent

3t � 1 � 0 and 3s � 1 � 0

t � �13

�. s � �13

�.

Therefore BF�� 13

� b�� Therefore AF�� 13

� c��

and FC�� 23

� b�� and FE�� 23

� c��

�BF��:�FC�� 1:2. �AF��:�FE�� 1:2.

Therefore F divides BC in the ratio 1:2 and AF in the

ratio 1:2.

17. a., b.

In quadrilateral AP � AQ andBP � BQ.

Let BA and PQ intersect at T.In ∆APB and ∆AQBAP � AQPB � QBAB is common therefore ∆APB � ∆AQB

and ∠PAB � ∠QAB.AP � AQAT is common. Therefore ∆APT � ∆AQTand PT � QT, hence AB bisects PQ and ∠ATP � ∠ATQ � x

2x � 180°x � 90°, hence AB is perpendicular to PQ.

orSince A is equidistant from P and Q, A is on theperpendicular bisector of PQ. Similarly B isequidistant from P and Q and B is on theperpendicular bisector of PQ therefore AB is theperpendicular bisector of PQ, i.e., AB ⊥ PQ andAB bisects PQ.

18.

MNPQ is a tetrahedron with MN�� ⊥ PQ��, MP�� ⊥ NQ��.

Show MQ�� ⊥ NP��.

Since MN�� ⊥ PQ��, MN�� · PQ�� 0

and MP�� ⊥ NQ��, MP�� · NQ�� 0.

P

N

Q

M

P

A

Q

TB

A B

F

D C E

→a

→a

→a2

→sc

→ tb→

(1 – t)b→ (1 – s)c


Now MQ�� MN�� NQ��

and NP�� NQ�� QP��

MQ�� · NP�� (MN�� NQ��) · (NQ�� QP��)

� MN�� · NQ�� MN�� · QP�� NQ�� · NQ��

NQ�� · QP��

� NQ�� · (MN�� NQ�� QP��), MN�� · QP�� 0

� NQ�� · MP��, MP�� MN�� NQ�� QP��.

Since MQ�� · NP�� NQ�� · MP�� 0, MQ�� ⊥ NP��.

Review Exercise

1. a. Since (2, 3) ≠ k(�4, 3), vectors (2, 3) and (�4, 3)may be used as basis vectors for a plane.

b. Let (3, �1) � a(2, 3) � b(�4, 3).Equating components gives:

2a � 4b � 3 ➀3a � 3b � �1 ➁

�32

� � ➀ � ➁: �9b � �121�

b � ��1118� and a � �

158�

hence (3, �1) � �158� (2, 3) � �

1118� (�4, 3).

2. a. Let u�� (3, 5, 6), v�� (6, 10, 12), w�� (�3, �5, 6).

Since v�� can be written as a linear combination of

u�� and w��, v�� 2u�� 0w��, the vectors u��, v��, and w�� are

linearly dependent.

b. Let a�� (5, 1, �1), b�� (6, �5, 2), c�� (3, 8, �2),

and d�� (�40, 39,�29). These vectors are linearly

dependent since four vectors in R3 are always

linearly dependent.

c. Let u�� (7, 8, 9), v�� (0, 0, 0), w�� (3, 8, 6).

Now au�� bv�� cw�� 0�� if a � c � 0 and b can

be a number other than zero, hence, u��, v��, and w�� are

linearly dependent.

d. Let u�� (7, �8) and v�� (14, 19). Since u�� ≠ kv��,

u�� and v�� are not parallel, hence are linearly

independent.

e. Let a�� (0, 1, 0), b�� (0, 0, �7), and c�� (7, 0,

0). Since a��, b��, and c�� are vectors along the y-axis,

z-axis, and x-axis respectively, the three vectors are

mutually perpendicular, not coplanar, and hence

linearly independent.

3. Since c�� t2a�� b�� and d�� (2t � 3)(a�� b��) are

linearly dependent, d�� kc��.

Therefore (2t � 3)(a�� b��) � k(t2a�� b��)

(2t � 3)a�� (2t � 3)b�� kt2a�� kb��

(b��t2 � 2t � 3)a�� (k � 2t � 3)b�� 0��.

Since a�� and b�� are linearly independent

kt2 �(2t � 3) � 0 ➁

and k �(2t � 3) � 0. ➀

From ➀ k � �(2t � 3).

Substitute into ➁ �(2t � 3)t2 � (2t � 3) � 0.

Factor (2t � 3)( � t2 � 1) � 0

2t � 3� 0 or � t2 � 1 � 0

t � �32

� t2 � �1

The only real solution is t � �32

� and from ➀ k � 0.

Therefore c�� and d�� are linearly dependent when t � �32

�.

We note that for t � �32

�, c�� 94

�a�� b��, and d�� 0��.

4. Expressing any of the vectors as a linear combination

of the other two gives

a�� 2b�� c�� p(2a�� b��) � q(a�� b�� c��)

from which

(1 � 2p � q)a�� (�2 � p � q)b�� (�1 � q)c�� 0.

Since a��, b��, and c�� are linearly independent,

1 � 2p � q � 0, ➀

�2 � p � q � 0, ➁

�1 � q � 0. and ➂

From ➀ and ➂, q � 1, p � 0.

From ➁ �2 � 0 � 1 � �1 ≠ 0.

p � 0, q � 1 does not satisfy equation ➁, therefore a��

� 2b�� c�� cannot be written as a linear combination

of 2a�� b�� and a�� b�� c��, i.e., they are not coplanar,

hence are linearly independent.


5. a. For ∆ABC with vertices A(0, 0), B(5, �6), C(2, 0),

the midpoint of AB is ��52

�, �3�, of AC is (1, 0), and

of BC is ��72

�, �3�. The position vector of the

centroid G of a triangle is given by

OG�� 13

�(OA�� OB�� OC��).

For ∆ABC, OG�� 13

�[(0, 0) � (5, �6) � (2, 0)]

� ��73

�, �2�.

Therefore the centroid is ��73

�, �2�.

b. For ∆ABC with vertices A(4, 7, 2), B(6, 1, �1),

C(0, �1, 4), the midpoint of AB is �5, 4, �12

��,

of AC is (2, 3, 3), and of BC is �3, 0, �32

��.

The centroid of ∆ABC is ��130�, �

73

�, �53

��.

6. Since OM�� 35

� ON�� 25

� OP��, M divides NP�� in the

ratio 2:3.

Since OM�� 45

� ON�� 15

� OQ��, M divides NQ�� in the

ratio 1:4.

Let NM � 2k, MP � 3k, then MQ � 8k and PQ�� 5k.

a. P divides NQ in the ratio 5:5 � 1:1.

b. Q divides NM in the ratio NQ:QM � 10:�8 � 5:�4.

7.

From ∆OAM, OM�� OA�� AM��

but AM�� 18

� AB��

and AB�� OA�� OB�� (from ∆OAB).

Therefore OM�� OA�� 18

� AB��

� OA�� 18

�(�OA�� OB��)

OM�� 78

� OA�� 18

� OB��.

8. a.

In ∆OMN, NM�� ON�� OM��

but ON�� 29

� OM�� 191� OQ��.

Therefore NM�� 29

� OM�� 191� OQ�� OM��

� �191� OM��

191� OQ��

� �191� (OM�� OQ��).

In ∆OMQ, OM�� OQ�� QM��.

Therefore NM�� 191� QM��

NM��QM�� and N, M and Q are collinear. We note

that �NM�� > �QM��, NM�� and QM�� have the same

direction; therefore we find Q between N and M,

and N divides MQ in the ratio 11:�2.

N

M

Q

O

AM

B

1

7

O

2 3

14

N M PQ


b. Since M divides NQ in the ratio 11:�9

OM�� 92

� ON�� 121� OQ��.

or

Since ON�� 29

� OM�� 191� OQ��

9ON�� 2OM�� 11OQ��

2OM�� 9ON�� 11OQ��

and OM�� 92

�ON�� 121�OQ��.

9.

In ∆ABC, let AQ�� c��, QB�� 6c��,

AP�� 3b��, PC�� 4b��,

BR�� (1 � s)a��, RP�� sa��,

CR�� (1 � k)d��, RQ�� kd��.

In ∆BQR (1 � s)a�� 6c�� kd�� ➀

In quad AQRP, c�� 3b�� 2a�� kd��.

Substitute in ➀: (1 � s)a�� 18b�� 6sa�� 6kd�� kd��

(1 � 7s)a�� 18b�� 7kd��. ➁

In ∆PCR: 4b�� sa�� (1 � k)d��

b�� 14

�sa�� 14

�(1 � k)d��.

A

Q

P

B C

R

→c

→6 c

3b→

→kd

→

sa

(1 – s)a→

(1 – k)d→ 4b

→

NQ

M2 9

11


Substitute in ➁: (1 � 7s)a�� 92

�sa�� 92

�(1 � k)d�� 7kd��

�1 � 7s � �92

�s�a�� 92

� � �92

�k � 7k�d��

�1 � �223�s�a��

92

� � �223�k�d��

a�� and d�� are linearly independent.

Therefore 1 � �223�s � 0 and �

92

� � �223�k � 0

s � �223� k � �

293�

BR�� 2213�a�� CR��

1243�d��

RP�� 223�a�� RQ��

293�d��

therefore �BR��:�RP�� 21:2. therefore �CR��:�RQ�� 14:9.

R divides BP in the ratio 21:2 and R divides CQ in the ratio 14:9.

10.

In parallelogram ABCD, let

BF�� 3b��, FC�� b��, AD�� 4b��, AE�� a��, EB�� 4a��, DC��

5a��, FK�� (1 � s)c��, KA�� sc��, EK�� kd��,

KD�� (1 � k)d��.

In ∆AEK, kd�� a�� sc��. ➀

In ∆ABF, 5a�� c�� 3b��

a�� 15

�c�� 35

�b��.

Substitute in ➀: kd�� 15

�c�� 35

�b�� sc��

kd�� 15

� � s�c�� 35

�b��. ➁

In ∆AKD, 4b�� sc�� (1 � k)d��

b�� 14

�sc�� 14

�(1 � k)d��.

A D

CFB

EK

→a

4a→

3b→

b→

kd→

sc→

(1 – k)d→

(1 – s)c→

Substitute in ➁:

kd�� 15

� � s�c�� 230�sc��

230�(1 � k)d��

�k � �230� � �

230�k�d��

15

� � s � �230�s�c��

��2230�k � �

230��d�� (�

15

� � �2230�s)c��.

Since c�� and d�� are linearly independent,

�2230�k � �

230� � 0 and �

15

� � �2230�s � 0

k � �233� s � �

243�

therefore EK�� 233�d��, FK��

1293�c��,

KD�� 2203�d�� KA��

243�c��

and �DK��:�KE�� 20:3. and �AX��:�KF�� 4:19.

Therefore K divides DE in the ratio 20:3 and AF in the

ratio 4:19.

11.

∆ABC is isosceles with AB�� c��, AC�� b��,

and �c�� b��.AD is a median, D is the midpoint of BC.

Therefore AD�� 12

�b�� 12

�c��

BC�� b�� c��.

Now AD�� · BC�� 12

��b�� c�� · �b�� c��

12

��b�� · b�� c�� · c��

12

��b��2 � �c��2.

But �b�� c��, therefore AD�� · BC�� 0 and AD�� ⊥ BC��.

The median to the base of an isosceles triangle is

perpendicular to the base.

12.

Given a circle with centre O, AB is a chord with

midpoint M.

Let OB�� b��, OA�� a��.

Since �OA�� and �OB�� are radii �a�� b�� and in ∆OAB,

∠OAB � ∠OBA � and the angle between AB�� and

a�� is 180° � .

Since M is the midpoint of AB��

OM�� 12

�a�� 12

�b��.

Now OM�� · AB�� 12

��a�� b�� · AB��

� �12

�a�� · AB�� 12

�b�� · AB��

� �12

��a��AB�� cos(180° � ) �

�12

��b��AB�� cos .

But �a�� b��.

Therefore OM�� · AB�� 0 and OM�� ⊥ AB��,i.e., a line through the centre of a circle and themidpoint of a chord is perpendicular to the chord.

A

O

B

M

180º

→a

→b

A

B D C

→c

→b


13.

∆ABC is isosceles with AB � AC. CE and BF

are medians.

Let BE�� EA�� c��

AF�� FC�� b��.

Since �2c�� 2b�� and �c�� b��.In ∆AEC, EC�� c�� 2b��

EC�� · EC�� (c�� 2b��) · (c�� 2b��)

� c�� · c�� 4c�� · b�� 4b�� · b��

� �c��2 � 4c�� · b�� 4�b��2.But �c�� b��

therefore �EC��2 � 5�b��2 � 4c�� · b��.

In ∆BAF, BF�� 2c�� b��

BF�� · BF�� 4c�� · c�� 4c�� · b�� b�� · b��

� 4�c��2 � 4c�� · b�� b��2.But �c�� b��BF��2 � 5�b��2 � 4c�� · b��

therefore �EC��2 � �BF��2 and �EC�� BF��.

The medians to the equal sides of an isosceles triangleare equal.

14.

ABCD is a quadrilateral. P, Q, R, and S are themidpoints of the sides as shown in the diagram.In ∆ABC, P and Q are the midpoints of AB and CB

therefore AC�� 2PQ��

and AC�� · AC�� 2PQ�� · 2PQ��

�AC��2 � 4�PQ��2. ➀

Similarly in ∆ADC, �AC��2 � 4�SR��2. ➁

By joining SP, BD, and RQ we have DB�� 2SP��

�DB��2 � 4�SP��2 ➂

and DB�� 2RQ��, �DB��2 � 4�RQ��2. ➃

Adding ➀, ➁, ➂, ➃:

�AC��2 � �AC��2 � �DB��2 � �DB��2

� 4�PQ��2 � 4�SR��2 � 4�SP��2 � 4�RQ��2.Divide by 2:

�AC��2 � �DB��2 � 2��PQ��2 � �SR��2 � �SP��2 � �RQ��2�.Since AC�� 2PQ�� 2SR��, PQRS is a parallelogram.We have seen that the sum of the squares of thediagonals of a parallelogram is equal to the sum of thesquares of the four sides. PR and SQ are the diagonals

of parallelogram PQRS, therefore

�PR��2 � �SQ��2 � �PQ�� SR��2 � �SP��2 � �RQ��2

and �AC��2 � �DB��2 � 2��PR��2 � �SQ��2�i.e., the sum of the squares of the diagonals of aquadrilateral is equal to twice the sum of the squaresof the line segments joining the midpoints of theopposite sides.

15.

a. AB�� · (QD�� 12

�AC��) � AB�� · (QD�� CE��)

� AB�� · (QD�� DF��)

(F and D are midpoints of BA and BC, therefore

DF�� 12

�CA�� CD��)

� AB�� · QF��

� 0 (since AB�� ⊥ QF��).

A

F E

Q

BD C

A

Q

B

P

C

R

D

S

A

B C

E F

→c

→c

→

b

→

b


b. AC�� · (QD�� 12

�AB��) � AC�� · (QD�� BF��)

� AC�� · (QD�� DE��)

� AC�� · QD��

� 0 (since AC�� ⊥ QE��).

c. Since AB�� · �QD�� 12

�AC�� 0 � AC�� · �QD�� 12

�AB��AB�� · QD��

12

�AB�� · AC�� AC�� · QD�� 12

�AC�� · AB��

AB�� · QD�� AC�� · QD�� 0

QD�� · (AB�� AC��) � 0

QD�� · CB�� 0.

d. Since CB�� · QD�� 0, CB�� ⊥ QD��. Q is common to the perpendicular bisectors QE, QF, and QD,hence the perpendicular bisectors are concurrent.

Chapter 6 Test

1. a. Three vectors u��, v��, and w�� are linearly independent.

Let �u�� 3, �v�� 2, and �w�� 1 and their

directions be along the x, y, and z axes respectively.

These vectors do not lie in the same plane since each is

perpendicular to the plane defined by the other two, they

are not coplanar hence are linearly independent.

b. The cross product of any two vectors u�� and v��, u�� v��, is a

vector perpendicular to both u�� and v��. Now if (u�� v��) · w��

� 0, w�� will also be perpendicular to u�� v�� hence parallel

to both u�� and v��. Thus, u��, v��, and w�� are coplanar. If

(u�� v��) · w�� ≠ 0, u��, v��, and w�� are linearly independent.

2.

P divides QR in the ratio 10:�3.

a. OP�� 37

� OQ�� 170� OR��.

b. R divides QP in the ratio 7:3

therefore OR�� 130� OQ��

170� OP��.

Q PR

103


3. a.

BA�� ra��

OB�� sb��

From the graph a�� (2, �4), b�� (8, 7), and

c�� (13, 2).

If c�� ra�� sb��, then

2r � 8s � 13 ➀

�4r � 7s � 2 ➁

➀ � 2 4r � 16s � 26 ➂

➁ � ➂ 23s � 28

s � �2283�.

Substituting in ➀, 2r � 8 ��2283�� 13

2r � �7253�

r � �7456�.

Then c�� 7456� a��

2283� b��.

B

A

O

→a

→c

→

b

4.

u��, v��, and w�� are coplanar

�u�� 5, �v�� 12, �w�� 18, and angles as shown in

the diagram.

From the head of AB�� u��, draw a line parallel to w��

meeting v�� at C.

Let AC�� av�� and

CB�� bw��.

Now u�� av�� bw��.

In ∆ABC, ∠BAC � 35°, ∠ACB � 180° � 55°

� 125°

and ∠ABC � 20°.

From the sine law

�si

an

�v2

��0°

� � �si

bn

�w�3

�

5

�°

� � �sin

�u1

��

2

�5°

�

a � ��v��u��

��s

s

in

in

1

2

2

0

5

°

°� b � �

�w��u�

�

�

��s

s

i

i

n

n

1

3

2

5

5

°

°�

a 0.1739. b 0.1945.

Therefore u�� 0.17v�� 0.19w��.

5. a.

F divides AP in the ratio 13:�8.

F divides PQ in the ratio 4:�3 � 8:�6.

b. P divides AG in the ratio 5:2.

6. a. A(�4, 2, �8), B(�1, �4, �2), P(1, �8, 2)

AP�� (5, �10, 10) � 5(1, �2, 2)

BP�� (2, �4, 4) � 2(1, �2, 2).

Since AP�� 52

�BP��, A, B, and P are collinear.

b. Let OP�� sOA�� tOB��

(1, �8, 2) � s(�4, 2, �8) � t(�1, �4, �2).

Equating components:

�4s � t � 1 ➀

2s � 4t � �8 divide by 2: s � 2t � �4 ➁

�8s � 2t � 2 divide by 2: �4s � t � 1 ➂

2 � ➀ � ➁: �9s � 6

s � ��23

�.

Substitute in ➀ �83

� � t � 1

t � �53

�.

Therefore OP�� 23

�OA�� 53

�OB��.

or

From a: AP�� 52

�BP��

Therefore P divides AB�� in the ratio 5:�2

and OP�� 23

�OA�� 53

�OB��.

7.

P

D

S

A

Q

BR

C

A B P

3 2

5 2 6

A P G F

A

B

C→v

→w

→u

20º

35º


Given quadrilateral ABCD with midpoints P, Q, R,

and S as shown. Join PS, AC, and QR.

In ∆PAC, P and S are midpoints of DA�� and DC��

therefore PS�� 12

�AC��

Similarly in ∆BAC: QR�� 12

�AC��

therefore PS�� QR�� and PQRS is a parallelogram. (In

a quadrilateral, if one pair of opposite sides are equal

and parallel, the quadrilateral is a parallelogram.)

8.

In ∆ABC, D and E are in AB and AC respectively so

that DE�� kBC��.

Let AD�� sc��, DB�� (1 � s)c��, AB�� c��

AE�� tb��, EC�� (1 � t)b��, AC�� b��.

In ∆ADE, DE�� sc�� tb��.

In ∆ABC, BC�� c�� b��.

Since DE�� kBC��

�sc�� tb�� k(�c�� b��) � �kc�� kb��

(k � s)c�� (k � t)b��.

c�� and b�� are linearly independent.

Therefore k � s � 0 and k � t � 0

s � k. t � k.

Now AD�� kc�� AE�� kb��

AD�� kAB��. AE�� kAC��.

A

B C

(1 – s)c

D

sc tb

E

(1 – t)b

→

→

→

→



Exercise 7.1

8. b. The vector equation of a line passing through

P�2, �34

�� with direction d�� 23

�, 6� � �23

�(1, 9)

is r� � �2, �34

�� t(1, 9).

9. a. x � �13

� � 2t, y � 3 � �23

�t.

The direction vector d�� (2, ��23

�) � �23

�(3, �1).

A direction vector with integer components is (3, �1). There is no point on this line with integercoordinates.

b. r� � ��13

�, �12

�� t��13

�, �14

��. The direction vector

d�� 13

�, �14

�� 112��4, 3�. A direction vector with

integer components is (4, 3). A point on the line with

integer coordinates is (1, 1) when t � 2.

c. r�� 12

�, 3� � t��12

�, 5�. The direction vector

d�� 12

�, 5� � ��12

� (1, �10). A direction vector

with integer components is (1, �10). A point on the

line with integer coordinates is (�1, 18) when

t � 3.

10. a. l1: x � 1 � 3t , y � 7 � 4t; l

2: x � 2 �4s,

y � �3s. The direction vectors are d��1

� (�3, 4)

and d��2

� (�4, �3). Since d��1

· d��2

� 0, the lines are

perpendicular.

b. l1: r� � (1, 7), � t(�3, 4); l

2: r� � (2, 0) � s(3, �4),

the direction vectors are d��1

� (�3, 4) and d��2

�

(3, �4). Since d��1

� �d��2, the lines are parallel.

Chapter 7: Lines in a Plane 91

Chapter 7 • Lines in a Plane

c. l1: r� � (1, 7) � t(�3, 4); l

2: r� � (2, 0) � s(4,

�3). The direction vectors are d��1

� (�3, 4) and d��2

� (4, �3). Since d��1

· d��2

� 0 and d��1

≠ kd��2, the

lines are neither perpendicular nor parallel.

11. The direction of r� � (1, 8) � t(3, 7) is d�� (3, 7).

A direction perpendicular to d�� is n�� (7, �3). An

equation of a line through (4, 5) with direction n�� is

r� � (4, 5) � t(7, �3).

12. a. x � 6, y � 1 � 7tSince x � 6, this line is parallel to the y-axistherefore does not intersect the y-axis. The lineintersects the x-axis at the point (6, 0).

b. r� � (�5, 10) � t(1, 5)The parametric equations are x � �5 � t,y � 10 � 5t.The line intersects the x-axis when y � 0, t � �2,and x � �7, i.e., the point (�7, 0). The lineintersects the y-axis when x � 0, t � 5, andy � 35, i.e., the point (0, 35).

y

x

35

–35 7 14 35

–21

y

4

–3

–2

4 6x

c. r� � (2, 3) � t(3, �1)The parametric equations are x � 2 � 3t, y � 3 � t. The line intersects the x-axis when y � 0, t � 3,and x � 11, i.e., the point (11, 0). The lineintersects

the y-axis when x � 0, t � ��23

�, and y � �131�, i.e.,

the point �0, �131��.

13. l1: r� � (3, 9) � t(2, 5)

x � 3 � 2t, y � 9 � 5twhen t � �1; x � 1, y � 4 and (1, 4) lies on l

1.

l2: r� � (�5, 6) � u(3, �1)

x � �5 � 3u, y � 6 � u when u � 2; x � 1, y � 4

and (1, 4) lies on l2.

The directions of l1

and l2

are d��1

� (2, 5) and d��2

�

(3, �1). The angle between the direction vectors is the

angle of intersection of these lines.

Now d��1

· d��2

� �d��1��d��

2� cos �

6 � 5 � �29� �10� cos �

cos � � ��29�

1

�10��

� � 87�.

The acute angle between these lines is 87�.

14. a. i) The line r� � (2, �6) � t(3, �4) has direction

d�� (3, �4). The direction of the positive x-axis

is i � (1, 0). The angle the line makes with the

x-axis is α. (a positive rotation about A). To use

the dot product we use direction AB�� (�3, 4).

Therefore AB�� · i � �AB��i� cos α

�3 � 5 cos α

cos α � ��35

�

α � 127�.The line makes an angle of 127� with the x-axis.

ii) The line r� � (6, 1) � t(5, 1) has direction vector

d�� (5, 1). We see from the diagram that α is

the acute angle between direction vectors

(5, 1) and (1, 0) therefore d�� · i � �d��i� cos α

5 � �26� cos α

cos α � ��

5

26��

α � 11�.The line makes an angle of 11� with the x-axis.

y

x–3

5

–2

5

α

→d

P(6,1)

yB

xA

P

�

α

→

d

5

–2

–2

5 10

y

x

92 Chapter 7: Lines in a Plane


b. Consider a line l that intersects the x-axis at

A(a, 0). Choose a point B(x1, y

1), y

1� 0 on l. The

angle of inclination of the line is ∠BAX � α.

The slope of line l is m � �y

x1

1

�

�

0

a� � �

x1

y

�1

a�.

Translate the line to the left a units

therefore A(a, 0) → A'(0, 0)B(x

1, y

1) → B'(x

1� a, y

1).

By definition tan α � �x

1

y

�1

a�

but �x

1

y

�1

a� � m (the slope of l)

therefore tan α � m.Note: if x

1� a, the slope is undefined and

α � 90�.

15. Point A(24, 96) with velocity vector v�� (85, �65)(units in km and km/h).

a. Parametric equations of the highway line are x � 24 � 85t, y � 96 � 65t.

b. The horizontal velocity component is 85 km/h. The

time taken to travel 102 km at 85 km/h is �1

8

0

5

2�

hours (1h 12 min).

c. When t � �1

8

0

5

2�, x � 126, y � 18. The coordinates

of P at that time will be (126, 18).

16. a. Parametric equations of a line are x � x0

� ab1,

y � y0

� bt.

Solving for t gives �x �

ax

0� � t, �y �

b

y0� � t.

Therefore �x �

a

x0� � �

y �

b

y0�.

b. (i) x � 5 � 8t, y � �3 � 5t.

Solving for t gives �x

�

�

8

5� � t and �

y �

5

3� � t.

A symmetric equation is x � ��

5

8� � �

y �

5

3�.

(ii) r� � (0, �4) � t(4, 1). A symmetric equation

is �4

x� � �

y �

1

4�.

c. A direction of the line through A(7, �2),

B(�5, �4) is AB�� (�12, �2) � �2(6, 1).

A symmetric equation of this line is

�x �

6

7� � �

y �

1

2�.

17. a. An equation of the line passing through A(7, 3)

with direction vector d�� (2, �5) is r� �

(7, 3) � t(2, �5). If t � �1, r�1

� (5, 8) then

P(5, 8) is on the line. If t � 5, r�2

� (17, �22),

then Q(17, �22) is on the line.

b. The line segment PQ is defined by the parametricequation x � 7 � 2t, y � 3 � 5t, for �1 t 5.

18. a. A direction vector of the line is PQ�� OP�� OQ��

� �p�� q��.

The vector equation of the line, passing through P

with direction PQ��, is r� � p�� t(�p�� q��)

r� � (1� t)p�� tq��.

y

Q

P

xO

y

xα α

x

y

' '

y

αx

A(a, 0)

y

α

A(a, 0)x

b. R is at P when t � 0 and at Q when t � 1.Therefore R is between P and Q for 0 < t < 1.

c. When t � 2, r� � �p�� 2q��

OS�� 2q��, SR�� p��

OPSR is a parallelogramPQ � QRR divides PQ in the ratio 2:�1.

d. If t � �1

2�, R will be the midpoint of PQ. Therefore

for t > �1

2�, R will be closer to Q than to P.

19. a. b.

l1: r�

1� (5, 2) � t(�3, 6), d��

1� (�1, 2)

l2: r�

2� (5, 2) � u(11, 2), d��

2� (11, 2).

Lines intersect at A(5, 2).

Let l3

represent the line bisecting the angle between

l1

and l2

and let the angle between l1

and l3, and l

2

and l3, be �. Let the direction vector of l

3be

m�� (m1, m

2).

m�� · d��2

� �m��d��2� cos �

11m1

� 2m��2

� �m�� 125� cos �

cos � � �11

5

m

�1

5�

�

�m�2

�

m

�2

�.

m�� · d��1

� �m��d��1� cos �

�m1

� 2m2

� �m��5� cos �

cos � � ��m

�1

5�

�

�m�2

��

m2

�.

Equating cos � gives

�11

5

m

�1

�

5��m�2

�

m

�2

� � ��m

�1

5�

�

�m�2

��

m2

�

11m1

� 2m2

� �5m1

� 10m2

16m1

� 8m2

m1

� �12

� m2·

Choose m2

� 2. Therefore m1

� 1 and the

direction of l3

is m�� (1, 2).

The vector equation of l3

is r� � (5, 2) � s(1, 2). To determine the bisector of the other angle weuse directions d��

1and �d��

2which gives equation

��1

5

1

�

m1

5�

�

�m��

2

�

m2

�� m

�1

5�

�

�m�2

��

m2

�

�11m1

� 2m2

� �5m1

� 10m2

6m1

� �12m2

m1

� �2m2.

Choose m2

� �1. Therefore m1

� 2 and the

direction of this line is (2, �1). The vector

equation of the second angle bisector is

l4: r� � (5, 2) � v(2, �1).

c. The direction of the two lines are (1, 2) and (2, �1). Since (1, 2) · (2, �1) � 0, the two lines are perpendicular.

Exercise 7.2

9. a. 5x � 3y � 15 � 0.

Solve for y: y � �53

� � 5.

Let x � 3t, y � 5t � 5

A vector equation is r� � (0, 5) � t(3, 5).Scalar equations are x � 3t

y � 5 � 5t.

A symmetric equation is �3x

� � �y �

55

�.

→d

� �

→m

A(5, 2)

2

→d1

1

4

2

→d2�

3

l

l

l l

PQ

S

R

O


b. �4x � 6y � 9 � 0.

Solve for y: y � �23

�x � �32

�.

Let x � 3t, y � 2t � �32

�.

A vector equation is r� � �0, ��32

�� t(3, 2).

Parametric equations are x � 3t

y � 2t � �32

�.

A symmetric equation is � .

10. Let P be a point not on the line l and D be on l, the footof the perpendicular from P. Choose a point A on lother than D. Now PAD is a right triangle where PA isthe hypotenuse. Therefore PA is the longest side andPA � PD. Therefore the shortest distance from a pointto a line is the perpendicular distance from the point tothe line.

11. The distance d from Q(3, �2) to each of the followinglines:

a. 3x �2y � 6 � 0

d ��Ax

�1

�

A2

B

�

y1

B��

2�C�

�

� �9 �

�

4

1

�

3�

6�

d � ��

7

13��.

b. �x �

23

� � �y �

74

� therefore 7x � 21 � 2y � 8

7x � 2y � 13 � 0

and d � �21

��

494

�

�

4�13

�

d � ��

12

53��.

c. r� � (�3, �7) � t��15

�, �16

��. A direction vector

m�� (6, 5) therefore a normal n�� (5, �6). A pointon the line is P(�3, �7).

Now d � ��PQ�

��

n��· n��

� �(6, 5)

�·

n�

(

�

5

�� 6)

�

� 0

therefore the point is on the line.

d. The distance from Q to x � �5 is 8.

12. Equation of the line is 6x � 3y � 10 � 0.

a. Point (4, 7)

d � �24

��

3

2

6

1

�

�

9�10

�

� �3�

35

5�� · �

��

5�5�

�

d � �7�

3

5��.

x = –5

x = –5 –5

3

2

2

Q (3, –2)

P

D

A

l

y � �32

�

�2

x�3


b. Point (4, �8)

d � ��24 �

3

2

�4

5�� 10��

� �3�

10

5��

d � �2�

3

5��.

c. Point (0, 5)

d � �15

3��

5�10

�

� �3�

5

5��

d � ��

35�

�.

d. Point �5, ��230��

d � �30 �

3�20

5��10

�

d � 0.

13. a. i) Given two lines l1

and l2. Let the direction

vector of l1

be d��1

� (a, b). Since l1�l

2, the

direction vector d��2

of l2

is a multiple of d��1. Let

d��2

� (ka, kb). Now the normal of l1

is n��1

�

(b, �a) and of l2

is n��2

– (kb, �ka). But n��2

�

(kb, �ka) � k(b, �a) � kn��1

therefore n��1�n��

2.

ii) Given two lines l1

and l2

having normals n��1

and

n��2. Let the normal of l

1be n��

1� (A, B) since

n��1�n��

2, n��

2� kn��

1, and n��

2� (kA, kB). The

direction of l1

and l2

will be m��1

� (B, �A) and

m��2

� (kB, �kA). Since m��2

� (kB, �kA) �

k(B, �A) � km��1. m��

2�m��

1and the lines l

1and l

2

are parallel, therefore two lines in a plane are

parallel if and only if their normals are parallel.

b. If two lines in a plane are perpendicular, their

normals are perpendicular. Let two lines l1

and l2

have direction m��1

and m��2. Let m��

1� (a, b). Since

l1

⊥ l2, m��

1· m��

2� 0, therefore m��

2� (�kb, ka).

A normal to l1

is (�b, a) and to l2

is (ka, kb). Now

(�b, a) · (ka, kb) � �kab � kab � 0, therefore the

normals are perpendicular.

If the normals of two lines are perpendicular, then

the two lines are perpendicular.

Let two lines l1

and l2

have normals n��1

� (A, B)

and n��2. Since n��

1⊥ n��

2, n��

1· n��

2� 0 and

n��2

� (kB, � kA). The normal to l1

is n��1

� (A, B),

therefore a direction is m��1

� (B, �A). Similarly a

direction of l2

will be m��2

� (kA · kB). Now

m��1

· m��2

� 0, therefore m��1

⊥ m��2

and the two lines

are perpendicular.

Therefore two lines in a plane are perpendicular if

and only if their normals are perpendicular.

14. a.

Given a line l intersecting the x-axis at A. The

angle of inclination of l is ∠BAX � α. Let the

direction vector of l be AB�� m�� (m1, m

2).

In ∆ABC, AE � m1

� �m�� cos αand EB � m

2� �m�� sin α.

Therefore a direction of the line is (cos α, sin α)

and a normal is (sin α, �cos α). The scalar

equation of a line is Ax � By � C � 0 where

(A, B) is a normal therefore the equation is

x sin α � y cos α � C � 0.

A

y

B

xE

l

→m

α


b. 2x � 4y � 9 � 0 has normal (1, 2) therefore adirection is (2, �1).

tan � � �m

m2

1

�

� ��12

�

� � 153�.The angle of inclination is 153�.

c. The equation will be x sin 120�� y cos 120� � D � 0.

sin 120� � ��

23�

�, cos 120�� 12

�

therefore we have �3�x � y � 2D � 0.

Now (6, �4) is on the line. 6�3� � 4 � 2D � 0,

2D � 4 � 6�3�and the equation is �3�x � y � 4 � 6�3� � 0.

15. a.

AN�� (x � 2, y � 2), BN�� (x � 8, y � 10)

AN�� · BN�� 0

(x � 2, y � 2) · (x � 8, y � 10) � 0

(x � 2)(x � 8) � (y � 2)(y � 10) � 0

x2 � 10x � 16 � y2 � 12y � 20 � 0

x2 � y2 � 10x � 12y � 36 � 0

(x2 � 10x � 25)� 25 � (y2 � 12y � 36)� 36 � 36� 0

(x � 5)2 � (y � 6)2 � 25.

Hence N lies on a circle with centre (5, 6) andradius 5.

b. The midpoint of AB is (5, 6), which is the centre ofthe circle. Since both A and B are on the circle, ABis a diameter of the circle.

16. a.

n�� is a normal to the line l. P(x, y) is a point on the

line and OP�� is the position vector of P. Rotate the

line about P until it passes through the origin. Now

the line l and the position vector OP�� are coincident.

Since n�� ⊥ l , it will now be perpendicular to OP��

and n�� · OP�� 0.

b. If the line goes through the origin then OP�� is a

direction vector of the line. n�� is normal to the line

therefore n�� · OP�� 0.

If n�� · OP�� 0 then n�� is perpendicular to OP��. But n��

is a normal to the line hence is perpendicular to l.

Since n�� is perpendicular to both OP�� and the line,

OP�� and the line are parallel. But P is a point on the

line, hence they are coincident and the line passes

through the origin.

Exercise 7.3

7. Given the points A(2, 3, �2) and B(4, �1, 5). The

midpoint of AB is M�3, 1, �32

��. The line passes through

C(0, �1, 1) and M, therefore a direction is

CM � �3, 2, �12

��.

Using direction vector (6, 4, 1) and point C(0, �1, 1),the parametric equations arex � 6t, y � �1 � 4t, z � 1 � t.

→n

P(x, y)

l

O

ly

5

–5 5x

–5

N(x, y)

A(2, 2)

B(8, 10)


8. A line through the origin and parallel to AB;

A(4, 3, 1), B(�2, �4, 3) has direction BA��

(6, 7, �2) and symmetric equation

�6x

� � �7y

� � ��

z2�.

9. a. line l1: r� � (1, 0, 3) � t(3, �6, 3) with direction

d��1

� 3(1, �2, 1) and l2: r� � (2, �2, 5) �

t(2, �4, 2) with direction d��2

� 2(1, �2, 1) since

d��1

� �32

�d��2, the lines are parallel. The symmetric

equation of l1

is

�x �

11

� � ��

y

2� � �

z �

13

�.

(2, �2, 5) is on l2. Check to see if it is on l

1.

�2 �

11

� � ��

�

22� � �

5 �

13

�.

Therefore the lines are parallel and distinct.

b. l1: r� � (2, �1, 4) � s(3, 0, 6); d��

1� 3(1, 0, 2).

l2: r� � (�3, 0, 1) � t(2, 0, 2); d��

2� 2(1, 0, 1).

Since d��1

� kd��2, the lines are not parallel nor the

same line.

c. l1: r� � (1, �1, 1) � s(6, 2, 0); d��

1� 2(3, 1, 0).

l2: r� � (�5, �3, 1) � t(�9, �3, 0);

d��2

� �3 (3, 1, 0).

Since d��1

� ��23

�d��2, the lines are parallel.

Symmetric equation of l1

is �x �

31

� � �y �

11

�; z � 1.

Check to see if (�5, �3, 1) is on l1.

��5

3� 1� � �

�31� 1� � �2; z � 1.

Since (�5, �3, 1) lies on l1

and l1

is parallel to l2,

l1

and l2

are the same line.

10. a. x � t, y � 2, z � �1: perpendicular to the yz-planepassing through (0, 2, �1),

b. x � 0, y � 1 � t, z � 1 � t: a line in the yz-planehaving y-intercept 2 and z-intercept 2.

c. x � �5, y � 2 � t, z � 2 � t represents a line inthe plane x � �5, a plane parallel to the yz-plane.In this yz-plane the line has equation y � z, a linepassing through (�5, s, s) for all s � R.

z

x

y

l

�5

(�5, s, s)

z

y

x

2

2

z

x

y

x = t, y = 2, z = –1


11. a. If a line in R3 has one direction number zero it will

be parallel to one of the coordinate planes; i.e., if

d�� (a, b, 0), the line is parallel to the xy-plane,

d�� (a, 0, c), the line is parallel to the xz-plane,

d�� (0, b, c), the line is parallel to the yz-plane.

b. If a line in R3 has two direction numbers zero, theline will be perpendicular to one of the coordinateplanes; i.e., if

d�� (a, 0, 0), the line is perpendicular to the yz-plane,

d�� (0, b, 0), the line is perpendicular to the xz-plane,

d�� (0, 0, c), the line is perpendicular to the xy-plane.

12. A line l1

passes through the point A(�6, 4, 2) and is

perpendicular to both

l1: �

�

x4� � �

y �

�610

� � �z �

32

� and

l2: �

x �

35

� � �y �

25

� � �z �

45

�.

A direction of l1

is d��1

� (4, 6, �3)

and of l2

is d��2

� (3, 2, 4). The direction of l1d��

1is

perpendicular to both l1

and l2. Therefore

d�� d��1

d��2

� (30, �25, �10) � 5(6, �5, �2). The

symmetric equation of a line through A(�6, 4, 3)

having direction (6, �5, �2) is �x �

66

� � �y�

�

54

�

� �z�

�

23

�.

13. a. The equation of a line l1

passing through C(0, 0, 2)

having direction vector d�� (3, 1, 6) is

r�� (0, 0, 2) � t(3, 1, 6). The parametric

equations are x � 3t, y � t, z � 2 � 6t. Check to see if A(�9, �3, �16) is on l byequating components 3t � �9, t � �3, 2 � 6t � �16t � �3 t � �3

Since t � �3 generates the point A, A(�9, �3,�16) is on the line l.Check B(6, 2, 14):3t � 6, t � 2, 2 � 6t � 14t � 2 t � 2

Since t � 2 generates the point B, B(6, 2, 14) is onthe line l.


b. The line segment AB is the set of points on the line x � 3t, y � t, z � 2 � 6t, for �3 t 2.

14. A line l has equation �x �

311

� � �y�

�

18

� � �z �

14

�.

The parametric equation of l is x � 11 � 3t,

y � �8 � t, z � 4 � t. A direction of l is d�� (3, �1, 1).

Let m be the required line passing through A(4, 5, 5)

and intersecting l at T. Since T is on l, represent its

coordinates as T(11 � 3t, �8 � t, 4 � t). Now a direction

of line m is AT�� (7 � 3t, �13 � t, �1 � t). Since l and

m are perpendicular, d�� · AT�� 021 � 9t � 13 � t � 1 � t � 011t � �33.

t � �3. Now AT�� (�2, �10, �4). Line m passes

through A(4, 5, 5) and has a direction (1, 5, 2). An

equation of m is r� � (4, 5, 5) � s(1, 5, 2).

15. a.

l is a line passing through P and having direction

vector d��. Q is a point, not on the line and FQ is the

distance from Q to the line. FQ�� ⊥ l. � is the angle

between PQ�� and d��.

From ∆PFQ, �FQ�� PQ�� sin �

� �PQ�� sin � ��d

d

�

�

�

��

�

� ��PQ��

��d

d�

�

�

�

�� sin ��.

Fl

Q

P �

→

d

T l

m

A(4, 5, 5)

d = (3, �1, 1)→

But �PQ�� d�� PQ��d�� sin �

therefore �FQ�� PQ��

�d��

�d��

�.

b. The distance from Q(1, �2, �3) to the line

r� � (3, 1, 0) � t(1, 1, 2). A point on the line is

P(3, 1, 0), therefore PQ�� (�2, �3, �3) and

d�� (1, 1, 2). PQ�� d�� (�3, 1, 1)

�PQ�� d�� 9 � 1�� 1� � �11�,

�d�� 1 � 1�� 4� � �6�.

Therefore the distance from Q

to the line is ��

1

6�1�

� � ��

666��.

c.

Two lines l1: r� � (�2, 2, 1) � t(7, 3, �4),

l2: r� � (2, �1, �2) � u(u, 3, �4).

The distance between two parallel lines is theperpendicular distance from a point on one of thelines to the other line.

A(�2, 2, 1) is a point on l1. The direction of l

2is

d��2

� (7, 3, �4) and a point on l2

is B(2, �1, �2).

Now BA�� (�4, 3, 3), BA�� d�� (�21, 5, �33).

Therefore �AF�� BA��

�d�

��d��

�

� ��

15

7

5

4�5�

�.

The distance between the two parallel lines is

��157545

�.

Exercise 7.4

4. a. r� � (�2, 0, �3) � t(5, 1, 3);

r� � (5, 8, �6) � u(�1, 2, �3).The parametric equations are:x � �2 � 5t x � 5 � uy � t y � 8 � 2uz � �3 � 3t z � �6 � 3uEquating components and rearranging gives:�2 � 5t� 5 � u 5t � u � 7 ➀

t � 8 � 2u t � 2u � 8 ➁�3 � 3t� �6 � 3u 3t � 3u � �3 ➂➀ � ➂ � 3: 4t � 8, t � 2, u � �3From ➁: t � 2u � 2 �2(�3) � 8.u � �3, t � 2 satisfies all three equations.Therefore the two lines intersect at the point (8, 2, 3).

b. line 1 line 2x � 1 � t x � 3 � 2uy � 1 � 2t y � 5 � 4uz � 1 � 3t z � �5 � 6uEquating components and rearranging terms:

1 � t � 3 � 2u t � 2u � 2 ➀1 � 2t � 5 � 4u 2t � 4u � 4 ➁1 � 3t � �5 � 6u 3t � 6u � 6 ➂

Each of equations ➀, ➁, and ➂ are equivalent.

Note that d��1

� (1, 2, �3), d��2

� (�2, �4, 6) �

�2(1, 2, �3) � �2d��1

therefore the lines are

parallel. Also a point on line 1 is (1, 1, 1) and it is

also on line 2 (u � 1) therefore the two lines are

coincident.

c. l1: r� � (2, �1, 0) � t(1, 2, �3);

l2: r� � (�1, 1, 2) � u(�2, 1, 1).

The parametric equations are:x � 2 � t x � �1 � 2uy � �1 � 2t y � 1 � uz � �3t z � 2 � uEquating components and rearranging gives:t � 2u � �3 ➀2t � u � 2 ➁3t � u � �2 ➂➁ � ➂: 5t � 0, t � 0 and u � �2.

From ➀ t � 2u � 0 � 4 � �4 � �3, the lines do

not intersect. Since d��1

� (1, 2, �3), d2

�

(�2, 1, 1) and d��1

≠ kd��2, the lines are not parallel.

Therefore the lines are skew.

A(�2, 2, 1)

F

l

l

B(2, �1, �2)

1

2



d. l1: (x, y, z) � (1 � t, 2 � t, �t);

l2: (x, y, z) � (3 � 2u, 4 � 2u, �1 � 2u).

Equating components and rearranging gives:t � 2u � 2t � 2u � 2t � 2u � 1

there is no solution to this system of equation hence

the lines do not intersect. d��1

� (1, 1, �1),

d��2

� (�2, �2, 2) � �2(1, 1, �1).

Since d��1

� �2d��2, the lines are parallel. Therefore

the lines are parallel and distinct.

e. l1: �

x �

43

� � �y �

12

�0 � z � 2;

l2: �

x�

�

32

� � �y �

21

� � �z�

�

12

�

The parametric equations are:x � 3 � 4t x � 2 � 3uy � 2 � t y � �1 � 2uz � 2 � t z � 2 � uEquating components and rearranging gives:4t � 3u � �1 ➀

t � 2u � �3 ➁t � u � 0 ➂

➂ – ➁: 3u � 3, u � 1, t � �1.Substitution into ➀: 4t � 3u � �4 � 3 � �1.u � 1, t � �1 satisfies all three equations.Therefore the two lines intersect at the point (�1, 1, 1).

5. a. l1: r� � (1, �1, 1) � t(3, 2, 1);

l2: r� � (�2, �3, 0) � u(1, 2, 3).

Equating components and rearranging gives:3t � u � �3 ➀

2t � 2u � �2 ➁t � 3u � �1 ➂

➀ � ➁ � 2: 2t � �2, t � �1, u � 0.t � �1, u � 0 also satisfies ➂. Therefore the twolines intersect at A(�2, �3, 0).

b. The direction of the two lines are d��1

� (3, 2, 1) and

d��2

� (1, 2, 3).

d��1

d��2

� (4, �8, 4).

A direction perpendicular to both given lines is

(1, �2, 1). The equation of the line passing through

A(�2, �3, 0) with direction (1, �2, 1) is

r� � (�2, �3, 0) � s(1, �2, 1).

6. l1: r� � (4, 7, �1) � t(4, �8, �4),

l2: r� � (1, 5, 4) � u(�1, 2, 3).

Equating components and rearranging gives:4t � u � �3 ➀

8t � 2u � �2 ➁4t � 3u � �5 ➂

➀ � ➁ ÷ 2: 8t � �4, t � ��12

�, u � �1 which also

satisfies ➂ therefore the two lines intersect at

(2, 3, 1). The directions are d��1

� (1, 2, �1) and

d��2

� (�1, 2, 3).

d��1

· d��2

� �1 � 4 �3 � 0. Therefore the two lines

intersect at right angles at the point (2, 3, 1).

7. x � 24 � 7t, y � 4 � t, z � �20 � 5t.For the x-intercept, both y � 0 and z � 0 which is truefor t � �4. Therefore the x-intercept is 24 � 7(�4) ��4.For the y-intercept, both x � 0 and z � 0 for the samet, which is not possible, therefore there is no y-intercept.Similarly there is no z-intercept.

8. Given the line 10x � 4y � 101 � 0 and a point A(3, �4). A normal to the line is n�� (5, 2) which is adirection for a line perpendicular to 10x � 4y � 101 � 0. A vector perpendicular to n�� is (2, �5).The equation of the line perpendicular to 10x �4y � 101 � 0 and passing through A(3, �4) is(2, �5) · (x � 3, y � 4) � 0

2x � 5y � 26 � 0.Now solving 10x � 4y � 101 ➀

2x � 5y � 261 ➁➀ � 5 ➁: 29y � �29

y � �1, x � �221�.

The point of intersection is ��221�, �1�.

9. Three lines l1, l

2, and l

3are in the same plane. Possible

intersections are:

a. All intersect in a common point.

1l

2l

3l

b. Pairs of lines intersect, but there is no commonintersection.

c. One line intersects two distinct parallel lines.

d. One line intersects two coincident lines.

e. The three lines are coincident.

10. Three lines l1, l

2, l

3are in space. The possible

intersections, in addition to those of question 9, are:

a. A line intersecting two skew lines.

l1

and l2

are skew. l3

intersects l1

and l2

at A and B.

b. Two parallel lines and the third intersecting one ofthe two.

l1�l

2, l

3intersects l

2at A.

11.

Given line l: r� � (7, �13, 8) � t(1, 2, �2), pointA(�5, �4, 2).The line m, through A, intersects l at right angles. LetT be the point of intersection of l and m. Since T is onl, we represent its coordinates by T(7 � t, �13 � 2t,8 � 2t).

AT�� (12 � t, �9 � 2t, 6 � 2t) is a direction of m.

Since l and m are perpendicular, their directions are

perpendicular, hence d�� · AT�� 0.(1, 2, �2) · (12 � t, �9 � 2t, 6 � 2t) � 0

12 � t � 18 � 4t �12 � 4t � 09t � 18

t � 2.

Now AT�� (14, �5, 2) and the equation of m is

r� � (�5, �4, 2) � s(14, �5, 2). The coordinates of

the point of intersection are T(9, �9, 4).

12. The line r� � (0, 5, 3) � t(1, �3, �2) with parametric

equations x � t, y � 5 � 3t, z � 3 � 2t, intersects the

sphere x2 � y2 � z2 � 6. (The centre of the sphere is

C(0, 0, 0) and its radius is �6�.) Substituting for x, y,

and z into the equation of the sphere:t2 � (5 � 3t)2 � (3 � 2t)2 � 6

t2 � 25 � 30t � 9t2 � 9 � 12t � 4t2 � 614t2 � 42t � 28 � 0

t2 � 3t � 2 � 0(t � 2)(t � 1) � 0

t � 2 or � 1

T

l

m

B (7, �13, 8)

d=(1, 2, �2)

A (�5, �4, 2)→

A

l3

l1

l2

A

B

l2

l1

l3

l1 l2 l3, , and

l 1

l 2 and l 3

l 3

l 2

l 1

l 1

l 2

l 3


therefore the line intersects the sphere at A(1, 2, 1)

and B(2, �1, �1). The midpoint of AB is ��32

�, �12

�, 0�which is not the centre of the sphere, therefore AB isnot a diameter of the sphere.

13.

Two lines: l1: r� � (2, �16, 19) � t(1, 1, �4);

l2: r� � (14, 19, �2) � u(�2, 1, 2). A line l

3passes

through the origin and intersects l2

at A and l1

at B.

Since B is on l1, we represent its coordinates as

B(2 � t, �16 � t, 19 � 4t).

Similarly the coordinates of A will be

A(14 � 2u, 19 � u, �2 � 2u).

Since O, A, and B are collinear

OB�� kOA��

(2 � t, �16 � t, 19 � 4t) � k(14 � 2u, 19 � u, �2 � 2u).Equating components: 2 � t � 14k � 2ku ➀

�16 � t � 19k � ku ➁19 � 4t � �2k � 2ku ➂

We solve by first eliminating ku:➀ � ➂: 21 � 3t � 12k ➃

➀ � 2 ➁: �30 � 3t � 52k ➄add: �9 � 64k

k � ��694�.

Substitute in ➃: 21 � 3t � 12k7 � t � 4k

t � 7 � 4 · �694�

t � �11261

�.

With t � �11261

�, we can find the coordinates of B hence

we have OB�� 11563

�, ��

11635�, �

�

11680��.

A direction for l3

is (17, �15, �20) and the equation

of l3

is r� � s(17, �15, �20).

NOTE: To determine the equation of l3

we require a

direction. Once t � �11261

� is established, a

direction is evident and further substitutions to determine u are not required. Uponsubstitution, one would find that u � 41.

14. a.

A line l with equation Ax � By � C � 0. N is on l

so that ON ⊥ l. The normal to l, n�� (A, B), is a

direction of the line m along ON. The equation of

m is r� � t(A, B); x � At, y � Bt.

Substituting in l: A2t � B2t � C � 0

t � �A2

�

�

C

B2�.

The coordinates of N are ��A2

�

�

AC

B2�, �

A

�2 �

BC

B2��.

b. ON�� A2

�

�

AC

B2�, �

A

�2 �

BC

B2��

�ON�� A

(

2

A

C2

2

�

�B

B2

2

)

C2

2

��

C

(

2

A

(A2

2

��

B

B2)

2)�

� ��A2

C

�

2

B2�

�ON�� A

�2

C

�

�

B�2��.

y

N

x

m

l

O

z A

y

xl3

B

l1

l2

O


15. Two skew lines, l1: (x, y, z) � (0, �1, 0) � s(1, 2, 1);

l2: (x, y, z) � (�2, 2, 0) � t(2, �1, 2). Let the line l

intersect l1

at A, l2

at B so that l ⊥ l1

and l ⊥ l2. Let the

coordinates of the intersection points be

A(s, �1 � 2s, s) and B(�2 � 2t, 2 � t, 2t).

BA�� (2 � s � 2t, �3 � 2s � t, s � 2t).

BA�� is perpendicular to l1

therefore BA�� · d��1

� 0

2 � s � 2t � 6 � 4s � 2t � s � 2t � 0

6s � 2t � 4, 3s � t � 2 ➀

BA�� is perpendicular to l2, therefore BA�� · d��

2� 0

4 � 2s � 4t � 3 � 2s � t � 2s � 4t � 02s � 9t � �7 ➁9 � ➀ � ➁: 25s � 25s � 1, t � 1.The coordinates of the points of intersection are A(1, 1, 1) and B(0, 1, 2).

16. The distance between two skew lines r� � OP�� td��1,

r� � OQ�� sd��2

is given by ��PQ�

�n

�

��

·

�n��

� where n�� d��1

d��2.

a. r� � (0, �2, 6) � t(2, 1, �1);

r� � (0, �5, 0) � s(�1, 1, 2).

The two points P(0, �2, 6), Q(0, �5, 0)

and directions d��1

� (2, 1, �1), d��2

� (�1, 1, 2).

Now PQ�� (0, �3, �6), n�� d��1

d��2

� (3, �3, 3)

��PQ�

�n

�

��

·

�n��

� � ��9

3

�

�

1

3�

8��

�

3

3�� 3�.

The distance between the lines is �3�.

b. x � 6, y � �4 � t, z � t; x � �2s, y � 5,

z � 3 � s. Two points P(6, �4, 0) and Q(0, 5, 3).

d��1� (0, �1, 1), d��

2� (�2, 0, 1).

Now PQ�� (�6, 9, 3),

n�� d��1

d��2

� (�1, �2, �2).

��PQ�

�n

�

��

·

�n��

� � ��

�6

1

�

�

18

4�

�

�

6

4�

��

1

3

8� � 6.

The distance between the lines is 6.

Review Exercise

2. a. A line through A(3, 9), B(�4, 2) has a direction AB��

� (�7, �7). An equation is r� � (3, 9) � t(1, 1).

b. A line passes through A(�5, �3) and is parallel to r�

� (4, 0) � t(0, 5). A direction is (0, 1) and an

equation is r� � (�5, �3) � s(0, 1).

c. A line passes through A(0, �3) and is perpendicular

to l: 2x � 5y � 6 � 0. A normal to l is (2, �5)

which is a direction of the required line. An

equation will be r� � (0, �3) � s(2, �5).

3. a. A line passes through A(�9, 8) and has slope

��23

�. A direction is (3, �2) and parametric

equations are x � �9 � 3s, y � 8 � 2s.

b. A line passes through A(3, �2) and is perpendicular

to l: r� � (4, �1) � t(3, 2). A direction of l is (3, 2);

a direction perpendicular to (3, 2) is (2, �3). An

equation of the line is x � 3 � 2s, y � �2 � 3s.

c. A line through A(4, 0), B(0, �2) has direction

AB�� (�4, �2)

� �2(2, 1).The line has equation x � 4 � 2t, y � t.

4. a. The line passes through A(2, 0, �3), B(�3, 2, �2).

A direction is AB�� (5, �2, �1) and a vector

equation is r� � (2, 0, �3) � t(5, �2, �1).

A

B

l

l1

l2

�


b. An x-intercept of �7 and a y-intercept of 4 means

the line passes through A(�7, 0, 0), B(0, 4, 0) and

a direction is AB�� (7, 4, 0). A vector equation is

r� � (�7, 0, 0) � t(7, 4, 0).

c. A line l, parallel to �x �

45

� � �y�

�

22

� � �z �

56

�,

passing through (0, 6, 0), has direction (4, �2, 5).

An equation of l is r� � (0, 6, 0) � t(4, �2, 5).

5. a. A line l passes through the origin and is parallel to

the line �x�

�

31

� � �y�

�

22

� � z � 3. A direction of

l is (3, 2, �1). Parametric equations for l are:

x � 3t, y � 2t, z � �t.

b. The line passes through A(6, �4, 5) and is parallelto the y-axis. A direction is (0, 1, 0). Parametricequations are x � 6, y � �4 � t, z � 5.

c. A line with z-intercept �3 with direction vector (1, �3, 6) passes through the point (0, 0, �3).Parametric equations are x � t, y � �3t,z � �3 � 6t.

6. a. A line l passes through A(�1, �2) and is parallelto 3x � 4y � 5 � 0. Line l will have equation 3x � 4y � c � 0. Since A � l, 3(�1) � 4(�2) � c � 0, c � �5 and the scalar equation is 3x � 4y � 5 � 0.

b. A line l passes through A(�7, 3) and isperpendicular to x � 2 � t, y � �3 � 2t. A normalof l is (1, 2) hence an equation is (1, 2) · (x � 7,y � 3) � 0, x � 2y � 1 � 0.

c. A line perpendicular to x � 4y � 1 � 0 will have a

direction vector d�� (1, 4). A vector perpendicular

to d�� is (4, �1). Therefore the equation of a line

through the origin with normal (4, �1) is

4x � y � 0.

7. a. A line l through A(6, 4, 0) and parallel to a line

through B(�2, 0, 4), C(3, �2, 1) has direction

BC�� (5, �2, �3) and parametric equations x � 6

� 5t, y � 4 � 2t, z � �3t.

b. Since (�4, m, n) � l, 6 � 5t � �4, t � �2; 4 � 2t � m, m � 8;

�3t � n, n � 6.

8. a. l1: r� � (2, 3) � t(�3, 1), d��

1� (�3, 1)

l2: r� � (�1, 4) � u(6, �2), d��

2� (6, �2)

� �2(�3, 1). Since d��2

� �2d��1, l

1and l

2are

parallel. The point (�1, 4) is a point on l1(t � 1),

therefore the two lines are coincident.

b. l1: x � 1 � 2t, y � �3 � t; d��

1� (2, �1).

l2: x � u, y � �

13

� � 2u; d��2

� (1, 2).

Since d��1

· d��2

� 0, the two lines are perpendicular.

c. l1: �

x �

21

� � �y �

14

�, z � 1; d��1

� (2, 1, 0).

l2: x � 4t, y � 1 � 2t, z � 6; d��

2� (4, 2, 0)

� 2(2, 1, 0). Since d��2

� 2d��1, l

1and l

2are parallel.

Since points on l1

are of the form (a, b, 1) and on l2

of the form (p, q, 6), there are no points common to

the two lines; hence the lines are parallel and

distinct.

d. l1: (x, y, z) � (1, 7, 2) � t(�1, �1, 1),

d��1

� (�1, �1, 1).

l2: (x, y, z) � (�3, 0, 1) � u(2, �2, �2),

d��2

� (2, �2, �2) � 2(1, �1, �1). Since d��1

� kd��2,

d��1

· d��2

� 0, the two lines are neither parallel nor

perpendicular.

9. Parametric equations of the line are x � �4 � 2t,y � 6 � t, z � �2 � 4t. If it meets the xy-plane, z � 0,

t � �12

�, and the point is (�3, �121�, 0).

If it meets the xz-plane, y � 0, t � 6, and the point is(8, 0, 22). If it meets the yz-plane, x � 0, t � 2, and the point is (0, 4, 6).

10. a. The symmetric equations of the line are

�x�

�

12

� � �y �

53

� and the scalar equation is

5x � y � 13 � 0.


b. The line 5x � 2y � 10 � 0 has normal (5, �2). A

direction is (2, 5) and a point on the line is (0, 5).

A vector equation is r� � (0, 5) � t(2, 5).

c. The line y � �34

�x � �12

� has slope �34

� and a direction

(4, 3). A point on the line is (2, 2) and a vector

equation is r� � (2, 2) � t(4, 3).

11. The parametric equations of the line are x � 12 � 3t,y � �8 � 4t, z � �4 � 2t.

a. Intersection with:xy-plane, z � 0, t � 2, the point is (6, 0, 0)xz-plane, y � 0, t � 2, the point is (6, 0, 0)yz-plane, x � 0, t � 4, the point is (0, 8, 4).

b. The x-intercept is 6 and is the only intercept.

c.

12. a. The line �x �

53

� � �y �

26

� � �z�

�

11

� has direction

d�� (5, 2, �1), �d�� 30�; direction cosines

cos � � ��

5

30��, cos � � �

�2

30��, cos δ � �

��

3

1

0��;

and direction angles � � 24�, � � 69�, δ � 101�.

b. The line x � 1 � 8t, y � 2 � t, z � 4 � 4t has

direction d�� (8, �1, �4), �d�� 9; direction

cosines cos � � �89

�, cos � � ��91�, cos δ � �

�94�;

and direction angles � � 27�, � � 96�, δ � 116�.

c. The line r� � (�7, 0, 0) � t(4, 1, 0) has direction

d�� (4, 1, 0), �d�� 17�; direction cosines

cos � � ��

417��, cos � � �

�117��, cos δ � 0;

and direction angles α � 14�, � � 76�, δ � 90�.

13. a. The parametric equation of the two lines are:l1: x� 4t, y � 3t, z � 2 � 4t and l

2: x � �4 � 4u,

y � 1 � u, z � �2u.Equating components and rearranging4t � 4u � �4 ➀3t � u � 1 ➁

4t � 2u � �2 ➂2 � ➁ � ➂: 10t � 0, t � 0, u � �1. Substitute in ➀: 4(0) � 4(�1) � �4 whichverifies. Therefore the lines intersect at (0, 0, 2).

b. Two lines x � t, y � 1 � 2t, z � 3 � t and

x � �3, y � �6 � 2u, z � 3 � 6u.Equating components gives t � �3.

�5 � � 6 � 2u, u � �12

�

6 � 3 � 6u, u � ��12

� � �12

�; therefore the two linesdo not intersect.

14. a. P(2, 1, 3), Q(0, �4, 7). The distance between these

points is �QP�� where QP�� (2, 5, �4) and

�QP�� 4 � 25� � 16�

� �45�.

The shortest distance between P and Q is 3�5�.

b. The distance from A(3, 7) to 2x � 3y � 7 � 0 is

given by d ��Ax

�1

�

A2

B

�

y1

B�

�

2�

C��

� ��6 �

�

21

13�

� 7��

d � ��

22

13��.

The shortest distance from the point to the line is

�22�

1313�

�.

z

y

x

6

4

8

(0, 8, 4)



c. From the point A(4, 0, 1) to the line,

r� � (2, �2, 1) � t(1, 2, �1). A direction of the

line is d�� (1, 2, �1) and a point on the line is

P(2, �2, 1).

Now PA�� (2, 2, 0), PA�� d�� (�2, 2, 2)

�PA�� d�� 2�3�, �d�� 6�.

The perpendicular distance from the point A to the

line is ��PA��

�d�

��d��

� � �2

�

�

6�

3�� 2�.

d. From the point A(1, 3, 2) to the line

�x

�

�

1

1� � �

y �

1

3� � �

z �

2

7�. A direction of the line is

d�� (�1, 1, 2). A point on the line is P(1, 3, 7).

Now AP�� (0, 0, 5), AP�� d�� (�5, �5, 0),

�AP�� d�� 5�2�, �d�� 6�.

The perpendicular distance from A to the line is

given by ��PA��

d�

�

d��

5

�

�

6�

2��

5�

3

3��.

15.

Let the foot of the perpendicular be A(�6 � 5t, �7 � 3t, �3 � 4t)

QA�� (�9 � 5t, �9 � 3t, �7 � 4t).

Since QA�� is perpendicular to the line, QA�� · d�� 0

therefore �45 � 25t � 27 � 9t � 28 � 16t � 050t � 100

t � 2.The coordinates of the foot of the perpendicular are(4, �1, 5).

Chapter 7 Test

1. A line through A(9, 2), B(3, 4) has direction

AB�� (�6, 2) � �2(3, �1).

a. A vector equation is r� � (9, 2) � t(3, �1).

b. Parametric equations are x � 9 � 3t, y � 2 � t.

c. Symmetric equations are �x �

3

9� � �

y

�

�

1

2�.

d. The scalar equation is x � 3y � 15 � 0.

2. A line l is perpendicular to 2x � 3y � 18 � 0

therefore its direction is d�� (2, �3). The y-intercept

of (x, y) � (0, 1) � t(�3, 4) is 1. The symmetric

equation of l is �2

x� � �

y

�

�

3

1� and the scalar equation is

3x � 2y � 2 � 0.

3. The line �x �

6

2� � �

y �

3

4� � �

z

�

�

3

2� has direction

d�� (6, 3, �3) � 3(2, 1, �1). Parametric equations

are x � 2 � 2t, y � 4 � t, z � �2 � t. For an

intersection with the xy-plane, z � 0, t � �2, and

x � �2, y � 2. The point in the xy-plane is A(�2, 2, 0).

For an intersection in the yz-plane, x � 0, t � �1, and

y � 3, z � �1. The point in the yz-plane is

B(0, 3, �1). The intersection with the xz-plane is

C(�6, 0, 2).

z C

A

y

B

x

A

l

Q (3, 2, 4)

d=(5, 3, 4)→

4. The line x � y � z � 2 has direction d�� (1, 1, 1).

A point on the line is A(0, 0, 2). P has coordinates

(1, �2, �3) and AP�� (1, �2, �5). AP�� d��

(3, �6, 3).

�AP�� d�� 9 � 36� � 9� � 3�1 � 4�� 1� � 3�6�.

�d�� 3�.

The perpendicular distance from P to the line is

��AP��

�d��

�d��

� � �3

�

�

3�

6�� 3�2�.

5. A line through (0, 0, 0) has direction angles � � 120�,

� � 45�.

Since cos2 � � cos2 � � cos2 � � 1,

cos2 � � cos2 120� � cos2 45� � 1;

cos 120� � ��

21�,

cos � � ��

1

2��

�

2

2��

cos2 � � �1

4� � �

1

2� � 1

cos2 � � �1

4�, cos � � �

1

2� or �

�

21�.

Now a direction is (1, �2, �2�) or (�1, �1, �2�).

Vector equations of the two lines are

r� � (1, �1, �2�)s and r� � (�1, �1, �2�)t.

6. Given l1: x � �2 � 5t, y � t, z� �3 � 3t

and l2: x � 5 � s, y � 8 � 2s, z � �6 � 3s.

Equating components and rearranging:5t � s � 7 ➀t � 2s � 8 ➁3t � 3s � �3 ➂➁ � ➂ � 3: �3s � 9, s � �3, t � 2from ➀: 5(2) �3 � 7, which verifies. The lines intersect at the point (8, 2, 3).

7. Given l1: x � �8 � t, y � �3 � 2t, z � 8 � 3t,

d��1

� (1, �2, 3) and

l2: x � 1 � 2s, y � �1 � s, z � 3s, d��

2� (2, 1, 3).

Since the direction d��1

� kd��2, the lines are not parallel.

Equating components gives:t � 2s � 9 ➀2t � s � �2 ➁3t � 3s � �8 ➂2 ➁ � ➀: 5t � 5, t � 1, s � �4.

From ➂, 3(1) �3(�4) � 3 � 12 � 15 � �8therefore the lines do not intersect. Since the lines are not parallel and do not intersect,the two lines are skew.

b. When t � �2, the coordinates of P1

are (�10, 1, 2).

c.

c. Let the coordinates of P2

be (1 � 2s, �1 � s, 3s).

Now P1P

2�� (11 � 2s, �2 � s, �2 � 3s).

Since P1P

2is perpendicular to l

2P

1P

2�� · d��

2� 0,

therefore 22 � 4s � 2 � s � 6 � 9s � 0

14s � �14

s � �1.

The coordinates of P2

are (�1, �2, �3).

l2

l1

P2

P1 (�10, 1, 2)

→d = (2, 1, 3)2



Exercise 8.1

7. Vector equations of planes:

a. Through A(�4, 5, 1), parallel to vectors a�� (�3,

�5, 3) and b�� (2, �1, �5). A vector equation is

r� � (�4, 5, 1) � s(�3, �5, 3) � t(2, �1, �5).

b. Contains two intersecting lines l1: r� � (4, 7, 3)

� t(1, 4, 3) and l2: r� � (�1, �4, 6) � s(�1, �1, 3).

A point on the plane is A(4, 7, 3) and two

directions are a�� (1, 4, 3) and b�� (�1, �1, 3).

A vector equation of the plane is

r� � (4, 7, 3) � s(1, 4, 3) � t(�1, �1, 3).

c. Contains the line r� � (�3, 4, 6) � t(�5, �2, 3)

and the point A(8, 3, 5). A second point on the

plane is B(�3, 4, 6). Two directions are a�� (�5,

�2, 3) and BA�� (11, �1, �1). A vector equation

is r� � (8, 3, 5) � s(�5, �2, 3) � t(11, �1, �1).

d. Contains parallel lines l1: r� � (0, 1, 3) � t(�6,

�3, 6) and l2: r� � (�4, 5, �4) � s(4, 2, �4).

A direction of the parallel lines is a�� (2, 1, �2).

Two points on the plane are A(0, 1, 3) and B(�4, 5,

�4), therefore a second direction is BA�� (4, �4,

7). A vector equation is r� � (0, 1, 3) � s(2, 1, �2)

� t(4, �4, 7).

e. Contains the three points A(2, 6, �5), B(�3, 1,

�4), C(6, �2, 2). Two directions are BA�� (5, 5,

�1) and AC�� (4, �8, 7). Choosing any of the

three points, a vector equation is r� � (2, 6, �5) �

s(5, 5, �1) � t(4, �8, 7).

8. Parametric equations of planes:

a. Through A(7, �5, 2), parallel to vectors a�� (4,

�1, 1) and b�� (�3, 4, 4). Parametric equations

are x � 7 � 4s � 3t, y � �5 � s � 4t, z � 2 � s

� 4t.

b. A plane contains the two intersecting lines

l1: r� � (5, 4, 2) � t(4, �2, 1) and

l2: r� � (7, 4, �7) � s(�3, 1, 4).

Since the plane contains both lines, two directions

of the plane will be a�� (4, �2, 1) and b�� (�3,

1, 4). Any point on l1

or l2

can be used, therefore

parametric equations of the plane arex � 5 � 4t � 3sy � 4 � 2t � sz � 2 � t � 4s.

c. Contains the line r� � (1, 3, �1) � t(2, 2, �5) and

the point A(8, 3, 5). A direction is a�� (2, 2, �5),

a second point is B(3, 5, �6), (when t � 1),

therefore another direction is BA�� (5, �2, 11).

An equation is x � 8 � 5s � 2t, y � 3 � 2s � 2t,

z � 5 � 11s � 5t.

d. Contains two parallel lines l1: r� � (3, 2, 2) � t(�9,

6, �6) and l2: r� � (1, 6, �6) � s(6, �4, 4). Two

points in the plane are A(3, 2, 2) and B(1, 6, �6).

Two directions are a�� (3, �2, 2) and BA��

(2, �4, 8) � 2(1, �2, 4). Parametric equations are

x � 3 � s � 3t, y � 2 � 2s � 2t, z � 2 � 4s � 2t.

e. Contains the three points A(2, 6, 5), B(�3, 1, �4),

C(6, �2, 2). From 7e, parametric equations are

x � 2 � 5s � 4t, y � 6 � 5s � 8t, z � �5 �

s � 7t.

9. a. A plane parallel to the yz-plane has directions

a�� (0, 1, 0) and b�� (0, 0, 1). This plane passes

through the point A(6, 4, 2). A vector equation is

r� � (6, 4, 2) � s(0, 1, 0) � t(0, 0, 1).

b. A plane passes through O(0, 0, 0), A(3, 3, 3), and

B(8, �1, �1). Two directions are OA�� (3, 3, 3)

� 3(1, 1, 1) and OB�� (8, �1, �1). An equation is

r� � s(1, 1, 1) � t(8, �1, �1).

Chapter 8: Equations of Planes 109

Chapter 8 • Equations of Planes

c. A plane contains the x-axis and the point A(�1,

�4, �7). Two points on the x-axis are O(0, 0, 0)

and B(1, 0, 0). Two directions are AO�� (1, 4, 7)

and OB�� (1, 0, 0). An equation is r� � s(1, 0, 0) �

t(1, 4, 7).

10. a. The three points A(2, 3, �1), B(8, 5, �5), and

C(�1, 2, 1) give directions AB�� (6, 2, �4) �

2(3, 1, �2) and AC�� (�3, �1, 2) � �1(3, 1, �2).

Since AB�� 2AC��, A, B, and C are collinear and

three collinear points do not define a unique plane.

b. The point P(8, �7, 5) is on the line r� � (4, 9, �3)

� t(1, �4, 2), (t � 4). Collinear points do not

define a unique plane.

11. The plane contains the line x � 7 � t, y � �2t,

z � �7 � t. A point on the plane will be A(7, 0, �7),

(t � 0), and a direction is a�� (�1, �2, 1) � �1(1,

2, �1). Since the plane does not intersect the z-axis,

the z-axis will be parallel to the plane, hence a second

direction will be b�� (0, 0, 1). A vector equation of

the plane is r� � (7, 0, �7) � s(0, 0, 1) � t(1, 2, �1).

Parametric equations are x � 7 � t, y � 2t, z � �7 �

s � t.

12. A plane has equations r� � (a, b, c) � s(d, e, f) �t(a, b, c). If s � 0 and t � �1, r � (0, 0, 0), hence the plane passes through the origin.

13. a.

A, B, and C are three points on the plane. Let two

directions be

AB�� a�� b�� and

AC�� a�� c��.

With point A, the vector equation of the plane

containing A, B, and C is

r� � a�� s(�a�� b��) � t(�a�� c��)

� (1 � s � t)a�� b��s � c��t

or r� � pa�� sb�� tc�� where p � 1 �s � t

or p � s � t � 1.

b. r� � (1 �s � t)a�� sb�� tc��, 0 � s � 1, 0 � t � 1.

Now if s � t � 0, r� � a��

s � 0, t � 1, r� � c��

s � 1, t� 0, r� � b��

s � t � 1, r� � �a�� b�� c��.

Now �a�� b�� c�� c�� a�� b��

� OC�� CD��, CD�� a�� b��

� OD��.

CD�� a�� b�� AB��, therefore ABCD is a

parallelogram and the fourth vertex D has position

vector OD�� a�� b�� c��. Hence the region in the

plane is all points in and on the parallelogram

whose vertices have position vectors a��, b��, �a�� b��

� c��, and c��.

14. a.

A line l has equation r� � r�0

� td��. r�0

is the position

vector of point R on the line. Q is a point not on l

and has position vector q��. Two directions of the

R

Q

l→

d

→r

→q

O

0

A

B

D

C

→a

→

b→c

�a +

b→

→

�a +

b→

→

c – a + b=�a + b + c

→ → →

→ → →

AB

C

O

→a

→

b

→c

110 Chapter 8: Equations of Planes

plane are d�� and RQ�� r�0

� q��.

The vector equation of this plane will be

r� � r�0

� l (�r�0

� q��) � td��

� (1 � l)r�0

� lq�� td��

r� � kr�0

� lq�� td��, 1 � l � k

or k � l � 1.

b. r� � kr�0

� lq�� td��, k � l � 1

r� � kr�0

� (1 � k)q�� td��, 0 � k � 1.

k � 0 gives the line r� � q�� td��. The line passes

through Q and has direction d��.

k � 1 gives the line r� � r�0

� td��. The line passes

through R and has direction d��.

Therefore the region of the plane determined by

0 � k � 1 is the region between and including the

two parallel lines through R and Q with direction d��.

Exercise 8.2

8. A plane contains the x-axis and the point A(4, �2, 1).

O(0, 0, 0) is on the x-axis, therefore a direction is

OA�� (4, �2, 1). A second direction is the direction

of the x�axis, i � (1, 0, 0). A normal is OA�� i � n��

� (0, 1, 2). P(x, y, z) is a point on the plane, hence

AP�� · n�� 0. The plane has equation

(x � 4, y � 2, z � 1) · (0, 1, 2) � 0y � 2z � 0.

9. A plane contains the intersecting lines

�x �

12

� � �2

y� � �

z �

3

3� and �

x

�

�

3

2� � �

4

y� � �

z �

23

�.

The common point is A(2, 0, �3). Two directions are

a�� (1, 2, 3) and b�� (�3, 4, 2). A normal to the

plane is n�� a�� b�� (�10, �11, 10) � �(10, 11,

�10). The scalar equation is AP�� · n�� 0

(x � 2, y, z � 3) · (10, 11, �10) � 0

10x � 11y � 10z � 50 � 0.

10. a. �1: x � 3y � z � 2 � 0, n��

1� (1, 3, �1).

�2: 2x � 6y � 2z � 8 � 0, n��

2� (2, 6, �2) �

2(1, 3, �1). Since n��2

� 2n��1, the planes are parallel.

Since (2, 6, �2, �8) � 2(1, 3, �1, �4) � 2(1, 3,

�1, �2), the two planes are distinct. Hence the

two planes are parallel and distinct.

b. �1: 2x � y � z � 3 � 0, n��

1� (2, 1, 1).

�2: 6x � 2y � 2z � 9 � 0, n��

2� (6, 2, 2) �

2 (3, 1, 1).

Since n��2

� kn��1, the two planes are distinct and they

intersect.

c. �1: 3x � 3y � z � 2 � 0, n��

1� (3, �3, 1)

�2: 6x � 6y � 2z � 4 � 0, n��

2� (6, �6, 2)

� 2(3, �3, 1). Since n��2

� 2n��1, the planes are

parallel. Now (6, �6, 2, �4) � 2(3, �3, 2, �2).

Therefore �2

� 2�1

and the planes are coincident.

d. �1: 2x � 4y � 2z � 6 � 0, n��

1� (2, �4, 2)

� 2(1, �2, 1).

�2: 3x � 6y � 3z � 9 � 0, n��

2� (3, �6, 3)

� 3(1, �2, 1).

n��2

� �2

3�n��

1, therefore the planes are parallel.

(3, �6, 3, �9) � 3(1, �2, 1, �3)

(2, �4, 2, �6) � 2(1, �2, 1, �3)

�2

� �23

��1, therefore the planes

are coincident.

11. a. �: 2x � y � 3z � 24 � 0. Let x � s, z � t.

Therefore y � �24 � 2s � 3t. A vector equation

is r� � (0, �24, 0) � s(1, 2, 0) � t(0, 3, 1).

b. �: 3x � 5z � 15 � 0. Solve for z: z � 3 � �3

5�x.

Let x � 5s, y � t, z � 3� 3s. A vector equation is

r� � (0, 0, 3) � s(5, 0, 3) � t(0, 1, 0).

12. �: 4x � y � z � 10 � 10. A normal to � is

n�� (4, 1, �1).

a. r� � (3, 0, 2) � t(1, �2, 2). A direction of the line

is d�� (1, �2, 2) since d�� · n�� 4 � 2 � 2 � 0.

The line is parallel to the plane. A point on the line

is A(3, 0, 2). Since A satisfies the equation of �,

the point is on the plane, hence the line lies on

the plane.


b. x � �3t, y � �5 � 2t, z � �10t. A direction of

the line is d�� (�3, 2, �10).

d�� · n�� 12 � 2 � 10 � 0, therefore the line

is parallel to the plane. A point on the line is

A(0, �5, 0), (t � 0). Since A does not satisfy the

equation of the plane, the line is not on the plane.

c. �x �

4

1� � �

y

�

�

1

6� � �

1

z�. A direction of the line is

d�� (4, �1, 1).

d�� · n�� 16 � 1 � 1 � 0. The line is not parallel

to the plane, hence it must intersect the plane at

some point.

13. a. �1: 2x � 3y � z � 9 � 0, n��

1� (2, 3, �1).

�2: x � 2y � 4 � 0, n��

2� (1, 2, 0).

The angle between n��1

and n��2

is where

n��1

· n��2

� �n��1��n��

2� cos

2 � 6 � �4 � 9�� 1� �1 � 4� cos

cos � ��14�

8

�5��

� 17.The angle between the planes is approximately 17.

b. �1: x � y � z � 1 � 0, n��

1� (1, �1, �1).

�2: 2x � 3y � z � 4 � 0, n��

2� (2, 3, �1).

n��1

· n��2

� 2 � 3 � 1 � 0, therefore n��1

⊥ n��2

and the

angle between the planes is 90.

14. a. l: x � 0, y � t, z � 2t, d�� (0, 1, 2)

�: 2x � 10y � 5z � 1 � 0, n�� (2, �10, 5)The line intersects the z-axis at (0, 0, 0).

The z-intercept is �1

5�.

Since d�� · n�� 0 � 10 � 10 � 0, the line is parallel

to the plane and since the z-intercept of the line is 0

and of the plane is �1

5�, the line lies below the plane.

b. �: x � 4y � 2z � 7 � 0, n�� (1, 4, �2)

d�� · n�� 0 � 4 � 4 � 0, the line is parallel to the

plane. The z-intercept of the plane is ��

2

7�.

Therefore the line lies above the plane.

15. a. P(x, y, z) is equidistant from A(1, 2, 3) and

B(4, 0, 1) therefore �AP�� BP�� and �AP��2 � �BP��2

therefore (x � 1)2 � (y � 2)2 � (z � 3)2 �

(x � 4)2 � y2 � (z � 1)2. Squaring and collecting

terms gives

�2x � 1 � 4y � 4 � 6z � 9 � �8x � 16 � 2z � 16x � 4y � 4z � 3 � 0.

b. 6x � 4y � 4z � 3 � 0 is a plane that has normal

coincident with AB�� and passes through the

midpoint of AB, i.e., AB�� (3, �2, �2),

n�� (6, �4, �4) � 2(3, �2, �2). n�� 2AB��. The

midpoint of AB is M��5

2�, 1, 2�

6��5

2�� 4(1) � 4(2) � 3 � 15 � 4 � 8 � 3 � 0.

M satisfies 6x � 4y � 4z � 3 � 0 therefore the

plane passes through the midpoint of AB.

16. a. B

CA

P

→

b

→n

→a

→r

→c

O


a��, b��, and c�� are position vectors of points A, B, and

C. ABC defines a plane with two directions

AB�� a�� b��

and AC�� a�� c��.

A normal to the plane is

AB�� AC�� (b�� a��) � (c�� a��)

� b�� c�� b�� a�� a�� c�� a�� a��.

But a�� a�� 0�� hence a normal is n�� b�� c�� b��

� a�� a�� c��. P(x, y, z) is a point on the plane and

r� is the position vector of P. AP�� a�� r�. Since

AP�� is in the plane and n�� is perpendicular to the

plane, AP�� · n�� 0.

(r� � a��) · (b�� c�� b�� a�� a�� c��) � 0

but �b�� a�� a�� b��, �a�� c�� c�� a��,

therefore the scalar equation of the plane through

A, B, and C is

(r� � a��) · (a�� b�� b�� c�� c�� a��) � 0.

b. A(8, 4, �3), B(5, �6, 1), C(�4, 1, 2). The position

vectors of A, B, and C are a�� (8, 4, �3),

b�� (5, �6, 1),

c�� (�4, 1, 2).

b�� c�� (�13, �14, �19)

�b�� a�� (�14, �23, �68)

�a�� c�� (�11, 4, �24),

r� � a�� (x � 8, y � 4, z � 3).From 16a(x � 8, y � 4, z � 3) · (�38, �33, �111) � 0

38x � 33y � 111z � 304 � 132 � 333 � 038x � 33y � 111z � 103 � 0.

17. �: 2x � 3y � kz � 0. The plane intersects the xy-plane when z � 0. Therefore 2x � 3y � 0. Let x � 3t, y � �2t. The line of intersection of the plane in the xy-plane isthe line x � 3t, y � �2t, z � 0. This line passesthrough the origin (i.e., t � 0 gives the point (0, 0, 0)).Since the equation of the line is independent of k, allplanes 2x � 3y � kz � 0, k � R, intersect thexy-plane in this line and the plane rotates about thisline as k varies.

18. The distance from P1(x

1, y

1, z

1) to the plane Ax � By

� Cz � D � 0 is given by

d � .

The distance from the origin to the plane is

d ��A2 �

�D

B��2 � C2��.

If n�� (A, B, C) is a unit vector, �A2 � B�2 � C2� � 1

and d � �D�.

Therefore if the coefficients A, B, and C are thecomponents of a unit normal, �D� will represent thedistance from the origin to the plane.

19. a, b, and c are the x-, y-, and z-intercept of a plane.These intercepts correspond to the points A(a, 0, 0),B(0, b, 0) and C(0, 0, c). Two directions of this plane

are BA�� (a, �b, 0) and CA�� (a, 0, �c).

A normal is BA�� CA�� n�� (bc, ac, ab).

The scalar equation is AP�� · n��

(x � a, y, z) · (bc, ac, ab) � 0bcx � acy � abz � abc � 0

divide by abc: �ax

� � �by

� � �cz

� � 1.

The distance from the origin to this plane is

1d � ––––––––––––––––

��a12� � ��b

12� � �

c�12��

or �1d

� � ��a12� � ��b

12� � �

c�12��

or �d1

2� � �a12� � �

b12� � �

c12�.

20. The scalar equation of the plane with x-, y-, andz-intercepts a, b, and c from question 19 is

�ax

� � �by

� � �cz

� � 1. Developing this equation using the

results of question 16:

�Ax1

� By1

� Cz1

� D��

�A2 � B�2 � C2�


Let the position vectors of the three intercepts be

a�� (a, 0, 0), b�� (0, b, 0), c�� (0, 0, c).

Now a�� b�� (0, 0, ab), b�� c�� (bc, 0, 0),

c�� a�� (0, ac, 0).

a�� b�� b�� c�� c�� a�� (bc, ac, ab)

r� � a�� (x � a, y, z)

and (r� � a��) · (a�� b�� b�� c�� c�� a��) � 0

(x � a, y, z) · (bc, ac, ab) � 0

bcx � acy � abz � abc � 0

Divide by abc: �ax

� � �by

� � �cz

� � 1 � 0

or �ax

� � �by

� � �cz

� � 1

represents the equation of the plane having a, b, and c as

x-, y-, and z-intercepts respectively.

Exercise 8.3

4. a. The plane r� � (6, �4, 3) � s(�2, 4, 7) � t(�7, 6, �3) has parametric equationsx � 6 � 2s � 7ty � �4 � 4s � 6tz � 3 � 7s � 3t.

(i) To intersect the x-axis, y � 0, z � 0 therefore 4s � 6t � 4 ➀7s � 3t � �3 ➁

➀ � 2 � ➁: 9s � �1

s � ��19

�, t � �2207�

x � 6 � �29

� � �12470

�

� �162 �

267

� 140�

x � �2287�.

The plane intersects the x-axis at ��2287�, 0, 0�.


(ii) To intersect the y-axis, x � z � 0 therefore2s � 7t � 6 ➀7s � 3t � �3 ➁

3 � ➀ � 7 � ➁: 55s � �3,

s � ��535�,

t � �4585�, and y � �

5565�.

The plane intersects the y-axis at �0, �5565�, 0�.

(iii) To intersect the z-axis, x � y � 0 therefore2s � 7t � 6 ➀2s � 3t � 2 ➁

➀ � ➁: 4x � 4, t � 1, s � ��12

�, and z � ��72

�.

The plane intersects the z-axis at (0, 0, ��72

�).

b. (i) For an intersection with the xy-plane, z � 0.

Therefore 3 � 7s � 3t � 0, t � 1 � �73

�s.

Let s � 3k then t � 1 � 7k. Substitute for x andy, hencex � 6 � 6k � 7 � 49kx � �1 � 55k;y � �4 � 12k � 6 � 42ky � 2 � 54k.The intersection with the xy-plane is the line

r� � (�1, 2, 0) � k(�55, 54, 0).

(ii) An intersection with yz-plane, x � 0. Therefore

6 � 2s � 7t � 0, s � 3 � �72

�t.

Let t � 2p, s � 3 � 7p. Substitute for y and zy � �4 � 12 � 28p � 12py � 8 � 16p;z � 3 � 21 � 49p � 6pz � 24 � 55p.

The intersection with the yz-plane is the liner� � (0, 8, 24) � p(0, 16, 55).

(iii)An intersection with the xz-plane, y � 0.

Therefore �4 � 4s � 6t � 0, s � 1 � �32

�t.

Let t � �2u then s � 1 � 3u. Substitute for x and zx � 6 � 2 � 6u � 14ux � 4 � 8u;z � 3 � 7 � 21u � 6u z � 10 � 27u.The intersection with the xz-plane is the liner� � (4, 0, 10) � u(8, 0, 27).

5. A line r� � (6, 10, 1) � t(3, 4, �1) has parametricequations x � 6 � 3ty � 10 � 4tz � 1 � t.

a. The line meets the xy-plane when z � 0. Thereforet � 1 and x � 9, y � 14. The point is (9, 14, 0).

b. The line meets the xz-plane at y � 0. Therefore

t � ��52

� and x � ��32

�, z � �72

�.

The point is ��32

�, 0, �72

��.

c. The line meets the yz-plane at x � 0. Thereforet � �2 and y � 2, z � 3. The point is (0, 2, 3).

6. a. A line parallel to the x-axis will intersect a planeperpendicular to the x-axis in one point.

b. A line parallel to the y-axis could intersect a planeparallel to the y-axis in an infinite number of pointsor in no points.

c. A line perpendicular to the z-axis could intersect aplane parallel to the z-axis is in one point, aninfinite number of points, or in no points.

7. The plane 3x � 2y � 7z � 31 � 0 has normal n�� (3, �2, 7), which is the direction of the line throughthe origin. An equation of the line is x � 3t, y � �2t,z � 7t. Solving with the plane gives

9t � 4t � 49t � 31 � 0, t � �12

�

and the point of intersection is ��32

�, �1, �72

��.

8. The plane 4x � 2y � 5z � 18 � 0 has normaln�� (4, �2, 5). A line through (6, �2, �2) with direction n�� has equation x � 6 � 4t, y � �2 � 2t,z � �2 � 5t. Solve with the plane4(6 � 4t) � 2(�2 � 2t) � 5(�2 � 5t) � 18 � 0

24 � 16t � 4 � 4t � 10 � 25t � 18 � 045t � �36

t � ��45

�.

The point of intersection is ��154�, ��

25

�, �6�.

9. a. 12x � 3y � 4z � 12 � 0. The x-, y-, andz-intercepts are 1, 4, and 3 respectively.

b. x � 2y � z � 5 � 0. The x-, y-, and z-intercepts are

5, ��52

�, and �5 respectively.

z

y

5

x

z

y

x

1

3

4


c. 2x � y � z � 8 � 0. The x-, y-, and z-interceptsare �4, 8, and �8 respectively.

d. 4x � y � 2z � 16 � 0. The x-, y-, and z-interceptsare 4, �16, and 8 respectively.

10. a. x � y � 4 � 0. The x-intercept is 4. They-intercept is 4. The intersection with thexy-plane:

x � u, y � 4 � u, z � 0xz-plane: x � 4, y � 0, z � syz-plane: x � 0, y � 4, z � t.

b. x � 3 � 0. The x-intercept is 3. The intersection with thexy-plane: x � 3, y � t, z � 0 xz-plane: x � 3, y � 0, z � s.

c. 2y � 1 � 0. The y-intercept is ��12

�. The intersection with the

xy-plane: x � t, y � ��12

�, z � 0

yz-plane: x � 0, y � ��12

�, z � s.

z

y

x

3

z

y

x

4

4

8

�16

4

z

y

x

x

y

z

�4

8

�8


d. 3x � z � 6 � 0. The x-intercept is 2. The z-intercept is 6. The intersection with the xy-plane: x � 2, y � t, z � 0 xz-plane: x � u, y � 0, z � 6 � 3uyz-plane: x � 0, y � s, z � 6.

e. y � 2z � 0. The y-intercept is 0. The z-intercept is 0. The intersection with the xy- and xz-plane is the x-axis: x � t, y � 0, z � 0; the intersection withthe yz-plane is x � 0, y � 2u, z � u.

f. x � y � z � 0. Thex x-, y-, and z-intercepts are 0.The intersection with thexy-plane: x � t, y � �t, z � 0 xz-plane: x � s, y � 0, z � syz-plane: x � 0, y � u, z � u.

11. Given the line l: �x �

3k

� � �y �

24

� � �z �

16

� and the

plane �: x � 4y � 5z � 5 � 0. The parametricequations of the line are: x � k � 3t, y � �4 � 2t,z � �6 � t. Substitute into the equation of the plane:k � 3t � 16 � 8t � 30 � 5t � 5 � 0

0t � 9 � k.

a. No value of k will give a unique value to t. Notethat the direction vector of the line is d�� (3, 2, 1),the normal of the plane is n�� (1, �4, 5) andn�� · d�� 3 � 8 � 5 � 0. Since the direction of theline is perpendicular to the normal, the line isparallel to the plane.

b. If k � 9, t � R and there will be an infinite numberof points. In this case the line is on the plane.

c. If k � 9, there will be no points of intersection.

12. See exercise 8.2, questions 19 and 20.

z

y

x

z

y

x

z

x

y

6

2

z

y

x

12

� –


Exercise 8.4

4. a. 3x � 7y � z � 12x � y � 2z � � 3The augmented matrix is

�.

b. �4x � 3y � 2z � 42y � 5z � 5The augmented matrix is

�.

c. x � 4z � 16y � 8z � �2The augmented matrix is

�.

d. 5y � 2z � 6x � 43z � 5y � 2x � �4The augmented matrix, with the coefficients of x, y, and z in the first, second, and third columnsrespectively, is

�.

5. a. � represents the system

x � 4z � 9,y � 6z � 4.

b. � represents the system

8x � 2y � 3z � �6,2x � 6y � 6z � 9.

c. � represents the system

5x � 10z � 8,3y � 4z � 6.

d. � represents the system

x � 4z � 0,y � 9z � 0.

6. a. The augmented matrix of the system is

�. R2

� R1 �

R1

� 2 � R2 �.

The final matrix corresponds to the equationsx � 15z � 10 or x � 10 � 15z

y � 4z � �3 y � �3 � 4z.The parametric equations of the line of intersectionresult when z is set equal to t. They are x � 10 �15t, y � �3 � 4t, z � t. The vector equation isr� � (10, �3, 0) � t(�15, 4, 1).

b. The augmented matrix of the system is

�. �2 � R1

� R2 �.

The final matrix corresponds to the equations

x � 4y � 3z � 5, 9y � �10, y � ��

910�.

Substituting into the first equation and setting

z � t, x � �490� � 3t � 5, x � �

59

� � 3t.

The parametric equation of the line of intersection

is x � �59

� � 3t, y � ��

910�, z � t. The vector

equation is r� � ��59

�, ��

910�, 0� � t(�3, 0, 1).

c. The augmented matrix of the system is

�. R1

� 2 � R2 �.


2x � 8y � 2z � 7, 4z � 1, z � �14

�.

Substituting into the first equation and letting y � t

2x � 9t � �12

� � 7, x � �143� � 4t.

The vector equation of the line of intersection is

r� � ��143�, 0, �

14

�� t(�4, 1, 0).

71

24

80

20

73

2�1

84

21

5�10

30

�49

10

50

36

�41

12

10�3

15�4

01

10

4�3

7�4

21

10

41

73

23

11

00

49

01

10

�86

�10�4

03

50

�69

3�6

�2�6

82

94

4�6

01

10

4�4

�23

55

6�2

16�2

4�8

01

10

�45

2�5

32

�40

12�3

1�2

�71

31


d. The augmented matrix of the system is

�.

3 � R1

� 4 � R2 �.

The final matrix corresponds to the equations 4x � 8y �3z � 6, �5z � 10, z � �2. Substituting into the first equation and lettingy � t, 4x � 8t � 6 � 6, x � 2t. The vector equation of the line of intersection is

r� � (0, 0, �2) � t(2, 1, 0).

e. The augmented matrix of the system is

�.

2 � R1

� 3 � R2 �.

R2

� (�5) �.

R1

� 2 � R2 �.

The final matrix corresponds to the equations 3x � 21, y � 3z � �8, x � 7.

Substituting z � t into the second equation,y � �8 � 3t. The vector equation of the line ofintersection is r� � (7, �8, 0) � t(0, 3, 1).

f. The augmented matrix of the system is

�.

5 � R1

� 3 � R2 �.

R2

� 46 �.

R1

� 8 � R2 �.

R1

� 3 �.

The final matrix corresponds to the equations 2x � z � 3, y � 0. Substituting x � t into the first equation,2t � z � 3, z � �3 � 2t. The vector equation of the line of intersection is

r� � (0, 0, �3) � t(1, 0, 2).

7. a. �.

�.

�.

z � �11554

�, y � �11056

�, x � ��

32053�.

The three planes intersect at the point

��32053�, �

11056

�, �11554

��.

b. � 4 � R1

� R2 �.3 � R

1� R

3

22 � R2

� 35 � R3 �.

The three lines in R2

are not concurrent.

c. �R

1� 2 � R

2 48 � R

1� 3 � R

3 5 � R

1� 6 � R

4

�10�34�5104

�4�12�19

10

315333

6000

102215

�9

�441

�5

3�6�7

2

6385

522

�111

8350

�100

52217

83522

�100

52

�2

832

�14

�3

520

154

�2�715

6130

2009 � R

2� 13 � R

3

5202

�2�7�6

6139

200

3 � R1

� R2

R1

� R2

5�5

3

�214

65

�3

262

�30

�10

01

20

90

�30

01

60

90

�30

81

60

90

�30

846

60

915

�3�5

8�2

610

21�8

0�3

01

30

5�3

�6�3

21

30

�540

�615

2�5

30

5�10

�6�9

23

32

610

�3�5

�80

40

6�2

�31

�86

4�3


�11 � R2

� 5 � R3

R2

� 5 � R4

�31 � R

3� 37 � R

4

�.

The four planes are not concurrent.

8. �1: A

1x � B

1y � C

1z � D

1� 0.

�2: A

2x � B

2y � C

2z � D

2� 0.

�1

and �2

are two nonparallel planes in space. Now

A1x � B

1y � C

1z � D

1� k(A

2x � B

2y � C

2z � D

2)

� 0 ➀

can be written as (A1

� kA2)x � (B

1� kB

2)y �

(C1

� kC2)z � D

1� kD

2� 0, which is of the form of

a plane. Any point P1(x

1, y

1, z

1) that satisfies the

equation of �1

and also �2

will be on the line of

intersection of �1

and �2; i.e., A

1x

1� B

1y

1� C

1z

1�

D1

� 0 and A2x

1� B

2y

1� C

2z

1� D

2� 0. Also P

1

satisfies ➀ since substituting gives

L.S. � A1x

1� B

1y

1� C

1z

1� D

1� k(A

2x

1� B

2y

1�

C2z

1� D

2)

� 0 � k(0)

� 0

� R.S.

Therefore all members of the family of planes

represented by ➀ will also pass through the line of

intersection of �1

and �2.

Note: If k � 0, we get the plane �1; however, no value

of k gives �2.

b. The equation of the family of planes passingthrough the line of intersection of the given planesis 3x � 4y � 7z � 2 � k(2x � 3y � 4) � 0. Sincethe required plane contains the origin, then (0, 0, 0)must satisfy the equation. Therefore, substituting(x, y, z) � (0, 0, 0), we get�2 � 4k � 0,

k � ��12

�.

Substituting into the family

3x � 4y � 7z � 2 � �12

�(2x � 3y � 4) � 0

4x � 5y � 14z � 0. The equation of the required plane is4x � 5y � 14z � 0.

c. The line with equation x � 2y � 3z can be written

as �6x

� � �3y

� � �32

�. Therefore a direction is d�� (6, 3, 2).

The equation of the family of planes passingthrough the intersection of the given planes is4x � 3y � 5z � 10 � k(4x � y � 3z � 15) � 0(4 � 4k)x � (�3 �k)y � (�5 � 3k)z � 10 � 15k � 0.

Each plane of the family has normal n�� (4 � 4k,

�3 � k, �5 � 3k). Since the line is parallel to the

required plane, d�� ⊥ n�� and d�� · n�� 0,(6, 3, 2) · (4 � 4k, �3 � k, �5 � 3k) � 0

24 � 24k � 9 � 3k � 10 � 6k � 015k � �5

k � ��13

�.

Substituting, we get

4x � 3y � 5z � 10 � �13

�(4x � y � 3z � 15) � 012x � 9y � 15z � 30 � 4x � y � 3z � 15 � 0

8x � 8y � 12z � 15 � 0Therefore, the equation of the required plane is8x � 8y � 12z � 15 � 0.

9. Given the plane �: r� � (�2, 1, 3) � s(5, �2, �2) �

t(�1, 0, 1). Two directions of � are a�� (5, �2, �2)

and b�� (�1, 0, 1). A normal to this plane is a�� b��

� (�2, �3, �2) � �1(2, 3, 2). Since the line l: r� �

(9, �1, �5) � p(2, �2, 2) is on the required plane, a

second direction will be (2, �2, 2) � 2(1, �1, 1).

10�34349570

�4�12

370

31500

6000

10�34349277

�4�12

3731

31500

6000

�1 � R3

R4

� (�2)

10�34

�349�554

�4�12�37�62

31500

6000


A normal to the required plane is (2, 3, 2) � (1, �1, 1)

� (5, 0, �5) � 5(1, 0, �1).

A point on the required plane is any point on the given

line (9, �1, �5).

The required plane has equation

(x � 9, y � 1, z � 5) · (1, 0, �1) � 0x � z � 14 � 0.

Exercise 8.5

5. The matrix forms of the given systems are:

a. �b. �c.

� �

6. The systems of equations from the given matrices are:a. x � 8

y � �6z � 3

b. x � 6z � 4y � 5z � �50z � 0

c. x � 0y � 00z � 1

7. The augmented matrix of the given system is

�.

Now

�.

1125

�45

4�4

�18

�60

42

200

R1

� 2 � R2

R3

� 4 � R1

11�7�1

44

�2

�6�318

218

121510

306

450

024

08

�6

�302

153

�210

512

�3

1�5

2

�21

�5

531


Interchange

�.

�.


4z � �25, z � ��

425�

14y � 6z � �152x � 6y � 4z � 11

Substituting gives 14y � 6��

425�� 15,

y � ��145�

2x � 6��

415�� 4 ��

�

425�� 11, x � �

247�.

The planes intersect in the point

��247�, �

�

415�, �

�

425��.

8. a. The augmented matrix of the system is

�.

Now

�.

R2

� R3

�.

The final matrix corresponds to the equations�4z � �16, z � 4

5y � z � 19, 5y � 4 � 19, y � 3x � 2y � z � 12, x � 6 � 4 � 12, x � 2The three planes intersect in the point (2, 3, 4). Thesolution is unique.

1219

�16

11

�4

250

100

121935

115

255

100

2 � R1

� R2

3 � R1

� R3

1251

11

�2

2�1

1

123

11�15�25

4�6

4

�6140

200

R2

� 3R

3� (�1)

11�45

25

4�18�4

�6420

200R

2and R

3

b. The augmented matrix of the system is

�.

Now

�.

There are no values satisfying the equation from R3,

i.e., 0z � 1. Therefore the three planes do not

intersect. Note that the normals,

n��1

� �12

�n��2

� �13

�n��3, are collinear and the three planes

are parallel and distinct.

c. The augmented matrix of the system is

�.

Now

�.

�.


x � y � z � 5z � 2

and 0x � 0y � 0z � 0. Substituting z � 2 into the first equation and lettingy � t gives x � t � 2 � 5, x � 7 � t. The threeplanes intersect in a line with vector equation

r� � (7, 0, 2) � t(�1, 1, 0). There are an infinitenumber of solutions.

d. The augmented matrix of the system is

�.

Now

�.

Each row represents the same equation, x � 2y �3z � 1, hence the three planes are coincident andthere are an infinite number of solutions.

e. The augmented matrix of the system is

�.

Now

�.

The third row corresponds to the equation 0x � 0y� 0z � 1 or 0z � 1. There are no values for thevariable that will satisfy the equation, thereforethere are no solutions to the system of equations.The planes x � y � 2z � 2 and 3x � 3y � 6z � 5are parallel and distinct; the plane x � y � 2z � 5intersects these two planes.

f. The augmented matrix of the system is

�.

Now

�.

There is no solution to the system. The two planes2x � 6y � 10z � 18 and x � 3y � 5z � 9 arecoincident. The other plane, x � 3y � 5z � 10, isdistinct and parallel to the coincident planes.

g. The augmented matrix of the system is

�.

Now

�.

�.

�9�2

0

�210

�320

100

R2

� 7R

2� R

3

9�14�14

�277

�31414

100

R2

� R1

R3

� 2 � R1

9�5

4

�253

�3118

112

1020

500

300

100

2 � R1

�R2

R2

� 2 � R3

10189

5105

363

121

2�3

1

240

120

100

R1

� R2

3 � R1

� R3

255

2�2

6

1�1

3

113

111

�3�3�3

�2�2�2

111

R1

� (�1)R

2� 4

�241

6�12�3

4�8�2

�241

520

�110

100

100

R2

� 2R

2� 2 � R

3

542

�121

100

100

2 � R1

� R2

R1

� R3

563

�1�4�2

121

121

411

200

�100

100

2 � R1

� R2

3 � R1

� R3

47

11

246

�1�2�3

123


The final matrix corresponds to the equations x � 3y � 2z � 9

2y � z � �20z � 0, z � 2t.

Substitute into the second equation2y � 2t � �2, y � �1 � t and x � 3(�1 � t) �2(2t) � 9

x � 3 � 3t � 4t � 9, x � 6 � t.

There is an infinite number of solutions. The three

planes intersect in a line with equation r� � (6, �1,

0) � t(1, �1, 2).

h. The augmented matrix of the system is

�.

Now

�.

�.

There is no solution. Since no two planes areparallel, their intersection forms a triangular prism.

i. The augmented matrix is �.

Now

�.

R2

� 5 � R3

�.

R3

� (�18) �

.

The final matrix corresponds to the equationsz � 0, 5y � z � 0, 5y � 0, y � 0

2x � y � z � 0, 2x � 0, x � 0There is a unique solution. The three planesintersect in a single point, the origin, (0, 0, 0).

9. The three planes �1: x � 2y � z � 0

�2: x � 9y � 5z � 0

�3: kx � y � z � 0.

We find the line of intersection between �1

and �2.

Subtracting these equations gives 11y � 4z � 0,

y� �141z�.

Let z � 11t, then y � 4t. Substitute to find x:x � 2(4t) � (11t) � 0, x � 19t. �

1and �

2intersect in

a line with equation x � 19t, y � 4t, z � 11t. For theplanes to intersect in a line, this line must lie on �

3.

Therefore k(19t) � 4t � 11t � 019kt � �7t

k � ��

197�, t � 0.

The planes intersect in a line when k � ��179�. The

equation of this line is r� � t(19, 4, 11).

Review Exercise

1. a. The line with equation x � z, y � 0 has direction

vector d�� (1, 0, 1). A plane perpendicular to the

x-axis has normal n�� (1, 0, 0). If the plane

contains the line then the direction of the line will

be perpendicular to the normal, i.e., d�� · n�� 0. But

d�� · n�� (1, 0, 1) · (1, 0, 0) � 1 ≠ 0. Therefore a

plane perpendicular to the x-axis cannot contain

the line x � z, y � 0.

b. A plane parallel to the yz-coordinate plane will

have normal parallel to the x-axis, n�� (1, 0, 0).

Equation of the plane passing through A(�4, 0, 5)

with normal n�� is (x � 4, y, z � 5) · (1, 0, 0) � 0;

x � 4 � 0 is a plane parallel to the yz-coordinate

plane and containing the point (�4, 0, 5).

2. a. The plane passes through A(�1, �1, 2) and is

parallel to the plane r� � (2, �1, 0) � s(5, 4, 2) �

t(0, 0, 1). Two directions of the plane are (5, 4, 2)

and (0, 0, 1). The vector equation of the plane is

r� � (�1, �1, 2) � s(5, 4, 2) � t(0, 0, 1).

Parametric equations are

x � �1 � 5s,

y � �1 � 4s,

z � 2 � 2s � t.

000

171

150

200

000

17

�18

150

200

000

17

�5

15

�1

200

R1

� 2 � R2

3 � R1

� 2 � R3

000

1�3

4

1�2

2

213

64

�1

230

110

100

R2

� 2R

2� R

3

68

�9

26

�6

12

�2

100

R1

� R2

3 � R1

� R3

6�227

2�412

1�1

5

113


b. The plane passes through A(1, 1, 0) and B(�2, 0,

3) and is parallel to the y-axis. The direction of the

y-axis is i � (0, 1, 0). A second direction is BA��

(3, 1, �3). A vector equation of this plane is r� �

(1, 1, 0) � s(0, 1, 0) � t(3, 1, �3). Parametric

equations are x � 1 � 3t, y � 1 � s � t, z � �3t.

c. The plane has x-, y-, and z-intercepts �2, �3, and

4 respectively. Therefore three points that the plane

passes through are A(�2, 0, 0), B(0, �3, 0), and

C(0, 0, 4). Two directions of the plane are AB��

(2, �3, 0) and AC�� (2, 0, 4) � 2(1, 0, 2). A

vector equation of the plane is r� � (0, 0, 4) �

s(2, �3, 0) � t(1, 0, 2) and parametric equations

are x � 2s � t, y � �3s, z � 4 � 2t.

d. The plane contains the point A(1, 1, 1) and the line

�3x

� � �4y

� � �5z

�. Since the plane contains the line, the

direction of the line, (3, 4, 5), is also a direction of

the plane. A point on the line is B(0, 0, 0), hence a

second direction is BA�� (1, 1, 1). A vector

equation of the plane is r� � s(1, 1, 1) � t(3, 4, 5)

and parametric equations are x � s � 3t, y � s �

4t, z � s � 5t.

e. The plane contains the two intersecting lines

r� � (3, �1, 2) � s(4, 0, 1) and r� � (3, �1, 2) �

t(4, 0, 2). Since the plane contains these lines, the

direction of the lines, (4, 0, 1) and (2, 0, 1), will be

the direction of the plane. A point on both lines is

(3, �1, 2). A vector equation of the plane is r� �

(3, �1, 2) � s(4, 0, 1) � t(2, 0, 1) and parametric

equations are x � 3 � 4s � 2t, y � �1, z �

2 � s � t.

3. a. The plane passes through A(1, 7, 9) and has normal

n�� (1, 3, 5). The scalar equation is AP�� · n�� 0,

(x � 1, y � 7, z � 9) · (1, 3, 5) � 0

x � 3y � 5z � 67 � 0.

b. The plane passes through the points A(3, 2, 3),

B(�4, 1, 2), and C(�1, 3, 2). Two directions of the

plane are CA�� (4, �1, 1) and BC�� (3, 2, 0). A

normal to the plane is CA�� BC�� (�2, 3, 11).

The scalar equation is

(x � 3, y � 2, z � 3) · (2, �3, �11) � 02x � 3y � 11z � 33 � 0.

c. The plane passes through the point A(0, 0, 6) and

parallel to the plane y � z � 5. The family of

planes parallel to y � z � 5 is y � z � D. Since

A(0, 0, 6) lies on this family, substituting gives

0 � 6 � D, D � 6. The required plane has

equation y � z � 6 or y � z � 6 � 0.

d. The plane contains the point A(3, �3, 0) and the

line x � 2, y � 3 � t, z � �4 � 2t. The direction

of the line, d�� (0, 1, �2), is also a direction of the

plane. A point on the line, B(2, 3, �4), gives a

second direction BA�� (1, �6, 4). A normal to the

plane is d�� BA�� (�8, �2, �1). The equation of

the plane is (x � 3, y � 3, z) · (8, 2, 1) � 0

8x � 2y � z � 18 � 0.

e. The plane contains the line r� � (2, 1, 7) � s(0, 1,

0). Therefore a point it passes through is A(2, 1, 7)

and a direction is a�� (0, 1, 0). Since it is parallel

to the line r� � (3, 0, 4) � t(2, �1, 0), a second

direction is b�� (2, �1, 0). A normal to the plane

is a�� b�� (0, 0, �2). The equation of the plane is

(x � 1, y � 1, z � 7) · (0, 0, 1) � 0; z � 7 � 0.

f. The plane contains the points A(6, 1, 0) and B(3, 0,

2). One direction is BA�� (3, 1, �2). It is also

parallel to the z-axis, therefore a second direction is

k � (0, 0, 1). A normal to the plane will be k � BA��

� (1, �3, 0) and the scalar equation is (x � 6,

y � 1, z) · (1, �3, 0) � 0, x � 3y � 3 � 0.

4. Given the planes �1: 3x � ky � z � 6 � 0 with

normal n��1

� (3, k, 1) and �2: 6x � (1 � k)y � 2z � 9

� 0 with normal n��2

� (6, 1 � k, 2).


a. If the planes are parallel, the normals will be scalar

multiples and n��1

� an��2.

Therefore (3, k, 1) � a(6, 1 � k, 2)

3 � 6a, k � a(1 � k), 1 � 2a

a � �12

� a � �12

�

Since a � �12

�, k � �12

�(1 � k)

2k � 1 � k

k � �13

�.

The planes are parallel for k � �13

�.

b. If the planes are perpendicular, their normals will

be perpendicular. Since n��1

⊥ n��2, n��

1· n��

2� 0,

(3, k, 1) · (6, 1 � k, 2) � 0

18 � k � k2 � 2 � 0

k2 � k � 20 � 0

(k � 5)(k � 4) � 0

k � 5 or k � �4.

The planes are perpendicular for k � 5 or �4.

5. A plane contains the parallel lines

l1: x � 1, �

y �

43

� � �32

� and l2: x � 5, �

y �

25

� � �z �

13

�.

A point on l1

is A(1, 3, 0) and on l2

is B(5, �5, 3).

A and B are also on the required plane, hence one

direction of the plane is AB�� (4, �8, 3). Since both

lines are on the plane, a second direction of the plane

is the direction of the line, d�� (0, 2, 1). A normal to

the plane is d�� AB�� (14, 4, �8) � 2(7, 2, �4).

The scalar equation of the required plane is

(x � 1, y � 3, z) · (7, 2, �4) � 0

7x � 2y � 4z � 13 � 0.

6. Since the required plane is perpendicular to �: x � 2y

� z � 3 � 0, the normal n�� (1, 2, �1) will be a

direction vector. Since it passes through the origin,

O(0, 0, 0), and A(2, �3, 2), a second direction is

OA�� (2, �3, 2). A vector equation of the required

plane is r� � s(1, 2, �1) � t(2, �3, 2).

7. A plane is parallel to vectors a�� 6k � (0, 0, 6) and

b�� i � 2j � 3k � (1, 2, �3). a�� and b�� will be two

directions of the plane and a normal will be a�� b��

(�12, 6, 0) � �6(2, �1, 0). The plane passes through

A(1, 2, 3), therefore the scalar equation will be

(x � 1, y � 2, z � 3) · (2, �1, 0) � 0

2x � y � 0.

8. A line passes through the origin, O(0, 0, 0), and the

point A(1, �3, 2). Since the line is perpendicular to

the plane, a normal will be OA�� (1, �3, 2). The

plane passes through A, therefore (x � 1, y � 3,

z � 2) · (1, �3, 2) � 0 and the scalar equation of the

plane is x � 3y � 2z � 14 � 0.

9. Two lines, l1: �

x �

21

� � �y �

31

� � �z�

�

11

�, direction d��1

� (2, 3, �1) and l2: �

x�

�

11

� � �y �

51

� � �z �

41

�, direction

d��2

� (�1, 5, 4) intersect at the point A(1, 1, 1). A

normal to the plane containing l1

and l2

is d��1

� d��2

�

(17, �7, 13). Now (x � 1, y � 1, z � 1) · (17, �7,

13) � 0 and the scalar equation of the plane

containing the intersecting lines l1

and l2

is 17x � 7y

� 13z � 23 � 0.

10. The line r� � (�4, �3, �1) � t(2, 8, 3) passesthrough the point A(2, 21, 8) (t � 3 will give the pointA). A point and two non-collinear directions define aunique plane. Since A is on the given line, only onedirection is known, hence the equation of a planecannot be determined.

11. The distance from a point P1(x

1, y

1, z

1) to a plane

Ax � By � Cz � D � 0 is given by

d ��Ax

�1

�

A

B

2

y

�

1�

B�2

C

�

z1

C

�

2�D�

�.


a. The distance from the point P1(7, 7, �7) to the

plane by �z � 5 � 0 is

d �

D � ��5437��.

b. Point P1(3, 2, 1) and the plane �: 3x � 2y � z �

10. The distance from P1

to � is

d ��9 �

�4

9

�

�

1

4��

1�10�

��

414��.

c. The line l: r� � (1, 3, 2) � t(1, 2, �1) has direction

d��1

� (1, 2, �1). The plane �: y � 2z � 5 has

normal n�� (0, 1, 2). Since d��1

· n�� 0, the line is

parallel to the plane. The distance between the line

l and the plane � will be the distance from a point

on l, A(1, 3, 2) to the plane.

Therefore d � ��3

�

�

1

4

�

�

4�

5��

�2

5��.

The distance between the line and the plane is ��2

5��.

d. The plane �1: x � 2y � 5z � 10 � 0 has normal

n��1

� (1, 2, �5) and the plane �2: 2x � 4y � 1�z

� 17 � 0 has normal n��2

� (2, 4, �10) � 2(1, 2,

�5). Since n��2

� 2n��1, the planes are parallel, hence

the distance between the planes is the distance

from a point on �1, say A(10, 0, 0), and �

2.

d ��4

�2

�

0

1

�

6�

1

�

7�

100��

2�1 �

3

4�� 25��

2�

3

30��.�

The distance between the planes is �2�

3

30��.

12. The scalar equation of the plane having x-intercept�1, y-intercept 2, and z-intercept 3 is

��

x1� � �

2y

� � �3z

� � 1 or 6x � 3y � 2z � 6 � 0.

The distance from A(1, �2, �2) to this plane is

d � �6

�

�

36

6

�

�

9�

4

�

�

4�

6� � �

272�.

13. A normal to the plane �: 4x � 2y � 5z � 9 � 0 is

n�� (4, �2, 5). An equation of a line through the

origin with direction n�� will be x � 4t, y � �2t, z �

5t. Substituting into � gives 16t � 4t � 25t � 9 � 0,

t � �495� � �

15

�. The normal through the origin intersects

� at the point ��45

�, ��

52�, 1�.

14. The x-, y-, and z-intercepts of the plane �: 4x �5y � z � 20 � 0 are �5, �4, and 20 respectively.

z

x

y

20

�4

�5

�0(7) � 6(7) � 1(�7) � 5��

�02 � 6�2 � (��1)2�


15. a. 2x � y � z � 3 � 0.The x-, y-, and z-intercepts are �

32

�, 3, and 3,respectively.

b. 3y � 4z � 24 � 0.The y-intercept is �8, the z-intercept is 6.

c. 3z � 9 � 0. The z-intercept is �3, and the plane is parallel tothe xy-plane.

d. r� � (4, �5, 0) � s(�12, 9, 8) � t(8, �7, �8).

Two directions are a�� (�12, 9, 8)

and b�� (8, �7, �8).

A normal is n�� a�� b�� (�16, �32, 12)

� �4(4, 8, �3)(x � 4, y � 5, z) · (4, 8, �3) � 0

4x � 8y � 3z � 24 � 0x-, y-, and z-intercepts are �6, �3, and 8respectively.

16. The line l: x � �5 � 3t, y � 3 � 4t, z � 1 � 5t

passes through the point A(�5, 3, 1) and has direction

d�� (�3, �4, 5). The plane �: 2x � y � 2z � 5 � 0

has normal n�� (2, 1, 2). Since d�� · n�� 6 � 4 � 10

� 0, d�� ⊥ n��, hence the line is parallel to the plane.

Since 2(�5) � 3 � 2(1) � 5 � 0, the point A is on

the plane. Since a point of the line is on the plane

and the line is parallel to the plane, the line lies

on the plane.

17. The plane �1: 2x � 6y � 4z � 3 � 0 has normal

n��1

� (2, �6, 4) � 2(1, �3, 2) and the plane �2:

3x � 9y � 6z � k � 0 has normal n��2� (3, �9, 6)�

3(1,�3, 2).

Since n��1

� �23

�n��2, the planes are parallel.

3�1

� 2�2

� 0, therefore 9 � 2k � 0, k � �92

�.

z

y

x

�3

– 6

8

�3

z

y

x

z

y

x

�8

6

3

3

32–

z

x

y


a. Since the planes are parallel, they will not intersect

for k � �92

�.

b. The planes will never intersect in a line.

c. If k � �92

�, the two planes are coincident, hence

intersect in a plane.

18. A plane passes through the points A(1, 0, 2),

B(�1, 1, 0), and has a direction a�� (�1, 1, 1).

a. A second direction is BA�� (2, �1, 2) and a

normal to the plane is a�� BA�� (3, 4, �1).

Now (x � 1, y, z � 2) · (3, 4, �1) � 03x � 4y � z � 1 � 0.

The scalar equation of the plane is 3x � 4y � z � 1 � 0.

b. A line through Q(0, 3, 3) perpendicular to the plane

has direction n�� (3, 4, �1).

An equation of this line is

r� � (0, 3, 3) � t(3, 4, �1).

c. The parametric equations of the line are x � 3t,y � 3 � 4t, z � 3 � t. Solve by substituting thesevalues into the equation of the plane.3(3t) � 4(3 � 4t) � (3 � t) � 1 � 0

t � ��143�. The perpendicular through Q intersects

the plane at A��1132

�, �2133�, �

4133��.

d. The distance from A to the plane is given by

d �

� ��

�26�

d � 0.The distance from A to the plane is 0, implies thatA is on the plane.

19. Plane �1: x � 2y � 7z � 3 � 0 has normal n��

1�

(1, 2, �7), and plane �2: x � 5y � 4z � 1 � 0 has

normal n��2

� (1, �5, 4). A direction of the line of

intersection of the two planes is n��1

� n��2

�

(�17, �11, �7) � �1(27, 11, 7). A plane through

A(3, 0, �4) perpendicular to the line of intersection of

�1

and �2

has n��1

� n��2

as a normal. Therefore

(x � 3, y, z � 4) · (27, 11, 7) � 027x � 11y � 7z � 53 � 0

is the equation of the plane through A(3, 0, �4) and perpendicular to the line of intersection of thegiven planes.

20. a. The family of planes passing through the line ofintersection of the planes x � y � z � 1 � 0 and 2x � 3y � z � 2 � 0 is x � y � z � 1 �k(2x � 3y � z � 2) � 0. To find the particularmember that passes through the origin set (x, y, z) � (0, 0, 0). Now �1 � 2k � 0

k � �12

�.

The particular plane is

x � y � z � 1 � �12

�(2x � 3y � z � 2) � 0

2x � 2y � 2z � 2 � 2x � 3y � z � 2 � 04x � y � z � 0.

b. A normal to the plane 4x � y � z � 0 is n��1

�

(4, �1, 1) and a normal to the plane x � z � 0 is

n��2

� (1, 0, �1). The angle between the planes

is the angle between the normals. Therefore

n��1

· n��2

� � n��1�� n��

2� cos

4 � 1 � �16 � 1� � 1� �2� cos

cos � �36

� � �12

�

� 60and the angle between the planes is 60.

�36 � 92 � 43 �13��

13

��3��1132

�� 4��2133��

4133�� 1�

��9 � 16� � 1�


21. Plane �1: r� � (4, 0, 3) � t(�8, 1, �9) � u(�1, 5, 7)

has directions a��1

� (�8, 1, �9), b��1

� (�1, 5, 7) and

normal n��1

� a��1

� b��1

� (52, 65, �39) � 13(4, 5, �3).

The scalar equation is (x � 4, y, z � 3) · (4, 5, �3) � 04x � 5y � 3z � 7 � 0.

Plane �2: r� � (�14, 12, �1) � p(1, 1, 3) � q(�2, 1, �1),

has directions a��2

� (1, 1, 3), b��2

� (�2, 1, �1) and

normal n��2

� a��2

� b��2

� (�4, �5, 3) � �1(4, 5, �3).

The scalar equation is

(x � 14, y � 12, z � 1) · (4, 5, �3) � 04x � 5y � 3z � 56 � 60 � 3 � 0

4x � 5y � 3z � 7 � 0Since the scalar equations of both planes is the same, theplanes are coincident.

22. a. x � 5y � 8 ➀5x � 7y � �8 ➁

5 � ➀ � ➁: 32y � 48

y � �32

�, x � �12

�.

The two lines in R2 intersect in the point ��12

�, �32

��.

b. �1: 2x � 2y � 4z � 5, n��

1� (2, �2, 4) � 2(1, �1, 2).

�2: x � y � 2z � 2, n��

2� (1, �1, 2).

Since n��1

� 2 n��2, the planes are parallel. Since

�1

� 2 � 2

� 1 � 0 the two planes will be distinct.

c. �1: 3x � 2y � 4z � �1 ➀

�2: 2x � y � z � �3 ➁

➀ � 2 � ➁: 7x � 6z � �7

x � �1 � �67

�z

Let z � 7t, x � �1 � 6t. Substitute into ➁�2 � 12t � y � 7t � �3

y � 1 � 5t.The two planes intersect in a line with equation

r� � (�1, 1, 0) � t(6, 5, 7).

d. x � 2y � 3z � 11 ➀2x � y � 7 ➁

3x � 6y � 8z � 32 ➂2 � ➀ � ➁: 3y � 6z � 15, y � 2z � 5.3 � ➀ � ➂: �z � 1, z � �1, y � 3, x � 2.The three planes intersect at the point (2, 3, �1).


e. �1: x � y � 3z � 4, n��

1� (1, �1, 3).

�2: x � y � 2z � 2, n��

2� (1, 1, 2).

�3: 3x � y � 7z � 9, n��

3� (3, 1, 7).

Since no two normals are collinear, no two planesare parallel.➀ � ➁: 2x � 5z � 6 ➃

➀ � ➂: 4x � 10z � 13 ➄2 � ➃ � ➄: 0z � �1

There is no solution and the planes intersect to forma triangular prism.

f. �1: x � 3y � 3z � 8

�2: x � y � 3z � 4

�3: 2x � 6y � 6z � 16

Since 2 � �1

� �3

� 0, �1

and �3

are coincident.

�1

� �2: 4y � 4, y � 1 Substitute into �

2

x � 1 � 3z � 4

x � 5 � 3z.

Let z � t, x � 5 � 3t.

The planes intersect in the line with equation

r� � (5, 1, 0) � t(�3, 0, 1).

g. The augmented matrix of the system is

�.

Now

�.

�.

The last row corresponds to the equation0z � 0. Let z � �5t (to avoid fractions)

then 5y � 3z � �10, y � �2 � �35

�z, y � �2 � 3t.

x � 2y � z � �3, x � 4 � 6t � 5t � � 3,x � 1 � t.Letting t � 0, y � � 2, x � 1, z � 0r� � (1, �2, 0) � t(�1, 3, �5).

�3�10

0

130

250

100

�1 � R2

R2

� R3

�310

�10

1�3

3

2�5

5

100

R1

� R2

2 � R1

� R2

�3�13

4

14

�1

27

�1

112

h. �1: 3x � 3z � 12.

�2: 2x � 2z � 8.

�3: x � z � 4.

The three planes are coincident with the planex � z � 4.

i. �1: x � y � z � � 3.

�2: x � 2y � 2z � �4.

�3: 2x � 2y � 2z � �5.

�1

and �3

are parallel and distinct. �2

intersects

both �1

and �3

in two parallel lines.

Chapter 8 Test

1. a. Two planes, with normals satisfying n��1

· n��2

� 0,will be perpendicular to each other and intersect ina line.

b. Two planes, with normals satisfying n��1

� n��2

� 0,will be parallel.

c. Three planes, with normals satisfying n��1

� n��2

· n��3

� 0, will be parallel to each other.

2. The plane �: 4x � y � z � 10 � 0 has normaln�� (4, 1, �1)

a. The line l: x � �3t, y � �5 � 2t, z � �10t has

direction d�� (�3, 2, �10). Since d�� · n�� 12

� 2 � 10 � 0, the line is parallel to the plane. A

point on the line, A(0, �5, 0), (t � 0) does not

satisfy the equation of the plane, therefore the line

does not coincide with the plane.

b. The line �x �

42

� � �y �

12

� � ��

z1� has direction

d�� (4, 1, �1) and passes through the point A(2, 2,

0). Since A satisfies the plane, and d�� n��, the line

intersects the plane, at right angles, at the point

A(2, 2, 0).

3. Three planes intersect in a point A.

Three planes intersect in a line.

The three planes are coincident, thus intersect in a plane.

Two planes are coincident and the third planeintersects them in a line.

�3

�2

�1

line of intersection

�2

�3

�1

line ofintersection

A


4. The plane r� � (0, 0, 5) � s(4, 1, 0) � t(2, 0, 2) hasparametric equation x � 4s � 2ty � sz � 5 � 2t.

a. For an intersection with the x-axis, y � z � 0,

therefore s � 0, t � ��

25� and x � �5 and the point is

(�5, 0, 0).

b. An intersection with the xz-coordinate plane, y � 0,and the line of intersection, will be r� � (0, 0, 5) �t(1, 0, 1).

5. The line x � y, z � 0 has direction d�� (1, 1, 0) and

passes through the origin, O(0, 0, 0). The required plane

passes through A(2, �5, �4), therefore a second

direction of the plane is OA�� (2, �5, �4) and a normal

is d�� OA�� (�4, 4, �7) � �1(4, �4, 7). The scalar

equation of the plane is 4x � 4y � 7z � 0.

6. Given the system of equations x � 2y � z � �3 ➀x � 7y � 4z � �13 ➁

2x � y � z � 4 ➂➀ � ➂: 3x � y � 1 ➃

➁ � 4 � ➂: 9x � 3y � 3, 3x � y � 1.Let x � t, y � 1 � 3t and from ➀t � 2 � 6t � z � �3

z � �5 � 5tThe three planes intersect in a line with equationr� � (0, 1, �5) � t(1, �3, 5).

7. a. The distance from the origin, O(0, 0, 0) to the plane �:3x � 2y � z � 14 � 0 is

d � ��9

��

1

4

4

�� 1�

� � ��

14

14�� 14�.

b. The distance from the point P(10, 10, 10) to the plane3x � 2y � z � 14 � 0 is

d � � ��26

14��.

�30 � 20 � 10 � 14��

�14�


c. A plane Ax � By � Cz � D � 0 divides R3 into

three regions. All points P1(x

1, y

1, z

1) satisfying the

inequality Ax1

� By1

� C1

� D 0 lie on the

same side of the plane. Those satisfying Ax1

� By1

� D 0 lie on the other side of the plane, and

those satisfying Ax1

� By1

� Cz1

� D � 0 lie on

the plane. Since the sign of Ax1

� By1

� Cz1

� D

is positive for P and negative for the origin, P does

not lie on the same side of the plane as the origin.


1. Choose two unit vectors, â � ��1

2��, �

�1

2��, 0�

and b� ��13�

�, ��13�

�, �13

��.

Now â � b � ��16�

�, ��16�

�, 0�

�â � b� � ��16

� � �16��

�1

3�� 1.

Therefore the cross product of unit vectors is notnecessarily a unit vector.

2. Question as posed in first printing of textbook ismeaningless. Use (u�� v��) � v��.

3. ∆ABC has coordinates A(2, 4), B(0, 0), C(� 2, 1). To

find the cos ∠ABC, we need BA�� (2, 4) and BC��

(�2, 1). Now BA�� · BC�� 4 � 4 � 0. Therefore

BA�� ⊥ BC��, ∠ABC � 90 and cos ∠ABC � 0.

4. The vector (0, 8) is a linear combination of (2, 4) and(�2, 1). Therefore (0, 8) � m(2, 4) � n(�2, 1).Equating components 2m � 2n � 0 ➀and 4m � n � 8 ➁Solve for m and n: ➀ � 2 � ➁: 5m � 8,

m � �8

5�, n � �

8

5�

and (0, 8) � �85

�(2, 4) � �85

�(�2, 1).

5. Given four points A(2k, 0, 0), B(0, 2k, 0), C(0, 0, 2k),and D(2l, 2l, 2l).The midpoint of AB is W(k, k, 0)The midpoint of BC is X(0, k, k)The midpoint of CD is Y(l, l, l � k)

and of DA is Z(l � k, l, l).

Now WX�� (�k, 0, k) � �k(1, 0, �1)

ZY�� (�k, o, k) � �k(1, 0, �1)

and WX�� ZY��.

Since WX�� ZY��, W, X, Y, and Z are four points of a

parallelogram, hence the four points W, X, Y, and Z are

coplanar.

6.

In ∆ABC, let BP�� PC�� a��,

AQ�� 5b��, QP�� 2b��.

Extend BQ to meet AC at R. BQR is a straight line.

Let BQ�� md��, QR�� (1 � m)d��, AC�� c��, therefore

AR�� kc�� and RC�� (1 � k)c��.

In ∆BQP: md�� a�� 2b��.

In ∆BRC: 2a�� d�� (1 � k)c��,

therefore md�� 1

2�d��

1

2�(1 � k)c�� 2 b��.

In ∆AQR: 5b�� kc�� (1 � m)d��

and 2b�� 2

5�kc��

2

5�(1 � m)d��.

Now md�� 1

2�d��

1

2�(1 � k)c��

2

5�kc��

25

�(1 � m)d��

d�� m � �1

2� � �

2

5��

2

5�m� � c��

1

2� � �

1

2�k � �

2

5�k�

��7

5�m � �

1

9

0�� d��

1

2� � �

1

9

0�k�c��.

Since c�� and d�� are linearly independent,

�7

5�m � �

1

9

0� � 0 and �

1

2� � �

1

9

0�k � 0

m � �1

9

4�, k � �

5

9�.

Therefore, if k � �5

9�, BQR is a straight line.

7.

Place the polygon in the Cartesian plane so that P1

is

at the origin and P1P

2�� is along the positive x-axis. The

interior angles of the polygon as 150.

∠AP2P

1� 90, therefore ∠AP

2P

3� 60 and ∠AP

3P

2

� 30. Let the magnitude of each side of the polygon

be 2, therefore in ∠AP2P

3, P

2P

3� 2, AP

2� 1, and

AP3

� �3�. Now P2P

3�� (�3�, 1). Similarly in

∆BP4P

3, P

3P

4� 2, P

3B � �3�, BP

4� 1, and P

3P

4��

(1, �3�).

Similarly, we have the following:

P1P

2�� (2, 0), P

4P

5�� (0, 2), P

5P

6�� (�1, �3�),

P6P

7�� (��3�, 1).

A

B

Y

P1 P2

P3

P4

P5

P6

P7

X

A

B CP→

a→a

→kc

R

(l – k) c→

b

→

2→dm

Q

→

b5

(l – m) d→


a. P1P

3�� y��, P

1P

2�� x��.

In ∆P1P

2P

3, P

2P

3�� x�� y��.

b. P1P

4�� P

1P

2�� P

2P

3�� P

3P

4�� mx�� ny��.

Now x�� P1P

2�� (2, 0), y�� P

1P

2�� P

2P

3��

� (2, 0) � (�3�, 1)

y�� (2 � �3�, 1).

P1P

4�� (2, 0) � (�3�, 1) � (1, �3�) � m(2, 0) �

n(2 � �3�, 1).


2 � �3� � 1 � 2m � (2 � �3�)n ➀

1 � �3� � n ➁

Substitute in ➁:

3 � �3� � 2m � (2 � �3�)(1 � �3�)

3 � �3� � 2m � 2 � 3�3� � 3

2m � �2�3� � 2

m � �1 � �3�

therefore P1P

4�� (�1 � �3�)x�� (1 � �3�)y��.

c. P3P

4�� P

3P

4�� P

4P

5�� P

5P

6�� P

6P

7�� mx�� ny��

(1, �3�) � (0, 2) � (�1, �3�) � (��3�, 1) �

m(2, 0) � n(2 � �3�, 1).

(��3�, 2�3� � 3) � m(2, 0) � n(2 � �3�, 1).


2m � (2 � �3�)n � ��3� ➀

n � 2�3� � 3 ➁

Substitute in ➀:

2m � (2 � �3�)(3 � 2�3�) � ��3�

2m � 6 � 7�3� � 6 � ��3�

2m � �12 �8�3�

m � �6 �4�3�

therefore P3P

7�� (�6 �4�3�)x�� (3 � 2�3�)y��.

8. a��, b��, and c�� are three linearly independent vectors. If

u�� 3a�� 2b�� c��, v�� 2a�� 4c��, and w�� a�� 3b��

� kc�� are coplanar then one of u��, v��, or w�� can be

written as a linear combination of the other two.

Say w�� pu�� qv��.

Now �a�� 3b�� kc�� p(3a�� 2b�� c��) �

q(�2a�� 4c)

(�1 � 3p � 2q)a�� (3 � 2p)b�� (k � p � 4q)c�� 0��

Since a��, b��, and c�� are linearly independent

�1 � 3p � 2q � 0 ➀, 3 � 2p � 0 ➁, and

k � p � 4q � 0. ➂

From ➁, p � �3

2�.

Substituting in ➀, �1 � �9

2� � 2q � 0

q � �1

4

1�.

Substituting for p and q in ➂: k � �3

2� � 11 � 0,

k � �1

2

9�.

If k � �1

2

9� then u��, v��, and w�� will be coplanar.

9.


DA�� CB�� a��

DC�� AB�� b��.

Diagonals DB and AC intersect at E and E divides

AC in the ration m:n.

Since E divides AC in the ratio m:n,

DE�� m �

n

n� a��

m

m

� n� b��.

A B

CD

E→a

→a

→

b

→

b

m

n


Also BE�� m �

n

n� BA��

m

m

� n� BC��

EB�� m �

n

n� AB��

m

m

� n� CB��

EB�� m �

n

n� b��

m �

n

n� a��.

Since D, E, and B are collinear, DE�� kEB��.

�m �

n

n� a��

m

m

� n� b�� k�m �

n

n� b�� k�m

m

� n� a��

��m �

n

n� � �

m

k

�

m

n�� a�� m

m

� n� � �

m

k

�

n

n�� b�� 0.

But a�� and b�� are linearly independent, therefore

�m �

n

n� � �

mk�

mn

� � 0 and �m

m

� n� � �

m

k

�

n

n� � 0

n � km � 0 ➀ m � kn � 0 ➁

➀ � ➁: m � n � k(m � n) � 0 and k � 1.

Since k � 1, DE�� EB�� and E is the midpoint of DB.

Substitute k � 1 into ➀: n � m � 0, m � n. Since

m � n, E divides AC in the ration m:m � 1:1 and E is

the midpoint of AC. Therefore the diagonals of a

parallelogram bisect each other.

10.

ABCD is a quadrilateral with AB�DC. AC and DBintersect at M. A line through M parallel to AB meetsAD and BC at P and Q respectively. Draw RMSperpendicular to AB. Since AB�DC, RMS will also beperpendicular to DC. In ∆ABC, MQ�AB. Therefore

�CC

QB� � �

CC

MA� � �

MAB

Q�.

Since SR is an altitude of ∆ABC, M divides SR in thesame ratio.

Therefore �CC

QB� � �

CC

MA� � �

MAB

Q� � �

SSMR� ➀

A R B

P

DS

MQ

C


Similarly in ∆ADB, PM�AB

and �DD

PA� � �

DD

MB� � �

PAMB� � �

SSMR�. ➁

From ➀ and ➁ �MAB

Q� � �

PAMB� �both equal �

SSMR��

and MQ � PM

and M is the midpoint of PQ.

11.

ABC is an isosceles triangle with AB � AC. Apexangle BAC is bisected by DA therefore ∠BAD �∠CAD. We are to show that AD ⊥ BC. In ∆ABD and∆ADC,

AB � AC∠BAD � ∠CAD.

DA is common, therefore ∆ABD � ∆ACDand ∠ACD � ∠ADC � x

∠BDC � 180therefore 2x � 180

x � 90and AD ⊥ BC, hence the bisector of the apex angle ofan isosceles triangle is perpendicular to the base.

12. Two lines l1: �

x �

18

� � �y �

34

� �z �

12

� has direction

d��1

� (1, 3, 1), and l2: (x, y, z) � (3, 3, 3) � t(4, �1, �1)

has direction d��2

� (4, �1, �1).

a. Since d��1

· d��2

� 0, the two lines are perpendicular.

A

B D C

✓ ✓

b. From l1: x� �8 � s, y � �4 � 3s, z � 2 � s

l2: x � 3 � 4t, y � 3 � t, z � 3 � t.

Equating components and rearranging gives thefollowing equationss � 4t � 11 ➀3s � t � 7 ➁

s � t � 1 ➂➁ � ➂: 2s � 6, s � 3 and t � �2. Substitute in➀: 3 � 4(�2) � 11 � R.S. The lines intersect atthe point (�5, 5, 5).

13. Given four points: O(0, 0, 0), P(1, �1, 3), Q(�1, �2,

�5), and R(�5, �1, 1). Now OP�� (1, �1, 3), OQ��

(�1, �2, 5), OR�� (�5, �1, 1). OP�� OQ��

(1, �8, �3) and OP�� OQ�� · OR�� 5 � 8 �3 � 0,

therefore OP�� , OQ��, and OR�� are coplanar, hence O(0,

0, 0) lies on the plane that passes through P, Q, and R.

14. A plane �, passes through P(6, �1, 1), has z-intercept�4 therefore passes through the point A(0, 0, �4),

and is parallel to the line �x �

32

� � �y �

31

� � ��

z1�.

Two directions of � will be AP�� (6, �1, 3) and the

direction of the line,

d�� (3, 3, �1). A normal to the plane is d�� AP��

(14, �21, �21) � 7(2, �3, �3). The scalar equation

of the plane is (x, y, z � 4) · (2, �3, �3) � 0

or 2x � 3y � 3z � 12 � 0.

15. The coordinates of a point on the line (x, y, z) � (�3,4, 3) � t(�1, 1, 0) is A(�3 � t, 4 � t, 3) and on theline (x, y, z) � (3, 6, �3) � s(1, 2, �2) is B(3 � s,6 � 2s, �3 � 2s).

AB�� (6 � t � s, 2 � t � 2s, �6 � 2s)

AB�� is parallel to m�� (2, �1, 3), therefore AB�� km��

and 6 � t � s � 2k ➀2 � t � 2s � �k ➁

� 6 � 2s � 3k ➂We solve for s and t:➀ � 2 � ➁: 10 � t � 5s � 0 ➃3 � ➁ � ➂: � 3t � 4s � 0 ➄

3 � ➃ � ➄: 30 � 11s � 0, s � ��3101�, t � �

43

�s � ��4101�.

The points are A��171�, �

141�, 3� and B ��

131�, �

161�, �

2171��.

16. The sphere (x � 1)2 � (y � 2)2 � (z � 3)2 � 9 has

centre C(1, 2, 3). The plane tangent to the sphere at

A(2, 4, 5), a point at one end of a diameter, will have

CA�� (1, 2, 2) as normal. Therefore (x � 2, y � 4, z � 5) · (1, 2, 2) � 0

x � 2y � 2z � 20 � 0 is therequired plane.

17. The line l: x � �1 � t, y � 3 � 2t, z � �t interestseach of the following planes.

a. �1: x � y � z � 2 � 0.

Substituting for x, y, and z:�1 � t � 3 � 2t � t � 2 � 0

0t � 2.

There is no intersection. The direction of the line is

d��1

� (1, 2, �1), a normal to the plane is n��1

�

(1, �1, �1), d��1

· n��1

� 0, hence the line is parallel

to the plane and distinct from the plane.

b. �2: �4x � y � 2z � 7 � 0.

Substituting for x, y, and z:4 � 4t � 3 � 2t � 2t � 7 � 00t � 0, t � R.

Note that the plane has normal n��2

� (�4, 1, �2)

and d�� · n��2

� 0, the line is parallel to the plane; in

fact, the line is on the plane. The intersection will be

r� � (�1, 3, 0) � t(1, 2, �1).

c. �3: x � 4y � 3z � 7 � 0.

Substituting for x, y, and z:�1 � t � 12 � 8t � 3t � 7 � 0

12t � �18

t � ��32

�.

The line intersects the plane at the point ��52

�, 0, �32

��.


18. Given the planes �1: 4x � 2y � z � 7

and �2: x � 2y � 3z � 3.

Solve to find the line of intersection.

Add �1

� �2: 5x � 2z � 10

x � 2 � �25

�z.

Let z � 5t, x � 2 � 2t.From �

2: 2 � 2t � 2y � 15t � 3

y � �12

� � �123�t.

The parametric equation of the line of intersection is

x � 2 � 2t, y � �12

� � �123�t, z � 5t.

For the intersection with the xy-plane, z � 0. Therefore

t � 0 and the point of intersection is �2, �12

�, 0�.

19. A plane �1

passes through A(2, 0, 2), B(2, 1, 1), and

C(2, 2, 4). Two directions of � are

AB�� (0, 1, �1)

and AC�� (0, 2, 2) � 2(0, 1, 1).

A normal to the plane will be n�� (0, 1, � 1) � (0, 1, 1)

� (2, 0, 0)

� 2(1, 0, 0).The equation of the plane � is (x � 2, y, z � 2) · (1, 0, 0) � 0

x � 2 � 0.A line l through P(3, 2, 1), Q(1, 3, 4) has direction

QP�� (2, �1, �3) and parametric equations

x � 3 � 2t, y � 2 � t, z � 1 � 3t. Solving the line

with the plane gives 3 � 2t � 2 � 0

t � � �12

�.

The coordinates of the point of intersection of the line

with the plane is �2, �52

�, �52

��.

20. a. To determine the line of intersection of the two planes

�1: 3x � y � 4z � �6 and

�2: x � 2y � z � 5, we solve.

2 � �1

� �2: 7x � 7z � �7, x � z � �1.

Let z � t, x � �1 � t substitute into �1

�3 � 3t � y � 4t � �6y � 3 � t

The parametric equations of the two planes arex � �1 � t, y � 3 � t, z � t.

b. To intersect the xy-plane, z � 0 therefore t � 0 andthe point is A(�1, 3, 0).To intersect the xz-plane, y � 0 therefore t � �3and the point is B(2, 0, �3).To intersect the yz-plane, x � 0 therefore t � �1and the point is C(0, 2, �1).

c. The distance between the xy- and xz-intercepts is

�AB��. AB�� (3, �3, �3), �AB�� 3�3�.

21. Since Q is the reflection of P(�7, �3, 0) in the plane

�: 3x � y � z � 12, PQ will be perpendicular to the

plane and the plane will bisect PQ. Let this midpoint

be R. A normal to the plane is n�� (3, �1, 1). The

line passing through PQ will have direction n�� and

parametric equations x � �7 � 3t,

y � �3 � t,

z � t.

Solving with the plane gives�21 � 9t � 3 � t � t � 12

t � �3101�.

This gives the coordinates of the midpoint of PQ,

R��1131�, ��

6131�, �

3101��.

�

Q

R

→n

P (�7, �3, 0)


Let Q have coordinates (a, b, c). Since R is themidpoint of PQ,

�a �

27

� � �1131�, �

b �

23

� � ��6131�, �

2c

� � �3101�

a � �11013

�, b � ��9131�, c � �

6101�

and the coordinates of Q will be ��11013

�, ��9131�, �

6101��.

22. Two planes 3x � 4y � 9z � 0 and 2y � 9z � 0intersect in a line. From the second plane we have

y � �92

�z. Let z � 2t, then y � 9t, substituting in the

first plane gives 3x � 36t � 18t � 0x � 6t.

Parametric equations of the line of intersection are

x � 6t, y � 9t, z � 2t. A direction of this line is

d�� (6, 9, 2). Now �d�� 36 � 8�1 � 4�

� �121�

�d�� 11.

A unit vector along d�� is d � ��161�, �

191�, �

121��.

A vector of length 44 that lies on this line ofintersection will be

44d � 44 ��161�, �

191�, �

121��

44d � (24, 36, 8).

23. The line through P(a, 0, a) with direction

d��1

� (�1, 2, �1) has equation l1: x � a � t

y � 2t

z � a � t.

l1

intersects the plane �: 3x � 5y � 2z � 0 at Q.

Solving l1

and � gives

3a � 3t � 10t � 2a � 2t � 0

5t � �5a

t � �a.

The point Q has coordinates (2a, �2a, 2a).

The line through P(a, 0, a) with direction

d��2

� (�3, 2, �1) has equation

l2: x � a � 3s

y � 2s

z � a � s.

l2

intersects � at R. Solving l2

and � gives

3a � 9s � 10s � 2a � 2s � 0

s � 5a.

The point R has coordinates (�14a, 10a, �4a)

RQ�� (16a, �12a, 6a).

Since RQ�� 3 we have

�(16a)2� � (��12a)2 �� (6a)2� � 3

256a2 � 144a2 � 36a2 � 9

a2 � �4936�

a � � ��

3436��

� � �2�

3109��.

The distance between Q and R will be 3 if

a � �2�

3109�� or a � �

2��

1309�

�.

24. Two lines L1: (x, y, z) � (2, 0, 0) � t(1, 2, �1)

L2

: (x, y, z) � (3, 2, 3) � s(a, b, 1).

To determine the intersection of L1

and L2

we equatecomponents then solve.x � 2 � t � 3 � sa, t � sa � 1 ➀y � 2t � 2 � sb, 2t � sb � 2 ➁z � �t � 3 � s, t � s � �3 ➂➂ � ➀: s � sa � �4

s(1 � a) � �4

s � �1

�

�

4a

�

Substitute in ➂: t �1

�

�

4a

� � �3

t � �1 �

4a

� � �3

t � �11

�

�

3aa

�

Substitute for s and t into ➁:

�21

�

�

6aa

� � �1

4�

ba

� � 2

2 � 6a � 4b � 2 � 2a8a � 4b

a � �12

�b

L1

and L2

will intersect whenever a � �12

�b.



25. x � 2y � 3z � 1 ➀2x � 5y � 4z � 1 ➁

3x � 5y � z � 3 ➂2 � ➀ � ➁: �y � 10z � 1

3 � ➀ � ➂: �8z � 0z � 0y � �1

Substitute into ➀: x � 2 � 1, x � 3.The solution to the system is x � 3, y � �1, z � 0.

26. �2x � y � z � k � 1 ➀kx � z � 0 ➁y � kz � 0 ➂

➀ � ➂: �2x � z � kz � k � 1 ➃2 � ➁ � k � ➃: 2z � kz � k2z � k(k � 1)

(k2 � k � 2)z � � k(k � 1)(k � 2)(k � 1)z � �k(k � 1).

a. (i) If k � 2, 0z � �4 and there will be no solution.

(ii) If k � 2, k � �1, z � ��k �

k2

� and the system

will have exactly one solution.

(iii) If k � �1, 0z � 0 and there will be an infinitenumber of solutions.

b. Since 0z � 0, let z � t, back substituting will givefrom ➂: y � z � 0, y � t, and from ➁: � x � z �0, x � t. The solution set is (x, y, z) � (t, t, t),which is the equation of a line passing through theorigin with direction (1, 1, 1).


Exercise 9.1

5. To determine points of intersection of the circle andthe line, solve the equations. Substitute y � 3x � 5 in x2 � y2 � 5 � 0.x2 � (3x � 5)2 � 5 � 0

10x2 � 30x � 20 � 0x2 � 3x � 2 � 0

(x � 1)(x � 2) � 0x � �1 or x � �2.

When x � �1, y � 2.When x � �2, y � �1.The points of intersection are A(�1, 2) and B(�2, �1).

The length of AB is

�(�1 �� 2)2 �� (2 � 1�)2� � �10�

6. The longest possible chord in a circle is the diameter.Since the centre is (0, 0) and the diameter passesthrough (1, �2), the equation of the line is y � �2x.The method of question 5 can now be repeated. Thisis unnecessary, of course. From the circle equation theradius is 4, so the length of the diameter is 8.

7. Setting y � 0 we have x2 � 6x � 0x � 0 or x � �6.

Setting x � 0 we have y2 � 2y � 0y � 0 or y � 2.

There are three intercepts. They are at the points (0, 0), (�6, 0), and (0, 2).

8. The circle equation can be rewritten

x2 � 6x � y2 � 2y� �6

(x � 3)2 � (y � 1)2 � 4.Then the centre of the circle is C (3, 1) and the radius of the circle is 2. From the diagram,

PT2 � PC2 � CT2

� [(5 � 3)2 � (7 � 1)2] � 4

� 36. The length of the tangent is 6.

9. Substituting (1, 5) in the circle equation,L.S. � 1 � 25 � 4 � 10 � 20 � 0R.S. � 0Then A is on the circle. If the opposite end of thediameter from A is B(m, n), by symmetry,m �(�2) � �2 � 1

m � �5 and n � 1 � 1 � 5

n � �3B has coordinates (�5, �3).

10. Let the centre be C(m, n). Then n � 2m � 4. Also C is equidistant from (8, 2) and (�2, �4). Then (m � 8)2 � (n � 2)2 � (m � 2)2 � (n � 4)2

or 5m � 3n � 12 � 0.Substituting n � 2m � 4, we obtain

m � 0, n � 4.

Now the radius is �(0 � 8�)2 � (4� � 2)2� � �68�.The circle equation is x2 � (y � 4)2 � 68.

y

x

B(m, n)

C(�2, 1)

A(1, 5)

y

T

x

P(5, 7)

C(3, 1)

Chapter 9: Proof Using Different Approaches 139

Chapter 9 • Proof Using Different Approaches

11. Let the centre be C(a, a) where a < 0. The radius is a.Then AT � 5 and CT � a. Now 52 � a2 � (a � 4)2 � (a � 2)2

then a2 � 4a � 5 � 0(a � 5)(a � 1) � 0

a � �5 or a � 1but a < 0, so a � �5.

The circle equation is (x � 5)2 � (y � 5)2 � 25.

12. The radius is the line segment joining C(0, 0) and R(3, �4). Hence the tangent is the line through R withslope perpendicular to CR.

The slope of CR is ��43

�, so the slope of PQ is �34

�.

The equation of PQ is

y � 4 � �34

�(x � 3)

or 3x � 4y � 25 � 0

Then P has coordinates ��235�, 0� and Q has coordinates

�0, ��245��.

The length of PQ is

��235��2��

245��2� � 25 ��

19

� � �1�16��

� 25��152��

� �11225

�.

13. Circle 1 is x2 � y2 � 9 and has centre (0, 0) andradius 3. Circle 2 is x2 � y2 � 12x � 6y � 41 � 0

or (x � 6)2 � (y � 3)2 � 4.This circle has centre (6, �3) and radius 2. The shortest distance between their centres is

�62 � (��3)2� � 3�5�, so the least distance between

the circles is 3�5� � 5.

14. To determine points of intersection, solve theequations.

x2 � y2 � 4x2 � y2 � 6x � �2

Subtracting, 6x � 6x � 1

Then y2 � 3 and y � � �3�.

Points of intersection are �1, �3�� and �1, ��3��. The length of the chord is 2�3�.

Exercise 9.2

3. Let the quadrilateral have coordinates P(0, 0),Q(2a, 0), R(2b, 2c), S(2d, 2e). The midpoint of PQ isW(a, 0), the midpoint of QR is X(a � b, c) themidpoint of RS is Y(b � d, c � e), and the midpoint

of SP is Z(d, e). The slope of WX is �bc

� and the slope

of YZ is �cb

�

�

ed

�

�

ed

� � �bc

�, so WX�YZ.

The slope of XY is�a �

c �

b �

(c(�

b �

e)d)

�� a

�

�

ed

�

and the slope of WZ is �a

�

�

ed

� so XY�WZ.

Therefore WXYZ is a parallelogram.

y

x(0, 0)

(6, �3)

C1

C2

y

xT

A(�4, 2)

C(a, a)

140 Chapter 9: Proof Using Different Approaches

4. Let the triangle have coordinates P(0, 0), Q(2a, 0),R(2b, 2c). The midpoint of PR is X(b, c) and themidpoint of QR is Y(a � b, c). Then the slope of XY is 0 and XY�PQ.

The length of XY is a � b � b � a � �12

� PQ.

The line joining midpoints of two sides is parallel toand equal to one-half of the third side.

5. Let the parallelogram have coordinates P(b, c),Q(0, 0), R(a, 0), S(a � b, c). It is given that PR � QS.

PR � �(a � b�)2 � c�2�QS � �(a � b�)2 � c�2�.

Then (a � b)2 � c2 � (a � b)2 � c2

(a � b)2 � (a � b)2

Therefore ab � 0 and a � 0 or b � 0.If a � 0 the parallelogram does not exist. Hence a ≠ 0. If b � 0 then P is on the y-axis and ∠PQR � 90°, so the parallelogram is a rectangle.Therefore, if the diagonals of a parallelogram areequal the parallelogram is a rectangle.

6. For the given circle the centre is C(h, k) and the radiusis r. If T is the point of contact of the tangent then

P1T2 � CP

12 � CT2

l2 � (x1

� h)2 � (y1

� k)2 � r2.

And l � �(x1

��h)2 ��(y1

��k)2 ��r2�.

7. Let the circle be x2 � y2 � r2, let AB be the diameteron the y-axis, and let CD have equation x � k. Choose point P(0, a) on diameter AB. Since C and Dare on the circle, the coordinates of C are

�k, �r2 � k�2�� and of D are �k, � �r2 � k�2��.Then PC2 � PD2 � k2 � �a � �r2 � k�2��2

�

k2 � �a � �r2 � k�2��2

� 2k2 � 2a2 � 2(r2 � k2)

� 2a2 � 2r2.

Since the expression does not contain k, PC2 � PD2 isindependent of the position of chord CD.

y

xP(0, a)

y

x

l( )

( )

y

x0, 0 a, 0

( ) (

(( )

)

)

y

x

y

x

S(2d, 2e) Y


8. Let quadrilateral PQRS have coordinates as shown.Then the midpoints of the sides are A(d, e), B(a, 0),C(a � b, c), and D(b � d, c � e).

The midpoint of AC is X��a � b2

� d�, �

c �

2e

��and the midpoint of BD is ��a � b

2� d�, �

c �

2e

��.

Since the same point is the midpoint of each line, thelines bisect each other.

9. Let C1

be the circle x2 � y2 � r2 and let C2

be the

circle (x � a)2 � y2 � k2.

Then C1

� C2

� x2 � y2 � r2 � [(x � a)2 � y2 � k2)]

� x2 � y2 � r2 � x2 � 2ax �

a2 � y2 � k2

� 2ax � r2 � a2 � k2

and C1

and C2

� 0 gives the line x � �r2 � a

2

2

a� k2

�.

From problem 6, the tangent from P1(x

1, y

1) to C

1� 0

has length

�x12 ��y

12 ��r2�

and the tangent from P1

to C2

� 0 has length

�(x1

��a)2 ��y12 ��k2�.

Since these are equal,

x12 � y

12 � r2 � (x

1� a)2 � y

12 � k2

or 2ax1

� r2 � a2 � k2

or x1

� �r2 � a

2

2

a� k2

�.

But this means that P1

is on the line C1

� C2

� 0.

Exercise 9.3

4. Method 1.Since D and E are midpoints of AB and AC,

DE � �12

� BC.

In ∆FBC, G and H are midpoints of FB and FC,

so GH � �12

� BC.

Then DE � GH.

Method 2.Since D and E are midpoints of AB and AC,

DE�� 12

� BC�� (from exercise 6.4, question 1).

Since G and H are midpoints of FB and FC,

GH�� 12

� BC��.

Then DE�� GH�� and �DE�� GH��.

5. a. Method 1.Since AD�BC, ∠EAD � ∠AEB and ∠BDA �∠DBC.Therefore ∆FAD ~ ∆FBE.

Then �FF

EA� � �

AB

DE� � �

21

�

or FE � �13

� AE.

y( )

y

xP(0, 0) 0


Draw FG ⊥ BC. Then ∠FGE � ∠ABE and∠FEG � ∠AEB.Therefore ∆FEG ~ ∆AEB.

Then �FA

GB� � �

FAE

E� � �

13

�

or �F2G� � �

13

� and FG � �23

�.

Method 2.Apply coordinates as in the diagram. The equationof BD is y � xThe equation of AE is y � �2(x � 1)Substituting x � y we obtain

y � �2y � 2

y � �23

�

The altitude from F is �23

�.

b. Since AD � 2 and BE � 1 and ∆FEB ~ ∆FAD,

�∆∆

FF

EAD

B� � �

12

2

2� � �14

�.

6. Method 1.Join BD. Choose a point in AB such that AP : PB �m : n.Now choose point Q in AD such that AQ : QD – m : n.

Then PQ�BD and PQ � �m

m� n� BD.

Now choose R in CD such that CR : RD � m : n andchoose S in BC such that CS : SB � m : n.

Then RS�BD and RS � �m

m� n� BD.

Therefore PQ � RS and PQ�RS, so PQRS is aparallelogram.

Method 2.This same approach can be used with vectors.

Method 3.Using Example 2 as a guide, choose points P, Q, R,and S in the sides of quadrilateral ABCD withcoordinates as in the diagram, such that AP : PB �CS : SB � CR : RD � AQ : QD � m : n.

Then P has coordinates ��mn�

dn

�, �m

n�

en

��Q has coordinates ��mm

b �

�

nnd

�, �m

mc �

�

nne

��,

R has coordinates ��mmb �

�

nna

�, �m

m�

cn

��S has coordinates ��m

n�

an

�, 0��mc �

mn�

e �

nne

�The slope of PQ is _____________ � �

bc

�.

�mb �

mn�

dn� nd

�

The slope of RS is �mm

bc� � �

bc

�.

The slope of PS is �nd

n�

ena

� � �d �

ea

�.

The slope of QR is �nd

n�

ena

� � �d �

ea

�.

Then PQ�RS and PS�QR, and PQRS is aparallelogram. This is a good problem to check usingGeometer’s Sketch Pad.

y

x

( (

(

)

)

)

y

x

( )


7. Method 1.

From P draw PX perpendicular to the plane ABCD.

Then XA2 � XC2 � XB2 � XD2 from an earlier proof.

Now using the Pythagorean Property repeatedly,

PA2 � PC2 � PX2 � XA2 � PX2 � XC2

� 2PX2 � XA2 � XC2

PB2 � PD2 � PX2 � XB2 � PX2 � XD2

� 2PX2 � xB2 � xD2

Then PA2 � PC2 � PB2 � PD2.

Method 2.Apply coordinates using 3 dimensions, letting A be(0, a, 0), B be (b, a, 0), C be (b, 0, 0), D be (0, 0, 0),and P be (p, q, r).

Then PA2 � PC2 � p2 � (q � a)2 � r2 � (p � b)2 �

q2 � r2

� p2 � q2 � 2r2 � (q � a)2 � (p � b)2

and PB2 � PD2 � (p � b)2 � (q � a)2 � p2 �

q2 � r2

� p2 � q2 � 2r2 � (p � b)2 �

(q � a)2.

Therefore PA2 � PC2 � PB2 � PD2.

8. a. Let ∆ABC have coordinates as in the diagram.Then M has coordinates (c, 0).

Now AB2 � AC2 � a2 � b2 � (a � 2c)2 � b2

� 2a2 � 2b2 � 4c2 � 4ac.

Also 2AM2 � 2MC2 � 2[(a � c)2 � b2] � 2c2

� 2a2 � 4ac � 2c2 � 2b2 � 2c2

� 2a2 � 2b2 � 4c2 � 4ac.

Therefore AB2 � AC2 � 2AM2 � 2MC2.


b. In ∆AMB, AB2 � AM2 � MB2 � 2AM · MB cos �.In ∆AMC, AC2 � AM2 � MC2 � 2AM · MCcos (180° � �).Noting that MB � MC andcos � � � cos (180° � �),AB2 � AC2 � 2AM2 � 2MC2.

Exercise 9.4

1. Let (x, y) be the coordinates of P. Let (a, a) be thecoordinates of point A and (b, 2b) be the coordinates ofpoint B.

Then x � �a �

32b

� and y � �a �

34b

�.

Solving these for a and b in terms of x and y,a � 2b � 3xa � 4b � 3y

b � �3y �

23x

�

And a � 6x � 3y

Now AB � 6 and AB � �(a � b�)2 � (a� � 2b)�2�

��125�x ��

92

�y�2� � (9x � 6y)2

Then �2245

� x2 � �1325

� xy � �841� y2 � 81x2 � 108 xy �

36y2 � 36

or 549x2 � 702xy � 225y2 � 144

or 61x2 � 78xy � 25y2 � 16.

A

B CM�

y

x

Note: The answer in the first printing of the textbookhas an incorrect coefficient of x2. It is recommendedthat students use Geometry Sketchpad to determinethe conic described by the equation.

2. Join BY and AX. Arc AB subtends ∠AYB and ∠AXB. Then ∠AYB � ∠AXB for all positions of X. Since XY is a diameter, ∠XBY � ∠XAY � 90°.Then ∠XPY � 90° � ∠AXB.But this is a constant value and ∠XPY is always thesame and stands on arc AB.Then the locus of P is part of a circular arc havingAB as a chord.

3. Let the square ABCD have coordinates as shown andlet P(x, y) be a point such that the squares of theperpendiculars from it to the four sides sum to k, k >1.Then the distance from P to AD is x and to BC is 1 � x.The distance from P to AB is y and to CD is 1 � y.

Then x2 � (1 � x)2 � y2 � (1 � y)2 � k

or 2x2 � 2x � 2y2 � 2y � 2 � k

or �x � �12

��2

� �y � �12

��2

� �12

�(k � 1).

This is a circle with centre ��12

�, �12

�� and

radius ��12

�(k �� 1)�.

4. Join AY and XB to intersect at T. Since XY is a diameter,

∠XAY � ∠XBY � 90°.Then ∠PAT � ∠PBT � 90°,

and ∠PAT � ∠PBT � 180°.Therefore PATB is a cyclic quadrilateral, and since A and B are fixed, P lies on a circle passing through A and B. Note that the intersections T of XB and AYalso lie on the circle.

5. From the centre of the circle with radius r draw OAand OB to meet the tangents from P. Then ∠PAO �∠PBO � 90° and since ∠APB � 90° and OA � OB,

PAOB is a square. Then PO � �2�r and the locus ofP is a circle with centre O and radius �2�r.

P B

AO

P

A

B

X YT

y

xA(0, 0) B(1, 0)

D(0, 1) C(1, 1)

P(x, y)

Y B

A

X

O P

y

A

x

P

y = x

y = 2xB


Review Exercise

10. The slope of PA is 2 greater than the slope of PB.

Slope PA � �yx

�

�

23

�.

Slope PB � �y �

x4

�.

Then �yx

�

�

23

� � �y �

x4

� � 2 or, simplifying,

3y � 2x2 � 12x � 12.This is the equation of a parabola that passes throughA and B and has axis of symmetry x � 6.

13. Rewrite the circle equations.

x2 � y2 � 4x � 18y � 60 � 0 becomes

(x � 2)2 � (y � 9)2 � 25.

The centre of this circle is (�2, 9).

x2 � y2 � 2x � 10y � 1 � 0 becomes

(x � 1)2 � (y � 5)2 � 25.

The centre of this circle is (1, 5).

Substituting (1, 5) in the first circle,

(1 � 2)2 � (5 � 9)2 � 32 � 42 � 25, and (1, 5) is on

the circle. Substituting (�2, 9) in the second circle,

(�3)2 � (4)2 � 25, and (�2, 9) is on the second circle.

14. The circle equation can be rewritten.x2 � y2 � 6x � 8y � 24

or (x � 3)2 � (y � 4)2 � 49. The centre is (3, 4) and the line from the centre to (5, �1) is perpendicular to and bisects the chord. The

slope of OA is �43

�

�

15

� � ��52

�.

The slope of XY is �52

�.

The equation of XY is y � 1 � �52

�(x � 5)

or 5x � 2y � 27 � 0.

15. Method 1.Let the centre be C(0, k). Then CO � CA

k � �22 � (�1 � k)�2�2k � 5

k � �52

�.

The equation is x2 � �y � �52

��2

� ��52

��2

or x2 � y2 � 5y � 0.

Method 2.The centre lies on the right bisector of OA, which is

the line with slope �2 and passing through �1, �12

��.

The equation of the right bisector is

y � �12

� � �2(x � 1)

or 4x � 2y � 5 � 0.

The y-intercept is �52

�, so the centre is �0, �52

�� and the

radius is �52

�.

16. Let A(a, 0) be the end of the diameter and let B(b, c)be any other point on the circle. Then P(X, Y) is the

midpoint of chord AB, where X � �a �

2b

� and Y � �2c

�.

Then b � 2X � a and c � 2Y.

y

A(2, 1)

xO(0, 0)

y

xA

O

X

Y

y

x

B(0, 4)

P(x, y)

A(3, �2)


But since B is on the circle, c � �a2 � b�2�� a2 � (�2X � a�)2�� 4aX �� 4X2�

Then rY2 – 4aX � 4X2

or X2 � y2 � aX � 0

This is the equation of a circle with centre ��a2

�, 0� and

radius �a2

�.

17. Let the vertices of ∆ABC have coordinates as in the diagram and let P(x, y) be a point such that

PA2 � PB2 � PC2 � k, a constant.

Then (x � a)2 � (y � b)2 � x2 � y2 � (x � C)2 �

y2 � k

or 3x2 � 3y2 � 2(a � c) x � 2by � a2 � b2 � c2 � k.

This is the equation of a circle.

18. Position the parallelogram so that diagonal AC lies onthe x-axis with the origin as its midpoint. Then thecoordinates of the parallelogram are A(�a, 0),B(�b, �c), C(a, 0), and D(b, c).

We are given that for a point P(x, y),

PA2 � PC2 � PB2

or (x � a)2 � y2 � (x � a)2 � y2 � (x � b)2 �

(y � c)2

or x2 � y2 � 2bx � 2cy � 2a2 � b2 � c2 � 0This is the equation of a circle with centre (b, c),which is vertex D.

Chapter 9 Test

1. a. The set is all points on the line x � 4.

b. The set is all points on the right bisector of the linejoining A and B. This is the line 2x � y � 6 � 0.

c. The set is all points on the circle with centre (�3, 2) and radius 5. This is the circle (x � 3)2 �(y � 2)2 � 25.

2. a. x2 � y2 � 2x � 6y � 3 � 0 can be written

(x � 1)2 � (y � 3)2 � 13. This is a circle with

centre (1, �3) and radius �13�.

b. This is the equation of a sphere with centre (1, �2, 3) and radius 3.

3. a. If P(x, y) divides AB in the ratio 1 : 4, then fromthe diagram R has coordinates (x, 7) and S hascoordinates (2, y).

Now ∆PAR ~ ∆BPS, so �AP

RS� � �

x2

�

�

3x

� � �14

�

and 5x � �10x � �2.

Also �RSB

P� � �

1y7�

�

7y

� � �14

�

and 5y � 45y � 9.

The coordinates of P are (�2, 9).

y

x

S

R

B(2, 17)

P(x, y)

A(–3, 7)

y

x

D(b, c)

A(�a, 0) C(a, 0)

B(�b, �c)

y

xB(0, 0) C(c, 0)

A(a, b)P(x, y)

y

B

xA(a, 0)

P(x, y)


b. From the diagram, if P(x, y) is the midpoint of aline segment joining O(0, 0) to any point A(4, y),then x � 2 always.The locus is the straight line with equation x � 2.

4. The right bisector of AB is y � 6. The right bisector ofAC is x � 3. The circle has centre (3, 6).

Its radius is �(3 � 1�)2 � (6� � 4)2� � �8�.

The required equation is (x � 3)2 � (y � 6)2 � 8.

5. Let the rulers be the axes. Then A has coordinates (a, 0) and B has coordinates (0, b) and AB � 10.From similar triangles,

�a �

xx

� � �14

�, so a � �54

�x

and �b �

yy

� � �14

�, so b � 5y.

Now since AB � 10, a2 � b2 � 100,

then �2156�x2 � 25y2 � 100

or �6x4

2

� � �y4

2

� � 1.

This is the equation of an ellipse.

6. From the diagram we have A(1, �3�), B(0, 0), and

C(2, 0). In P(x, y), such that PA2 � PB2 � PC2 � 11,

we obtain (x � 1)1 � (y � �3�)2 � x2 � y2 �

(x � 2)2 � y2 � 11

or 3x2 � 3y2 � 6x � 2�3�y � 8 � 11

or x2 � 2x � y2 � �23

� �3�y � 1

or (x � 1)2 � �y � ��33�

��2� �

73

�.

This is the equation of a circle with centre �1, ��33�

��and radius ��

73

��.

7. Let the centre of the circle be P(x, y) and coordinatize

as in the diagram. The midpoint of the chord AB is

M (c, t � k).

Then t � k � y and t � y � k (k a constant).

Since CO � CA,

x2 � y2 � (x � c)2 � (y � t)2

or 2cx � c2 � t2 � 2ty

� c2 � (y � k)2 � 2(y � k)y

� c2 � y2 � 2ky � k2 � 2y2 � 2ky

� c2 � y2 � k2.

The equation of the locus of P is

2cx � �y2 � (c2 � k2), the equation of a parabola.

y

x

M

A(c, t)

C(x, y)

x = c

B(c, t, �2, k)O(0, 0)

y

xB(0, 0) C(2, 0)

P(x, y)

A(1, 3)

y

B(0, b)

P(x, y)

xA(a, 0)

y x = 4

A(4, y)

x

P(x, y)

O(0, 0)



Exercise 10.1

11. a. The result of the draw will form three-digitnumbers of the form 100a � 10b � c,a, b, c � {1, 2, 3, … , 9} and a ≠ b ≠ c.

b. Since the ball labelled 4 is selected first, the secondball can be any of the remaining eight, and for eachof these, the third ball can be any of the remainingseven, giving 8 � 7 � 56 subsets. Therefore n(A)� 56. A similar argument applies when the secondand third balls drawn are 4s, hence n(B) � 3 � 56� 168.

12. a. The three-digit number will be of the form 100a � 10b � c,a, b, c � {1, 2, 3, … , 9}.

b. With the first ball labelled 4, the second ball can beany of the balls labelled 1 through 9, and for eachof these, the third ball can be any of the ballslabelled one through nine, hence n(A) � 9 � 9 � 81. The number of elements in U is n(U

1) �

9 · 9 · 9 � 729. (We get this from the argument thatthe first ball can be any of the nine. Since it is nowreplaced, the second can be any of nine as will thethird after replacement of the second.)Now the number of elements in U, where no 4 is drawn, will be 8 · 8 · 8 � 512. (Each of the first,second, and third draws can be any of the ballsnumbered 1, 2, 3, 5, 6, 7, 8, 9). The number of three-digit numbers when 4 is selected will be n(B) � 93 � 83 � 217.n(U

1) > n(U) since U is a subset of U

1.

13. U � {1, 2, 3, … , 1000}.

a. The integers divisible by 7 in factored form are 7(1), 7(2), 7(3), … , 7(142). 7 � 142 � 994, which is the lowest multiple of 7less than 1000. n(E) � 142.

b. The perfect squares are 12, 22, 32, … , 312. 312 � 961 and n(F) � 31.

c. The integers divisible by 3 are 3(1), 3(2), 3(3), … , 3(333) therefore n(G) � n(u) � n(G�)

� 1000 � 333n(G) � 667.

d. Counting the number of integers ending in 9 wehave from each of 10 rows, 10 such integers:

9, 19, 29, … , 99 → 10109, 119, 129, … , 199 → 10209 → 10

. .. .. .909, 919, … , 999 → 10therefore n(H�) � 100 and

n(H) � n(U) � n(H�)� 1000 � 100

n(H) � 900.

14. Letters A, B, and C are placed in envelopes a, b, and c.Let the ordered triple (x, y, z) mean letter A is inenvelope x, B in y, and C in z. The universal set willbe U � {(a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b),(c, b, a)}.No letters to the correct person will be the subset {(b, c, a,), (c, a, b)}.

15.

Exercise 10.2

5. If A and B are disjoint, then n(A � B) � n(A) � n(B). If A and B have common elements, n(A � B) ≠ 0,then n(A � B) < n(A) � n(B).Therefore n(A � B) � n(A) � n(B).

8. With A and B disjoint, there are no elements commonto both A and B. The number of elements in the unionof these two sets is the sum of the number of elementsin both sets.

y

x1

V

V

(1, 1)1

Chapter 10: Introduction to Counting 149

Chapter 10 • Introduction to Counting

.

11. Let the committee with two girls be P, one boy andone girl Q, and two boys R.P � {AB, AC, BC}Q � {AD, AE, AF, BD, BE, BF, CD, CE, CF}R � {DE, DF, EF}u � P � Q � R.

13.

The labelling of a point represents i � j � c.

c. All of the subsets in b are disjoint.

d. The union of the subsets in b includes all of theelements in P.

15. Consider n(A � B) ≠ 0.

16. U � (A � B) � (A � B�) � (A� � B) � (A� � B�).

Exercise 10.3

5. The ten numbers that start with 7 are 70, 71, 72, … ,77, 78, 79, and those that end in 7 are 17, 27, 37, … ,77, 87, 97. The 77 is in both sets was counted twice.

6. U: two-digit integers.V: contain at least one 5.W: contain at least one 6.The complement of V � W, V��W� is the set of two-digit integers that do not contain the digits 5 or 6.

n(u) � 9 � 10 � 90.

n(V��W�) � 7 � 8 � 56 (the first digit is any one of1, 2, 3, 4, 7, 8, 9, and for each of these the seconddigit can be any one of 0, 1, 2, 3, 4, 7, 8, 9).

Therefore n(V � W) � n(U) � n(VVV��W�)� 90 � 56

n(VV � W) � 34.

7. Let A be the number of integers divisible by 3.

A � {3(1), 3(2), 3(3), … , 3(10)}therefore n(A) � 10

therefore n(A�) � n(U) � n(A)� 30 � 10

n(A�) � 20.

8. Of 90 students, 42 are girls, therefore 48 are boys. Also37 take Business 101, of which 42 � 19 � 23 aregirls, hence 14 are boys. 14 boys take Business 101.

10. Since the Ei, i � 0, 1, 2, 3, are disjoint sets and

U � E0

� E1

� E2

� E3

then n(U) � n(E0) � n(E

1) � n(E

2) � n(E

3).

12. There are 365 days in a non-leap year. 365 � 7(52) � 1, which gives 52 � 2 � 104 days that fall on a weekend. The extra day can fall on any day ofthe week. If it falls on a Saturday or Sunday there will be 104 � 1 � 105 days falling on a weekend,which is the maximum number possible.

13. a. The numbers divisible by 5, set A, are A � {5(1), 5(2), 5(3), … , 5(200)}

therefore n(A) � 200.

b. The numbers divisible by 7, set B, are B � {7(1), 7(2), 7(3), … , 7(142)}

therefore n(B) � 142.

c. Numbers divisible by both 7 and 5 are divisible by 35, set C, C � {35(1), 35(2), 35(3), … , 35(28)}

n(C) � 28 � n(A � B).

A B A B

A B A BA B A B

A � B A � B

BA

4

3

2

1

5 6 7 8

4 5 6 7

3 4 5 6

2 3 4 5

j

i1 2 3 4

150 Chapter 10: Introduction to Counting

d. n(A � B) � n(A) � n(B) � n(A � B)� 200 � 142 � 28� 314.

The number of numbers divisible by neither 5 nor 7,

n(A��B�) � 1000 � 314, is 686.

e. Of the 200 numbers divisible by 5, 28 are also divisibleby 7.Therefore there are 200 � 28 � 172 numbers divisibleby 5 but not divisible by 7.

14. The number of integers divisible by 7, the set A, is n(A) � 142 (from 13(c)). The integers divisible by 13,set B,

B � {13(1), 13(2), 13(3), … , 13(76)}therefore n(B) � 76.

Now A � B � {91(1), 91(2), 91(3), … , 91(10)}and n(A � B) � 10n(A � B) � n(A) � n(B) � n(A � B)

� 142 � 76 � 10� 208.

The number of integers divisible by 7 or 13 is 208.

17. Let A be the set of integers divisible by 2, n(A) � 500.B is the set of integers divisible by 3, therefore n(B) � 333.C is the set of integers divisible by 5, therefore n(C) � 200.

Now A � B � {6(1), 6(2), 6(3), … , 6(166)}and n(A � B) � 166.

A � C � {10(1), 20, 30, … , 10(100)}and n(A � C) � 100.

B � C � {15(1), 15(2), 15(3), … , 15(66)}and n(B � C) � 66.

A � B � C � {30(1), 30(2), 30(3), … , 30(33)}and n(A � B � C) � 33.Now n(A � B � C)= n(A) � n(B) � n(C) � n(A � B) � n(A � C) �

n(B � C) � n(A � B � C)= 500 � 333 � 200 � 166 � 100 � 66 � 33= 734.734 integers are divisible by 2, 3, or 5.

18. Define n(E1) so that

n(Ei) � n(E

1) � n(E

2) � n(E

3) � … � n(E

n)

n(EiE

j) be the sum of all possible intersecting pairs

Ei� E

j, 1 � i, j � n, i < j.

n(EiE

jE

k) be the sum of all possible intersecting triples

Ei� E

j� E

k, i � 1, k � n, i < j < k, and so on.

Then n(Ei� E

j� E

k� … � E

n)

= n(Ei) � n(E

iE

j) � n(E

iE

jE

k) � n(E

iE

jE

kE

l) � … � (�1)n

– 1 n(EiE

jE

k… E

n).


Exercise 10.4

4. The number of letters available for postal codes isn(X) � 26 � 7 � 19. The number of digits is n(x) � 10. There are 19 choices for the first X and foreach of these there are 10 choices for the first x,19 for the second X, 10 for the second x, 19 for thethird X and 10 for the third x. Then the number of postal codes is 10 · 19 · 10 · 19 · 10 · 19 � 193 � 103

� 6 859 000.

5. 3 10 10 � 10 10 10 10. The first digit can be any oneof 3, 5, or 6, 3 choices. For each of these, the seconddigit can be any of the 10 digits. Similarly, theremaining digits can be any of 10. Hence from theproduct rule there will be 3 � 106 seven-digittelephone numbers starting with a 3, 5, or 6.

6. 2 4 3 2 1 1The first person on the left can be either of the two tabled people, 2 choices, and the position of theextreme left then is filled with the other of the two tallest people. The second can be filled with any of the four remaining people, the third with 3,fourth with 2 and fifth in 1. Hence there are 2 · 4 · 3 · 2 · 1 · 1 � 48 possible arrangements.

7.

There are 6 choices in which the first box can befilled, for each of these the second box can be filled in 5 ways, the fourth in 4 ways, fifth in 3 ways, sixthin 2 ways, and the seventh in 1 way. Therefore thenumber of anagrams is 6 · 5 · 4 · 3 · 2 · 1 � 720.

8. a. b. Since there are only three letters, C, A, and T,once the first letter is chosen, there are only twochoices for the second letter and then one for thethird letter. From the product rule, the number ofdifferent “words” is 3 · 2 · 1 � 6.

9. The product rule implies order. Choosing a pizza withmushrooms, sausage, and onions is the same had wechosen sausage, mushrooms, and onions. Repetition isincluded in the 504 pizzas. Each selection of 3 itemsgives rise to 6 arrangements, hence there would be 504 � 6 � 84 different pizzas possible.

10. a. A possible answer sheet is T T F F F T F T T T.

b. Since each question can be either True or False,there will be 210 � 1024 different answer sheets.

i e

11. Since there are five possibilities for each question,there are a possible 57 � 78 125 possible answersheets. Since there are 30 000 students, every answersheet could be different.

12. Consider a, e, i, o, and u as the only vowels for thisexercise.

a. U � {aab, pqr}A � {eez, ibc}B � {pqr, pss}C � {xyz, mno}

b. From the product law we have:n(U) � 26 · 26 · 26 � 17 576n(A) � 5 · 26 · 26 � 3380n(B) � 5 · 5 · 5 � 125n(C) � 26 · 25 · 24 � 15 600

c. n(A � B) � 0, since the acronym must start with a vowel and the only letters that can be used are p, q, r, s, and t, none of which are vowels.

d. A � B represents the acronyms that start with a vowel or are made up using only the letters p, q, r, s, and t.

e. n(A � B) � n(A) � n(B) � n(A � B)� 3380 � 125 � 0� 3505.

f. The number of acronyms in which all letters aredifferent is 26 · 25 · 24 � 15 600. The number ofacronyms that use one letter at least twice is n(u) � 15 600 � 1976.

13. Letters a, b, c, d, e, f are rearranged in n(u) � 6 · 5 · 4 · 3 · 2 · 1 � 720 ways. The number of wordsthat being with a is determined by the product rule as1 · 5 · 4 · 3 · 2 · 1 � 120. Therefore the number ofwords that do not begin with a is 720 � 120 � 600. Or the first letter any of b, c, d, e, f and the productrule gives 5 · 5 · 4 · 3 · 2 · 1 � 600.

14. The first letter is any of the given 6, the second any of 5, the third any of 4, and the fourth any of theremaining 3. From the product rule the number offour-letter words is 6 · 5 · 4 · 3 � 360. If the wordbegins with a, then the second letter is any of theremaining 5, the third any of 4, and the fourth any of 3, giving 1 · 5 · 4 · 3 � 60 words starting with a.

15 a. 9 · 10 · 10 · 10. The first digit can be any of 1, 2, 3,4, 5, 6, 7, 8, or 9, 9 choices. The second, third, andfourth places can be any of the 10 digits, hencethere are 9 · 10 · 10 · 10 � 9000 such integers.

b. The units digit is either 7 or 8, 2 choices. Hencefrom the product rule we have 9 · 10 · 10 · 2 � 1800 integers ending in 7 or 8.

c. If there are no repeated digits, then the second digithas 9 possibilities, the third 8, and the fourth 7,giving 9 · 9 · 8 · 7 � 4536 integers.

d. The number with repeated digits will be 9000 � 4536 � 4464.

16. There will be 26 � 26 � 10 � 62 symbols available.

a. _ _ _ _ _ _ _ _ Each position can be any of the 62symbols (with repetition) hence there will be 628

possible passwords.

b. 10 _ _ _ _ _ _ 10 The first and last position have 10choices. The remaining positions can be any of the62 symbols, hence there are 626 � 102 passwords.

c. With no repeated digits, there will be 6261 · 60 · 59 · 58 · 57 · 56 · 55 different passwords.

d. The number of passwords with no 9 is 618.Therefore with at least one 9 there will be 628 � 618 passwords.

17. a. Represent each divisor as an ordered pair,i.e., (a, b) � 2a 3b. With 0 � a � 2 and 0 � b � 1 (0, 0) � 1, (1, 0) � 2, (2, 0) � 4 (0, 1) � 3, (1, 1) � 6, (2, 1) � 12.

b. There are six sequences that can be formed asshown in (a) as ordered pairs.

c. 12 has 6 divisors.

d. 144 � 24 · 32

Integer divisors of 144 can be written in the form 2a · 3b where 0 � a � 4, 0 � b � 2. Thereare 5 possible values of a, the first member of thesequence; and for each of these there are 3 possiblevalues for b, the second member of the sequence;hence there will be 5 � 3 � 15 integer divisors of 144.

e. For odd divisors, there can be no even factors,hence the sequence has 0 as its first term, thesecond term is either 0, 1, or 2; hence there are 3odd divisors.

18. a. 64 800 � 25 · 34 · 52.Here we form three-term sequences where there are 6 choices for the first term (0, 1, 2, 3, 4, and 5); and for each of these, 5 choices for the second termand 3 for the third. Hence there are 6 · 5 · 3 � 90 integer divisors of 64 800.

b. For even divisors, the first term of our sequencemust be one of 1, 2, 3, 4, or 5; 5 choices, hencethere are 5 · 5 · 3 � 75 even divisors of 64 800.



19. a. n � 2a3b5c a � 1, b � 1, c � 1.The first term of the three-term sequence has (a � 1) choices; for each of these, the second term has (b � 1), and the third term has (c � 1).Hence there are (a � 1)(b � 1)(c � 1) integerdivisors of n.

b. For an even divisor, a two must be included.Therefore the first term of the sequence has achoices and there will be a(b � 1)(c � 1) evendivisors of n.The fraction of even divisors is

� a �

a1

.

20. a. The first term of the sequence can be filled in mways; and for each of these, the second in (m � 1) ways, and for each of these the third in (m � 2) ways. From the product rule there will be m(m � 1)(m � 2) sequences of length three.

b. Since the symbols can be repeated, the number ofsequences will be m · m · m � m3.

21. In a sequence of length r and the product rule therewill be r factors in the product, the last one being [m � (r � 1)] � (m � r � 1). Hence the number ofsequences will be m(m � 1)(m � 2) … (m � r � 1).With repetition allowed there will be mr sequences of length r.

22. a. Since the five people can have birthdays on thesame day, the number of sequences will be 3655.

b. If the birthdays are distinct, the number ofsequences will be 365 · 364 · 363 · 362 · 361.Percent having distinct birthdays will be

� 0.972864

� 97.29%.

c. With two or more birthdays on the same day, wehave 100% � 97.29% � 2.71%.

23. a. The number of sequences that are possible is 365n.With different birth dates, the number of sequenceswill be 365 · 364 · 363 · … · (365 � n � 1).With two or more birthdays on the same day, thefraction is

1 � .

b. Now 1 � > 12

.

Therefore < 12

for n � 22, the fraction is 0.5243.

Exercise 10.5

2. Sequence of length 6 using the 10 digits and 26letters.

If the letter is in the final position, there will be 26 · 10 · 9 · 8 · 7 · 6 passwords. The number ofpasswords will be the same if the letter is in thesecond, third, fourth, fifth, or sixth position. Thereforethe number of passwords is 26 · 10 · 9 · 8 · 7 · 6 � 6� 4 717 440.

3. The number of binary sequences of length n is 2n. Thenumber of binary sequences with length of at most 5will be the sum of the sequences having length 1, 2, 3,4, and 5, i.e., 21 � 22 � 23 � 24 � 25

= 2 � 4 � 8 � 16 � 32= 62.

4. 2 2 2 2 2

The number of binary sequences of length 6 thatbegins with a zero is 25. Similarly, the number ofsequences of length 7 and 8 beginning with zero willbe 26 and 27.Therefore the sum is 25 � 26 � 27

= 25 (1 � 2 � 4)= 25 � 7= 224.

5. a. The number of one-letter words is 4; two-letterwords, 4 � 3 � 12; three-letter words, 4 · 3 · 2 � 24; and four-letter words, 4 · 3 · 2 · 1 � 24.The total number of words is 64.

b. Number of words ending in a:one-letter words � 1; two-letter words, 3 · 1 � 3; three-letter words, 3 · 2 · 1 � 6; and four-letterwords, 3 · 2 · 1 · 1 � 6.The number of words ending in a is 16.

Or 14

of the words end in each of

a, b, c, or d. Therefore 14

of 64 � 16

words end in a.

0

365 · 364 · 363 · … · (365 � n)

365n

365 · 364 · 363 · … · (365 � n)

365n

365 · 364 · 363 · … · (366 � n)

365n

365 · 364 · 363 · 362 · 361

3655

a(b � 1)(c � 1)(a � 1)(b � 1)(c � 1)

c. Number of words that do not contain an a:one-letter words � 3; two-letter words, 3 · 2 � 6; three-letter words, 3 · 2 · 1 � 6; and four-letterwords � 0.The number that do not contain an a is 15.Therefore the number of words that do contain an a is 64 � 15 � 49.

6. a. If repetition of letters is allowed,one-letter word � 4; two-letter word, 42 � 16; three-letter word 43 � 64; and four-letter word,44 � 256.The number of words that can be created will be340.

b. Number of words that end in a:one-letter: a 1;two-letter: _ a, 4 � 1 � 4;

three-letter: _ _ a, 4 � 4 � 1 � 16; and four-letter: _ _ _ a, 43 � 1 � 64.

The number of words ending in a is 85.

c. Number of words that do not contain a:one-letter: _ 3;two-letter: _ _ 3 � 3 � 9;

three-letter: _ _ _ 33 � 27; andfour-letter: _ _ _ _ 34 � 81.

The number of words with no a is 120. Therefore the number of words containing an awill be 340 � 120 � 220.

7. The rectangular arrays that are possible are 1 by 10, 2by 5, 5 by 2, and 10 by 1. In each array the numbers 1 to 10 can be arranged in 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 � 3 628 800 ways.There will be 3 628 800 � 4 � 14 515 200 ways ofstoring the numbers.

8. The number of three-letter words possible is 263.The number of three-letter words that do not containeither a, e, i, o, u, or y is 203.Therefore the number of three-letter words that containat least one of a, e, i, o, u, or y is 263 � 203 � 9576.Similarly, the number of four-letter words would be 264 � 204 � 296 976 and five-letter words would be 265 � 205 � 8 681 376. The number of three-, four-,or five-letter words will be 9576 � 296 976 � 8 681 376 � 8 987 928.

9. To determine the number of 5s, we look at the numberof terms that are divisible by 5x, x � 1.x � 1, 5, 5(2), 5(3), … , 5(200): when x � 1,there are 200 5s.x � 2, 25, 25(2), 25(3), … , 25(40): when x � 2,there are an additional 40 5s.x � 3, 125, 125(2), 125(3), 125(4), … , 125(8): when x � 3, there are an additional 8 5s.x � 4, 625: when x � 4, there is 1 additional 5.Therefore the total number of 5s in the set is 200 � 40 � 8 � 1 � 249.

10. Let P be the set of sequences starting with A.Therefore P � {ABCDEF, ABCDFE, …}n(P) � 1 � 5 · 4 · 3 · 2 · 1 � 120Q is the set ending with F.Therefore Q � {ABCDEF, BACDEF, …}n(Q) � 5 · 4 · 3 · 2 · 1 · 1 � 120P � Q � {ABCDEF, ACBDEF, …}n(P � Q) � 1 · 4 · 3 · 2 · 1 · 1 � 24Therefore the number of sequences that start with A orend with F is n(P � Q) � n(P) � n(Q) � n(P � Q))

� 120 � 120 � 24� 216.

11. Represent the set of integers with n distinct digits byA

n, 1 � n � 3.

Now n(A1) � 9

n(A2) � 9 · 9 � 81. (The first digit can be any of

{1, 2, 3, … , 9}: 9 choices, and for each of these thereare 9 choices for the second digit: one of the 8 notchosen for the first digit plus 0).n(A

3) � 9 · 9 · 8 � 648.

The four-digit integers must start with 1 (since we arelooking for integers between 1 and 2000), the second,third, and fourth digits are chosen from the remaining9, 8, and 7 digits.

Therefore we have 1 · 9 · 8 · 7 � 504 allowable four-digit integers.Therefore there are 9 � 81 � 648 � 504 � 1242positive integers between 1 and 2000 inclusive havingdistinct digits.

Let On

represent the set of integers with n distinctdigits that will be odd. 1 � n � 3O

1� {1, 3, 5, 7, 9}, therefore n(O

1) � 5

n(O2) � 8 · 5 � 40. (The last digit is odd, the first can

be chosen by the remaining 8 digits. The first digitcannot be zero.)n(O

3) � 8 · 8 · 5 � 320.

8 � 7 � 4. The first digit must be 1, the units

digit can be any one of {3, 5, 7, 9}, 4 choices, thesecond digit and the third digit can be any one of theremaining 8 and 7 digits.

1


Therefore there will be 224 allowable odd four-digitintegers. Total number of odd integers is 5 � 40 � 320 � 224 � 589.

Fraction that is odd is 1528492

.

12. a. Choosing Cat there are 3 · 2 · 1 � 6 sequences,Mouse there are 5 · 4 · 3 � 60 sequences, and inGoldfish there are 8 · 7 · 6 � 336 sequences. There will be 6 � 60 � 336 � 402 sequences oflength three.

b. Ending in S, choosing Cat there are O, Mouse thereare 4 · 3 · 1 � 12, and choosing Goldfish there are7 · 6 · 1 � 42. There are 0 � 12 � 42 � 54sequences of length three ending in S.

c. Start with a vowel: from Cat there are 1 · 2 � 1 � 2, Mouse there are 3 · 4 · 3 � 36, and Goldfish there are 2 · 7 · 6 � 84. In all, 122 sequences of length three start with a vowel.

d. The number of sequences of length three that do not contain O will be: from Cat, 6; Mouse,4 · 3 · 2 � 24; and from Goldfish, 7 · 6 · 5 � 210;and in total 240. The number of sequencescontaining O will be 402 � 240 � 162.

13. a. Let Cn

represent the set of sequences of length n. n � 1. ThereforeC

1� {0, 1, 2, … , 9} and n(C

1) � 10

C2

� {10, 11, 12, … , 99} and n(C2) � 102

C3

� {100, 101, … , 999} and n(C3) � 103

and so on to n(C10

) � 1010.

n(U) � 10 � 102 � 103 � … � 1010.There is a geometric sequence with

n(U) � a(

rrn

�

�

11)

, a � 10, r � 10, n � 10

therefore n(U) � 10(101

9

0 � 1)

n(U) � 11111111110.

b. Let An

represent the set of sequences of length nhaving unique digits. n � 1.

A1

� {0, 1, 2, … , 9}, n(A1) � 9

n(A2) � 10 � 9 � 90 (first digit any of the 10

available digits, for each of these we can choose

any of the remaining 9 digits). Similarly

n(A3) � 10 · 9 · 8 � 720

n(A4) � n(A

3) � 7 � 5040

n(A5) � n(A

4) � 6 � 30 240

n(A6) � n(A

5) � 5 � 151 200

n(A7) � n(A

6) � 4 � 604 800

n(A8) � n(A

7) � 3 � 1 814 400

n(A9) � n(A

8) � 2 � 3 628 800

n(A10

) � n(A9) � 1 � 3 628 800.

Therefore n(A) is the sum of the above.

n(A2) � 9 864 100.

c. Let En

represent the set of sequences of length nthat contain no zeros. Therefore

E1

� {1, 2, 3, … , 9} and n(E1) � 9

n(E2) � 9 � 9 � 81,

n(E3) � 93,

n(E4) � 94,

n(E5) � 95, … ,

n(E10

) � 910

n(E) is the sum of the above.

Therefore n(E) � 9 � 92 � 93 � … � 910

� 9(910

8� 1)

n(E) � 3 922 632 450n(B) � n(U) � n(E)n(B) � 7 188 478 660.

14. a. The number of sequences of length 3 will be r · r · r � r3.The number of sequences of length 3 that do notcontain an A will be (r � 1)3. Therefore thenumber of sequences containing at least one A is r3 � (r � 1)3

� r3 – (r3 � 3r2 � 3r – 1)

� r3 – r3 � 3r2 – 3r � 1

� 3r2 � 3r � 1.

b. If no repetition of symbols is allowed, the numberof sequences of length 3 will be r(r � 1)(r � 2).Excluding A, there will be (r � 1)(r � 2)(r � 3)sequences of length 3.Therefore the number of sequences containing atleast one A is

r (r � 1)(r � 2) � (r � 1)(r � 2)(r � 3)

= (r � 1)(r � 2)[r � (r � 3)]

= 3(r � 1)(r � 2)

= 3r2 � 9r � 6.


15. Represent the set of arithmetic sequences (x, y, z),

x < y < z, of length 3 with common difference n by An.

Therefore A1

� {(1, 2, 3), (2, 3, 4), (3, 4, 5), … ,

(7, 8, 9)}

therefore n(A1) � 7

A2

� {(1, 3, 5), (2, 4, 6), (3, 5, 7), … ,

(5, 7, 9)}


A3

� {(1, 4, 7), (2, 5, 8), (3, 6, 9)}


A4

� {(1, 5, 9)}, n(A4) � 1.

Now n(A1) � n(A

2) � n(A

3) � n(A

4) � 16.

But each sequence in An

can be reversed, therefore the

number of sequences of length 3 that forms an

arithmetic progression is 16 � 2 � 32.

16. Let An

represent the set of binary sequences of

length n.

Therefore n(A1) � 2

n(A2) � 22

n(A3) � 23 and so on, to n(A

k – 1) � 2k – 1

and n(Ak) � 2k.

Now the number of binary sequences of length less

than k is n(A1) � n(A

2) � n(A

3) � … � n(A

k – 1)

� 2 � 22 � 23 � … � 2k – 1

This is a geometric series with

a � 2, r � 2, n � k � 1

therefore n(A1) � n(A

2) � n(A

3) � … � n(A

k – 1)

� 2(2k – 1 � 1)

� 2k � 2

� n(Ak) � 2.

Review Exercise

3. a. U � {A3A2K1, N5C2P2, …}A � {N2R3N1, N4L2N7, …}B � {A3K5B8, N8H8T8, …}C � {X1Y3A7, P3A4K2, …}D � {N5C2R2, A4B5N3, …}

b. There are 26 � 7 � 19 letters and 9 digits that canbe used with repetitions, thereforen(U) � 19 · 9 · 19 · 9 · 19 · 9 � 193 · 93

� 5 000 211.

(The first, third, and fifth positions can be any ofthe 19 letters, and for each of these the second,fourth, and sixth positions can be any of the 9digits. From the product rule we have (9 � 19)3

possible postal codes)n(A) � 1 · 9 · 19 · 9 · 19 · 9 � 192 · 93

� 263 169.(The first letter must be N, one choice, the othersfollow as for n(U).)n(B) � 19 · 9 · 19 · 9 · 19 · 9 � 263 169.(The last position must be 8, one choice.)n(C) codes using the letter A.Let X be the set not using the letter A, therefore

n(X) � 18 · 9 · 18 · 9 · 18 · 9� 183 · 93.

Therefore n(C) � n(U) � n(X)� 193 · 93� 183 · 93

� 93(193 � 183

n(C) � 748 683.The letter N can be in the first position. There willbe 18 letters that can then be placed in the thirdand fifth positions, giving 1 · 9 · 18 · 9 · 18 · 9 postal codes starting with N.

If N is in the third position there will be 18 · 9 · 1 · 9 · 18 · 9 postal codes, and if N is fifth,there are 18 · 9 · 18 · 9 · 1 · 9 postal codes, thereforen(D) � 182 · 93 � 3

� 708588.

c. The number of postal codes starting with N orending with 8 is n(A � B).Now n(A � B) � n(A) � n(B) � n(A � B)

n(A � B) � 1 · 9 · 19 · 9 · 19 · 1.(The first and last positions have one choice each.The second and fourth can be any of the 9 allowable digits and the third and fifth can be any of the allowable 19 letters.)Therefore n(A � B) � 192 · 93 � 193 · 92 � 192 · 92

n(A � B) � 789 507.

4. a. U � {012, 509, …}A � {035, 246, …}B � {048, 572, …}C � {037, 146, …}

b. n(U) � 10 · 9 · 8 � 720.(Choice of 10 digits in the first position, and foreach of these, 9 digits for the second position and 8 for the final position.)n(A) � 5 · 9 · 8 � 360.(The first digit is even, so can be any of the 5; thesecond digit can be any of the remaining 9; and the third has 8 choices.)n(B) � 9 · 8 · 5 � 360.


(The last digit must be even, therefore there are 5choices for the last position. The first position has9 possibilities and the second has a choice of 8.)The number of sequences containing only odddigits is 5 · 4 · 3 � 60. Therefore the number of sequences containing an even digit is n(C) � n(u) � 60n(C) � 660.

c. A � B � {064, 274, …}n(A � B) � 5 · 8 · 4 � 160.(The first and last digits must be even, hence thereare 5 and 4 choices for these positions. There willbe 8 possible digits for the middle position.)

5. The number of binary sequences of length 6 that startwith 1 and end in zero is 24 (1 _ _ _ _ 0).Similarly, the number starting with zero and ending in 1 is 24 (0 _ _ _ _ 1), and the number of sequencesstarting and ending with 1 is 24 (1 _ _ _ _ 1).Therefore the number of sequences starting or endingwith one is 24 � 24 � 24 � 3 � 24.The number of binary sequences of length 6 is 26.

Therefore the required fraction is 3 �

2624

� 34

.

(The student’s argument that 12

of the sequences start

with 1 includes sequences of the form 1 _ _ _ _ 0 and

1 _ _ _ _ 1, and that 12

of the sequences end in 1

includes the sequences of the form 0 _ _ _ _ 1 and 1 _ _ _ _ 1. The form 1 _ _ _ _ 1 is included twice,

which occurs in 14

of the sequences. Hence he must

subtract 14

from his sum of 1, giving the correct

fraction of 34

.)

6. Local telephone numbers consist of 7 digits, the firstselected from the digits 2 to 9, 8 choices. Theremaining 6 digits can be any of the 10 numbers fromthe set {0, 1, 2, … , 9}, therefore the number ofusable telephone numbers is 8 � 106.

The number of telephone numbers ending in 99 willbe 8 � 104 (8 · 10 · 10 · 10 · 10 · 1 · 1; first digitselected from 8 available, last two digits must be 9,1 choice for each, the other 4 digits can be selectedfrom the 10 available with repetitions allowed).Fraction of numbers ending in 99 is

88�

�

1100

4

6 � 1100.

7. Let A represent the set of integers that contain thedigit 7 and A

nthe set of integers of length n that

contain the digit 7.Therefore A

1� {7} and n(A

1) � 1

A2

� {17, 27, 37, … , 67, 70, 71, … , 79, 87, 97}n(A

2) � 6 � 10 � 2 � 18.

The number of three-digit integers is 9 · 10 · 10 � 900.The number of three-digit integers that do not containa 7 is 8 · 9 · 9 � 648.Therefore n(A

3) � 900 � 648 � 252.

Therefore n(A) � 1 � 18 � 252n(A) � 271.

8. a. The number of three-digit PIN numbers is 103; the number of four-digit PIN numbers is 104;therefore the number of PIN numbers available is 103 � 104 � 11 000.

b. The number of three-digit PIN numbers startingwith 2 is 1 · 10 · 10 � 100 and four-digit PINnumbers starting with 2 is 1 · 10 · 10 · 10 � 1000.Therefore the number of PIN numbers starting with2 is 100 � 1000 � 1100.

c. The number of PIN numbers of length 3 that do nothave a 2 is 9 · 9 · 9 � 93 and of length 4 is 94,therefore the number of PIN numbers that do nothave a two is 93 � 94 � 93 (1 � 9)

� 10 · 93.The number of PIN numbers having at least onetwo will be 11000 � 10 · 93

� 10(1100 � 729)� 10(371)� 3710.

Chapter 10 Test

1. U � {1, 2, 3, … , 999}

a. A is a subset of U whose elements are not amultiple of 5.

b. A � {5, 5(2), 5(3), … , 5(199)}therefore n(A) � 199.

n(A�) � n(U) � n(A)� 999 � 199

n(A�) � 800.

2. The Product Rule: If the first of two tasks can be donein p ways, and for each of these ways, the second taskcan be done in q ways, then together the two tasks canbe done in p · q ways.


3. a. A � B is the set of six-letter words ending in -idor -ic.A � B is the set of six-letter words ending in -idand -ic. (This is not possible, hence n(A � B) � 0.)

b. n(A � B) � n(A) � n(B)n(A) � 4 · 3 · 2 · 1 · 1 · 1 � 24n(B) � 4 · 3 · 2 · 1 · 1 · 1 � 24

therefore n(A � B) � 48.(For n(A), since the last two letters must be -id,there is only one choice for the last two positions.The first letter can be any of the 4 remaining andfor each of these the second, third, and fourth areselected from the remaining 3, 2, and 1 letters,hence n(A) � 4 · 3 · 2 · 1 · 1 · 1

� 24.The argument for n(B) is similar.)

4. If A is first, the second letter can be any of theremaining 6 letters (B, C, D, E, F, G), and for each of these selections the third letter is any of theremaining 5.Therefore the number of words starting with A is 1 · 6 · 5 � 30.Similarly, if B is first there will be 30 words.Therefore the number of arrangements with A or Bfirst will be 60.

5. Binary sequences starting and ending with 1 will be ofthe form 1 _ _ _ 1, of which there are 23. Starting andending with 0 will be of the form 0 _ _ _ 0, of whichthere are 23.Therefore the number of binary sequences of length 5that start or end in the same number is 23 � 23 � 16.

6. a. A � {2(1), 2(2), 2(3), 2(4), 2(5), … , 2(25)}B � {5(1), 5(2), 5(3), … , 5(10)}Elements common to both sets is A � B � {10, 20,30, 40, 50}. Stating n(A � B) � 25 � 10 � 35 hasincluded 5 integers that have been counted twice.

b. n(A � B) � n(A) � n(B) � n(A � B)� 25 � 10 � 5

n(A � B) � 30.

7. The password is a sequence of four letters from thealphabet with repeated letters allowed.

a. The number of passwords is 26 · 26 · 26 · 26 � 264

� 456 976.(Each position can be any one of the 26 letters.)

b. If the letters are unique then the first position canbe any one of 26 letters, and for each of these thesecond letter can be any one of the remaining 25letters, and for each of these the third letter can beany one of the remaining 24 letters, and the fourthletter can be selected from the remaining 23 letters.Therefore the number of four-letter passwords is 26 · 25 · 24 · 23 � 358 800.

c. The number of passwords with no a is 254.Therefore the number with at least one a is 264 � 254 � 66 351.

8. The paths will be of 4 forms. There is one way ofgoing directly from A to E.In A _ E, the second letter can be any one of 3 letters,therefore there are 3 paths of this form. In A _ _ E, thesecond position can be any one of 3 letters and foreach of these the third position is any one of the 2remaining letters. Therefore there are 3 � 2 � 6 paths of the form A _ _ E.A _ _ _ E will yield 3 · 2 · 1 � 6 different paths of thisform, therefore the number of different paths from A to E is 1 � 3 � 6 � 6 � 16.

9. The final digit must be even. There are fourpossibilities, 2, 4, 6, 8. If the final digit is 2 or 4, thenthe first digit can be any of 5, 6, 7, 8, 9, and there are5 · 7 · 6 · 2 possible integers. If the final digit is 6 or8, then there are only four possible choices for thefirst digit, and there are 4 · 7 · 6 · 2 possible integers.The total number of integers is 5 · 7 · 6 · 2 � 4 · 7 · 6 · 2� 756.

10. Let A represent the set of integers between 1 and 1000that do not contain a 7, and A

nrepresent the set of

n digit numbers that do not contain a 7.

Now A1

� {2, 3, 4, 5, 6, 8, 9}

therefore n(A1) � 7.

n(A2) � 8 � 9 � 72. (The first digit can be any of the

digits from the set {1, 2, 3, 4, 5, 6, 8, 9}, and for each

of these the second digit can be any of the digits from

the set {0, 1, 2, 3, 4, 5, 6, 8, 9}.)

Similarly n(A3) � 8 · 9 · 9 � 648.

Therefore n(A) � n(A1) � n(A

2) � n(A

3)

� 7 � 72 � 648

n(A) � 727.



Exercise 11.1

2. P(n, 2) � �(n �

n!2)!

�

� �n(n �

(n1�

)(n2)

�

!2)!

�

� n(n � 1)

P(n � 1, 2) � �((nn

�

�

11))!!

�

� �(n �

(1n)(�

n)(1n)!� 1)!

�

� (n � 1)(n)

�P(

Pn

(�

n,12,)2)

� � �(nn(�

n �

1)1(n))

� � �nn

�

�

11

�.

5. The number of six-digit integers is the number ofsequences of length 6 formed from the 9 digits using each once at most. Hence there are P(9, 6) � 9 · 8 · 7 · 6 · 5 · 4 � 60 480 integers.

a. The first digit is a 6 and the remaining 5 digits for a sequence of length 5 are formed from theremaining 8 digits. Hence there are P(8, 5) � 8 · 7 · 6 · 5 · 4 � 6720.

The fraction is �660742800

� � �19

�.

b. The last digit must be one of 2, 4, 6, or 8. This can be chosen in 4 ways. For each of these, the first 5 digits form a string of 5 digits from theremaining 8. Hence there are 4 � P(8, 5) evenintergers. The fraction is

�4 �

P(P9,

(86,)5)

� � �49

�.

c. The first digit must be one of 1, 3, 5, 7, or 9. Thiscan be chosen in 5 ways. For each of these, the last5 digits can be arranged in P(8, 5) ways. Hencethere are 5 · P(8, 5) integers.Starting with an odd digit, the fraction is

�5

P·(P9(,85,)5)

� � �59

�.

d. The number of six-digit integers having the firstand last digits even is 4 · 3 · P(7, 4).The fraction here will be

�4 · 3

P(·9P,(57), 4)

� � �16

�.

Therefore the fraction of numbers that start or endin an odd digit is

1 � �16

� � �56

�.

e. The 9 can go into any of the 6 positions and the remaining 8 digits form a five-digit string in P(8, 5) ways.Therefore the fraction of numbers that contain a 9 is

�6 �

P(P9,

(86,)5)

� � 6 � �19

�

� �23

�.

f. The seven digits, 1, 2, 3, 4, 5, 6, 7, can be arrangedto form a string of 4 digits in P(7, 4) ways. Onesuch string is 6 1 4 5. Now the 8 can be placedamongst these digits in 5 ways. One of these mightbe 8 6 1 4 5. For each of these, the 9 can be placedin any of 6 places, e.g., 8 6 1 4 9 5. Hence for eacharrangement of the four-digit string P(7, 4), there are 5 � 6 � 30 ways of placing the 8 and 9. Hence there are 30 · P(7, 4) integers that includeboth 8 and 9. The fraction is

�30

P·(P9,

(76,)4)

� � �152�.

g. The number of integers that do not contain 8 and 9is a six-digit string formed from the remaining 7digits. Hence there are P(7, 6) integers that do not contain both 8 and 9. Therefore the number of integers that contain 8 or 9 or both will be P(9, 6) � P(7, 6).The fraction will be

1 � �PP

((79,,

66))

�

= 1 � �112�

= �1112�.

Chapter 11: Counting Methods 159

Chapter 11 • Counting Methods

h. Consider first digit 4, the second digit less than 6.There are 4 possibilities for the second digit (one of 1, 2, 3, or 5), and for each of these theremaining four-digit string, formed from 7 possibledigits, will be P(7, 4). This gives us 4 � P(7, 4)allowable integers with first digit 4. To be less than460 000, the first digit can be one of 1, 2, or 3, andfor each of these the remaining five-digit string,formed from 8 possible digits, will be P(8, 5).Hence there are 3 � P(8, 5) allowable integers lessthan 400 000.Therefore the number of integers less than 460 000 is 4 � P(7, 4) � 3 � P(8, 5) � 23 520.Therefore the fraction of numbers less than 460 000 is

�178�.

6. The eight tallest can be arranged in 8! ways in the lastrow. For each of these, the remaining 16 people can beplaced in the 16 remaining positions in 16! ways.Hence there will be 8! � 16! different arrangements.

7. The first integer can be any of a possible 9 integers(chosen from 1, 2, 3, … , 9). For each of these the lastinteger can be any of 10 possible (chosen from 91, 92,93, … , 100). For each of these 9 � 10 possible firstand last integers the middle integers form a string of98 integers in 98! ways. Hence the number of listspossible is 9 � 10 � 98!.

8. The number of sequences starting with 0 will be astring of length 5 formed from the remaining 9 digitsin P(9, 5) ways. Similarly, with last digit 9, there willbe P(9, 5) strings. But these both include the stringsthat start with 0 and end with 9, of which there are P(8, 4). Hence the number of sequences start with 0 orend with 9 will be 2 � P(9, 5) � P(8, 4) � 28 560.

9. Represent the novels by Ni, 1 � i � 6 and the math

books by Mj, 1 � j � 6.

a. The six novels can be arranged in 6! ways, and foreach of these, the six math books can be arrangedin 6! ways. Hence, with novels on the left and mathbooks on the right, there will be 6! � 6! sucharrangements.

b. If a novel is first, we have NMNM … NM giving 6! � 6! arrangements, and with a math book first,MNMN … MN gives 6! � 6! arrangements.Therefore there will be 2 � 6! � 6! arrangementswhere novels and math books alternate.

c. The book on the left can be any one of six differentnovels. For each of these the remaining 11 bookscan be arranged in 11! ways. Hence there are 6 � 11! possible arrangements.

d. The book on the left can be any of six differentnovels, and for each of these the book on the rightcan be any one of the six different math books. The remaining 10 books can be arranged betweenthe two end books in 10! ways, hence there are 36 � 10! arrangements.

e. If N represents the set with a novel on the left andM the set with a novel on the right, then n(N � M) = n(N) � n(M) � n(N � M)

= 6 � 11! � 6 � 11! � 36 � 10!= 12 � 11! � 36 � 10!= 12 � 10! [11 � 3]= 96 � 10!.

The number of arrangements with a novel in theleft end of the shelf or a math book on the rightend (n(N � M)) is 96 � 10!.

10. n(u) � P(n, 4). The set A is a sequence starting with aparticular symbol �, the remaining string of 3elements formed from n � 1 elements will give n(A)� P(n � 1, 3).The set B is a sequence having two particular symbols� and � side by side in order � �. The remainingstring of two elements formed from n � 2 elementsgive P(n � 2, 2) arrangements. The � � can beinserted before the two elements, between them, orafter them, hence there are 3 positions for the � �.Therefore n(B) � 3 � P(n � 2, 2) With � � in any order gives rise to n(C) � 2 � 3 � P(n � 2, 2)

� 6 � P(n � 2, 2).

11. n(U) � P(n, r). Set A is a sequence starting with aparticular �, the remaining string of r � 1 elementsformed from n � 1 elements will give n(A)� P(n � 1, r � 1).The set B is a sequence having two particular symbols� and � side by side in order � �. The remainingstring of r � 2 elements formed from n � 2 elementsgives P(n � 2, r � 2) arrangements. The � � can beinserted amongst the P(n � 2, r � 2) arrangements in r � 1 positions, hence n(B) � (r � 1) � P(n � 2,r � 2).With order � � or � �, n(C) � 2n(B)therefore n(C) � 2(r � 1) � P(n � 2, r � 2).

12. P(n, r � 1) � �(n �

nr!� 1)!�, P(n, r) � �

(n �n!

r)!�

�P(

Pn,

(nr,�

r)1)

� � �(n �

nr!� 1)!� � �

(n �

n!r)!

�

=

= n � r.

(n � r)(n � r � 1)!��

(n � r � 1)!

160 Chapter 11: Counting Methods

Since n and r are both integers, n � r will be aninteger, hence

�P(

Pn,

(nr,�

r)1)

� is an integer.

13. U: A string of 5 cards formed from 40 cards.Therefore n(U) � P(40, 5).

A: A string of 5 cards formed from 10 cards of thesame colour will give P(10, 5) sequences. Sincethere are 4 colours n(A) � 4 � P(10, 5).

C: The number of sequences of 5 cards in the samecolour is P(10, 5). Now each card in this sequencewill have a different number. Each of these cardscould be any of the 4 colours. Therefore there are45 � P(10, 5), 5 card sequences with no cardhaving the same number. Therefore n(C) � P(40, 5) � 45 � P(10, 5), the number of 5 cardsequences such that at least 2 cards have the samenumber. n(C) � 47 995 200.

B: The first card can be any of the four 2s and foreach of these the last card can be any of the four8s. The remaining 3 cards form a string of 3 chosenfrom the 38 cards available in P(38, 3) ways.Therefore n(B) � 4 � 4 � P(38, 3) � 16 P(38, 3).

14. The number of possible arrangements of the 144integers in the 12 � 12 array is 144!. Place the 12perfect squares in increasing order along the maindiagonal then arrange the remaining 132 integers in132! ways in the remaining spaces. Hence the fraction

will be �113424!!

�.

15. The number of arrangements of the 10 blocks is 10!.Consider the A and B as one block AB, and there willbe 9! arrangements of these 9 blocks. But the ABcan be written as BA, hence there will be 9! morearrangements with A and B adjacent. Therefore with A and B not adjacent, the number of arrangements will be 10! � 2 � 9! � 9! (10 � 2)

� 8 � 9!.

16. Since 10 � 2 � 5, we need to determine the numberof integers in the product of 20! that contain a factorof 5. These are 5, 10, 15, and 20; i.e., the four 5scombine with four 2s to give the largest power of 10to be 104.

17. The number of sequences of length r that can beformed from n symbols is P(n, r). The first symbolcan be any of the n different symbols. Now for each ofthese, the remaining n � 1 symbols will form a stringof length r � 1 in P(n � 1, r � 1) ways. Thereforethere will be n � P(n � 1, r � 1) sequences of length r that can be formed from n symbols, hence P(n, r) � n � P(n � 1, r � 1).

18. L.S. � P(n, r)

� �(n �

n!r)!

�

R.S. � n � P(n � 1, r � 1)

� n �

� �(n �

n(1n

�

�

r1�

)!1)!

�

� �(n �

n!r)!

�

� L.S.Therefore P(n, r) � n � P(n � 1, r � 1).

19. First form a sequence of length k. From the nelements this can be done in P(n, k) ways. Now foreach of these we have n � k elements to form a stringof r � k elements in P(n � k, r � k) ways. Thereforewe have P(n, k) � P(n � k, r � k) sequences oflength r formed from n elements; i.e., P (n, r) � P(n, k) � P(n � k, r � k), 1 � k � r � 1.

20. In the prime factorization of 100!, we need todetermine the number of times the factor 5 occurs.Now single multiples of 5 give 5(1), 5(2), 5(3), … ,5(20), or 20 factors of 5. But factors of 25 yieldadditional 5s, namely 25, 50, 75, 100. Therefore thereare 24 factors of 5 and k � 24.

Exercise 11.2

3. a. A _ _ _ _ _ begin with A, the remaining 11symbols, used as often as we like can be arrangedin n11 ways.

b. AA _ _ _ _ _ begin with AA, the remaining 10symbols can be arranged in n10 ways.

c. The number of arrangements with no As is (n � 1)11 and the number of arrangements with no restrictions is n11. Therefore the number ofarrangements with at least one A is n11 � (n � 1)11.

(n � 1)!��[(n � 1) � (r � 1)]!


4. Sequences of length 10 using 2 symbols, say a and b,is 210. But this includes the sequence with only a’sand only b’s, hence there are 210 � 2 sequences usinga’s and b’s. Similarly there will be 210 � 2 using only a’s and c’s. Therefore there are 3 � (210 � 2)sequences using only 2 symbols.

5. a. The binary sequence begins with 1, hence theremaining r � 1 positions can be either a 0 or a 1. Therefore there are 2r – 1 bit strings starting with 1.

b. A bit string of length r, starting and ending in 1,will have r � 2 positions to be filled with either 1or 2. Hence there will be 2r – 2 such sequences.

c. If B represents the set of bit sequences beginningwith 1, E the bit sequences ending in 1, then n(B � E) � n(B) � n(E) � n(B � E)

� 2r – 1 � 2r – 1 � 2r – 2

� 2r – 2 [2 � 2 � 1]n(B � E) � 3 � 2r – 2.

7. Place two plain tiles in the x positions and two bluetiles in the y positions. Now each of the remainingfour positions has a choice of two different tiles,hence there will be 24 different patterns. Changing the colours on x and y and rotating the table through180° gives the same patterns as above hence there are 24 � 16 different patterns possible.

8. a. The first three letters can be arranged, withrepetition, in 263 ways, and for each of these waysthe 10 digits can be arranged in 103 ways. Therewill be 263 � 103 different licence plates.

b. If all symbols are different the letters form asequence of 3 from 26 letters in P(26, 3), and foreach of these, digits can be arranged in P(10, 3)ways. Hence there are P(26, 3) � P(10, 3) differentlicence plates.

c. The number of licence plates with at least onesymbol repeated is 263 � 103 � P(26, 3) � P(10, 3).

d. The letters can be arranged in 263 � 26 · 25 · 24ways where at least one letter is repeated and foreach of these, the digits can be arranged in 103 � 10 · 9 · 8 ways where at least one digit isrepeated. These will be (263 � 26 · 25 · 24) (103 � 10 · 9 · 8) � 553 280 licence plates whereat least one of the digits and at least one of theletters are repeated.

9. There will be 52 variable names of length 1. (26 lower-case letters and 26 upper-case letters.) If the variablename has length 2, the first position in any one of the52 letter symbols and for each of these the secondposition can be any one of 62 symbols (52 letters and10 digits) giving 52 � 62 variable names of length 2.Therefore there will be 52 � 52 � 62 � 52 � 63 � 3276 variable names of length 1 or 2.

10. The number of possible last three symbols in a passwordis 623, therefore it will take 623 � 20 seconds or

�6620

3

�

�

6200

� � 1324 hours.

11. The number of passwords available of length 1 is 10,of length 2 is 102, of length 3 is 103, … , of length r is10r. The total number of passwords available is the sum of the geometric series 10 � 102 � 103 � … � 10r,

which equals �10(10

9

r � r)�. But the number of

passwords required is 1253 � 1000, therefore

�10(10

9

r � 1)� � 1 253 000, 10r � 1 127 701 and

r � 6.05, hence the minimum value of r will be 7.

12. The number of ways of drawing three numbers withreplacement after each draw is 1003. The number ofways of drawing three numbers with replacement aftereach draw (excluding the 1) is 993. Hence the numberof ways that a 1 is drawn at least once, in which caseyou win at least one prize, will be 1003 � 993 � 29 701.

13. The number of possible sequences with 8 flips is 28.The number of sequences with heads on the first andlast flip will be 26. Similarly if the first and last flipsturn out to be tails there will be 26 such sequences.Hence the number of sequences with the same resulton the first and last flips is 2 � 26 � 27. This will

occur �22

7

8� � �12

� of the time.

14. Since each switch has three different positions, thefour switches can be arranged in 34 different ways.

x x

y y

xx

yy

rotate180º


15. a. When each bit in the binary sequences is 1, wehave the decimal number 25 � 24 � 23 � 22 � 21

� 20 � 63, which is the largest decimal number.

b. If the bit sequence ends in a 1, the sum of thepowers of 2 will be odd, therefore these sequenceswill represent the odd numbers from 0 to 63.

c. The smallest decimal number will be 25 � 20 � 33and the largest will be 63, hence we have the oddnumbers from 33 to 63.

16. Writing 1000 in terms of powers of 2 we have 1000 � 64 � 36� 64 � 32 � 4 � 26 � 25 � 22

� 1 � 26 � 1 � 25 � 0 � 24 � 0 �23 � 1 � 22 � 0 � 21 � 0 � 20

Therefore 1000 written as a bit sequence is 1100100.Hence r must be of length 7.

17. a. Consider a bit sequence where 1 means a particularinteger is included and 0 means a particular integer is excluded. Writing the nine integers in order,123456789, a nine-bit sequence will give a particularsubset from the nine integers. For example100011101 will be the subset {1, 5, 6, 7, 9}. Sincethere are 29 bit sequences, there will be 29 subsets.Notice that the bit sequence 000000000 representsthe subset where no integer is chosen. This is the nullset which is a subset of all sets.

b. The subsets that contain 1 will be the bit sequencebeginning with 1. Hence the number of subsetscontaining 1 will be 28.

c. The number of subsets that contain 1 or 2 will beequivalent to the bit sequence starting with 10, 01,and 11; i.e., a subset containing 1 but not 2,containing 2 but not 1, and containing both 1 and 2.The number of bit sequences starting with 10 is 27,similarly for 01 and 11. Hence the number ofsubsets containing 1 or 2 is 3 � 27.

18. The number of possible sequences with the ball beingreplaced after each draw is 496. The number ofsequences with six different ball numbers is P(48, 6).

Now �P(4

499,66)

� � 0.72741.

Hence approximately 72.7% of possible sequenceshave six different ball numbers.

19. The number of sequences of length r with unlimitedrepetition is nr. The number of sequences with allterms different is P(n, r), hence the required fraction is

�P(n

n,r

r)�.

20. The number of sequences of length r 2 with nsymbols where there is no repetition is P(n, r), andwith replacement is nr. The number of sequencescontaining exactly one A, without replacement, isrP(n � 1, r � 1) (the A can be placed in any ofr positions and for each position of A the remainingr � 1 positions can be filled from n � 1 symbols inP(n � 1, r � 1) ways) and with replacement isr · (n � 1)r – 1. Hence we are to show that

�rP(n

P�

(n1,,rr)� 1)

� > �r · (n �

nr1)r – 1

�.

L.S. � r · �((nn

�

�

1r))!!

� · �(n �

n!r)!

�

� �r(n

n�

!1)!

�

� �nr

�

R.S. �

� �nr

� · ��n �

n1

�� n �

n1

�� … ��n �

n1

��to r � 1 factors of �

n �

n1

�.

Since �n �

n1

� < 1, the R.S. < �nr

�

Therefore�rP(n

P�

(n1,,rr)� 1)

� > �r(n �

nr1)r – 1

�.

21. If the length of the sequence is even, then the first

�2r

� positions can be filled in n�2r

�ways. The remaining

�2r

� positions must match with the first, hence can be

filled in one way. Therefore there are n�2r

�palindromes if

r is even.

If the length of the sequence is odd, the first �r �

21

�

positions can be filled in n�r +

21

�ways and the remaining

r � 1 positions can be filled in one way. Hence there

are n�r +

21

�palindromes if r is odd.

r (n � 1)(n � 1) … (n � 1) to (r � 1) factors��

n(n)(n)(n) … (n) to r factors


22. The number of sequences of length 1 will be n, of

length 2 will be n2, of length 3 will be n3, … and of

length r � 1 will be nr – 1. The number of sequences

of length r is nr. Now the number of sequences of

length less than r is

n � n2 � n3 � … � nr – 1 � �n(n

n

r –

�

1 �

11)

�.

�n(n

n

r –

�

1 �

11)

� � �n �

11

� (nr � n) < nr for n 2.

Hence n � n2 � n3 � … � nr – 1 < nr and the totalnumber of sequences with length less than r is smallerthan the number of sequences of length r.

23. If the number of elements in T is n(T) � n and thenumber of elements in S is n(S) � r then the numberof sequences of length r from n elements is nr. Hencethe number of functions that can be constructed froma set S to a set T is (n(T))n(S).

24. a. Each sequence defines a unique subset and eachsubset is defined by a sequence, hence there is aone-to-one correspondence between the possiblesubsets and sequences of length 6.

b. The number of sets that can be formed is 26 � 64.

Exercise 11.3

5. a. The number of sequences of length 4 from 10digits is P(10, 4) � 5040.

b. The number of subsets of size 4 will be � � and

for each of the subsets there will be 4! sequences.

c. From part a, the number of sequences is P(10, 4),which is equal to the number of sequences from

part b, hence � � � 4! � P(10, 4)

and � � � �P(1

40!, 4)�.

104

104

104


6. The two girls can be selected in � � ways, and

for each of these, the two boys can be selected in

� � ways. Hence there are � � � � � � 2970

ways of selecting two girls and two boys.

7. Since A and Z are included, we must select 4 moreletters from the remaining 26 � 2 � 24. Hence there

will be � � � 10626 subsets of 6 letters so that

A and Z are included.

8. � � represents the number of subsets of length r.

P(n, r) represents the number of sequences of lengthr. The elements in each subset can be arranged in r!

sequences, hence P(n, r) � �. Equality holds for

r � 1.

9. a. The number of subsets of:

i) 2 blocks chosen from the 12 is n(u) � � � � 66.

ii) 2 red chosen from 3 is n(A) � � � � 3.

iii) 1 red and 1 yellow is n(B) � � � � � � � 12.

iv) 2 blocks of the same colour could be 2 red,

� �, 2 blue, � �, or 2 yellow, � �.

Therefore n(C) � � � � � � � � � � 19.

b. The number of subsets of a different colour will be66 � 19 � 47.

10. a. The number of committees will be � � � 462.

b. The three women can be selected in � � ways and,

for each of these, the two men can be selected in

� � ways, thus forming � � � � � � 200 committees.52

63

52

63

115

42

52

32

42

52

32

41

31

32

122

nr

nr

244

102

122

102

122

13. The number of samples of five components from a

box of 100 is � �. If three are defective, the number

of samples of five where no component is defective

will be � � hence the number of samples that

contain at least one defective component is

� � � � �. The fraction that contains at least one

defective is � � � � �

� 1 � �9927!5!!

� � �9150!05!!

�

� �= 1 � �

19050··9949··9938

�

= �342635470

� ~ 0.144.

14. The two odd digits can be selected in � � � 10 ways,

and for each of these the two even digits can be

selected in � � � 6 ways, giving 60 subsets with

two odd digits and two even digits. The digits in eachsubset can be arranged to form 4! sequences, hencethere are 60 � 4! � 1440 sequences of length 4.

15. a. The subsets with at least two red blocks: {R1, R2,

A; R1, R2, B; R1, R2, C; R1, R3, A; R1, R3, B;

R1, R3, C; R2, R3, A; R2, R3, B; R2, R3, C; R1,

R2, R3}. There are 10 such subsets.

b. The � � ways of selecting the red blocks gives the

subset {R1, R2; R1, R3; R2, R3}. The � � ways of

selecting the remaining block gives either a red,

A, B, or C. If the first two blocks are R1, R2, the

third block could be R3; if the first two blocks

were R1, R3, the third could be R2; and if the first

two blocks were R2, R3, the third could be R1,

hence the set of blocks R1, R2, R3, was counted

3 times.

41

32

42

52

1005

975

1005

975

1005

975

1005

c. The number of committees with no men is � � � 6.

The number of committees with no women is

� � � 1. Therefore the number of committees with

at least one man and one woman is 462 � 6 � 1� 455.

d. The number of committees that have both Ron and

Enzo will be � � � 84. (From the remaining nine

people, select three to join Ron and Enzo.) Hencethe number of possible committees where Ron and Enzo are not on the same committee is 462 � 84 � 378.

11. a. The number of subsets of six balls selected from

the 49 is � �.

b. If 49 is to be included, then five balls must be

selected from the remaining 48, giving � �subsets.

c. There are 24 even-numbered balls, hence there are

� � subsets having only even-numbered balls.

d. From 24 even-numbered balls and 25 odd-

numbered balls, there will be � � � � � subsets

having three even- and three odd-numbered balls.

12. a. The number of different samples of 60 studentsselected from a school population of 1200 students

will be � �.

b. The number of samples of 15 students from 300

students in Grade 9 is � �. Similarly the number

of samples of 15 students from Grades 10, 11, and

12 will also be � �. Hence the total number of

subsets is �� 4.

c. The 60 Grade 12 students can be chosen in

� � ways.30060

30015

30015

30015

120060

253

243

246

485

496

93

55

65


16. a. The number of different samples of five light bulbs

from a carton of 100 bulbs is � �.

b. Since there are 97 with no defects, the number of

samples with no defects is � �.

c. Selecting one defective bulb in � � ways and four

non-defective bulbs in � � ways gives � �� samples with exactly one defective bulb.

d. Selecting two defective bulbs in � � ways and

three non-defective bulbs in � � ways gives

� �� samples with exactly two defective bulbs.

e. Selecting three defective bulbs in � � ways and two

non-defective bulbs in � � ways gives � �� samples with exactly three defective bulbs.

f. The number of samples with no defective bulbs is

� � � � �� . The � � subsets includes the

samples of five bulbs with 0, 1, 2, or 3 defects, hence

� � � � ��

� �� 17. a. A regular n-gon has n verticies. The number of line

segments that join 2 vertices is � �. Since there are

n sides, the number of diagonals will be

� � � n � �(n �

n!2)!2!� � n

= �n(n

2� 1)� � n

= �n2 � n

2� 2n�

= �n(n

2� 3)�.

b. If n is odd, no diagonal passes through the centre

of the n-gon. If n is even, then �n2

� diagonals pass

through the centre.

18. a. Since each group must have at least one star, thereare 10 places between the stars in which to place

the two vertical bars. This can be done in � � �

45 ways. Hence the three groups can be formed in45 ways.

b. Let x represent the number of stars in the firstgroup, y the number of stars in the second group,and z the number of stars in the third group. In theexample, x � 2, y � 5, z � 4, hence x � y � z � 11. This representation will give 45 solutions to theequation where x, y, and z are positive integers.

19. a. Consider 21 stars. Insert three vertical bars in 20spaces to give four groups. The number of stars ineach group, from left to right, will give the values

of x, y, z, and w. The number of groups is � � �

1140, hence there are 1140 solution to the equationwhere x, y, z, and w are positive integers.

b. If x � 1, 20 stars are to be divided into three groups

by inserting two vertical bars in 19 spaces in � �ways, assigning positive integer values to y, z, and w.

Similarly, if x � 2 there will be � � solutions and

if x � 3 there will be � � solutions. The number of

different solutions is � � � � � � � � � 460.

c. If x � y � 1, then z � w � 19 and the 19 stars must be divided into two groups to assign values to z

and w. This can be done in � � ways, hence there

are 18 solutions when x � y � 1.Similarly, when x � y � 2, z � w � 17 and thereare 16 solutions.When x � y � 3, z � w � 15 and there are 14solutions, etc., till x � y � 9 and z � w � 3 andthere are 2 solutions.Therefore the number of solutions that have x � ywill be 18 � 16 � 14 � 12 � … � 2 � 90.

181

172

182

192

172

182

192

208

102

n2

n2

33

972

32

973

31

974

30

975

1005

1005

30

975

975

972

33

972

33

973

32

974

32

974

31

974

31

975

1005


20. If x � y � z � 11, x 0, y 0, z � 0 implies thatsome of the groups will be empty. This can beaccommodated by adding three stars, dividing the 14 stars into three non-zero groups and, once values have been assigned to x, y, and z, decrease each oneby 1 (i.e., remove one star from each group). E.g. xxxx�x�xxxxxxxxx, here x � 4, y � 1, z � 9.Removimg one star from each group assigns x � 3,y � 0, z � 8, which is a solution to x � y � z � 11.

Hence there are � � solutions where x, y, and z are

non-negative integers. The number of solutions to

x1

� y1

� z1

� 14 is � � � 78.

21. Solving x � y � z � w � 21 for positive integers,we divide 21 stars into four groups by inserting threevertical bars in the spaces between the stars. Toaccommodate x �2 allows for three additionalvalues of x, namely �2, �1, and 0, hence add threestars and once values have been assigned to x, add �3to the assigned value. Similarly for y �1, add twostars and then add �2 to the value of y and for z 0,add one star then add �1 to the value of z. Thenumber of solutions will be equivalent to the numberof ways of inserting three vertical bars amongst the

27 stars (21 � 3 � 2 � 1) in � � � 2600 ways.

Hence there are 2600 solutions.

Exercise 11.4

3. a. MISSISSAUGA has 11 letters, of which there are 4 Ss, 2 Is, 2 As, 1 M, 1 G, and 1 U. The number of anagrams is the number of arrangements of the 11 letters of which there are four alike of onekind, two alike of another kind, and two alike of a third kind. Hence the number of anagrams is

�4!

121!!2!

� � 415 800.

b. Starting with an S, the number of anagrams will be the number of arrangements of the remaining

10 letters in �3!

120!!2!

� � 151 200.

c. Start and end with an S leaves 2 Ss, 2 Is, 2 As, 1 M,

1 G, and 1 U, giving �2!

92!!2!� � 45 360 anagrams.

d. Consider the two Is as one, say an X. Now the 4 Ss, 2 As, 1 M, 1 G, 1 U, and 1 X can be arranged in

�41!02!!

� � 75 600 ways. Now in each of these ways

replace the X with II, giving anagrams with the Istogether.

e. The number of anagrams starting with an A is

�41!02!!

�, an I is �41!02!!

�, and with a U is �4!

120!!2!

�. Hence

the number of anagrams starting with a vowel is 75 600 � 75 600 � 37 800 � 189 000.

4. The 12 bulbs, two bulbs of six different colours can be

arranged in �(122!)!6� � 7 484 400.

5. The number of possible outcomes of eight flips of acoin is 28. The number of outcomes with four heads

and four tails is �48!4!!

� � 70. In �7208� � �

13258

� of the

possible outcomes of the flips of eight coins result inexactly four heads.

6. The number of binary sequences of length 5 is 25 � 32. The number of sequences with no 1s is 1

and with one 1 is � � � 5. Hence the number of

sequences with two or more ones is 32 � 5 � 1 � 26.

7. In the first row, the two pines, two cedars, and one

spruce can be arranged in �25!2!!

� � 15 ways. In the

second row the trees can be arranged in 15 ways.Hence the trees can be planted in two rows in 15 � 15 � 225 ways.

8. DESCARTES has nine letters with 2 Ss, 2 Es, and theother 5 letters all different. The number of anagramsending in S will be the number of arrangements of 8letters of which 2 are Es and the other 7 are different.

This can be done in �82!!� � 20 160 ways.

5126

3

132

132


9. a. Since each switch can be in any one of twopositions, on or off, there are 27 � 128configurations.

b. From the seven switches, select four that can be

turned on; this can be done in � � ways, hence 35

configurations have four switches that are on.

c. There is one configuration with zero switches onand seven with one switch on, hence there are 128 � 8 � 120 configurations with at least twoswitches turned on.

d. Three of switches 2 through 7 must be turned on.

This can be done in � � ways, hence there are 20

configurations with exactly four switches on,including switch 1.

10. The number of strings with three 1s in the first fiveterms and three 1s in the last five terms is

� � · � � � 100.

The number of bit strings of length 10 with exactly

six 1s is � � � 210. The required fraction is

�120000

� � �1201�.

11. a. For each digit there are two choices. Either it isincluded within the subset or it is not included.Hence the number of subsets will be 210 � 1024.

b. The number of subsets that can be formed from thedigits 0, 1, 2, 3, 4, 5, 6 is 27 � 128.

c. The number of subsets that do not contain 0 or 9 is28. Hence the number of subsets containing 0 or 9is 210 � 28 � 768.

12. a. We choose four persons to receive treatment A in

� � ways. We then choose four of the remaining

12 to receive treatment B in � � ways, then

four to receive treatment C in � � ways, and the

final four to receive treatment D in � � ways. The

number of different orders is � ��

1126!4!!

� �81!24!!

� �48!4!!

� �04!4!!

�

� �(146!)!4�

� 63 063 000.

b. Each of four groups is defined. Within group 1 thefour treatments can be administered in 4! ways.This is also true for each of the other groups. Thetotal number of assignments is (4!)4 � 331 776.

13. a. For a palindrome the only position for a non-repeated letter is in the middle. But there are threenon-repeated letters, so there are no palindromes.

b. In MISSISSIPPI we have one M, four Is, four Ssand two Ps. If the M is placed in the middleposition, there must be five letters before it, ofwhich two are I, two are S, and one is P. This can

be done in �25!2!!

� � 30 ways. The last five must

then be in the reverse order of these. There are 30 palindromes.

14. Since the dice are different, there are 66 possibleoutcomes.

a. If there are no repeated values, there are 6!outcomes possible. The number of outcomes withat least one repeated value is 66 � 6! The fraction

of outcomes this gives is �66 �

666!

� � 0.9846.

b. Two 2s, two 4s, and two 6s can occur in �2!

62!!2!�

ways. As a fraction of all possible this is

�23

6�

!66� � 0.0019.

c. Three dice to have odd values can be chosen in

� � ways. Now, each of these has three choices,

1, 3, or 5. Similarly, each of the remaining threedice have three choices, 2, 4, or 6. Then the

number of sequences is � � 36. As a fraction of all

sequences, this is � � 36

� �156�.

66

63

63

63

44

84

124

164

44

84

124

164

106

53

53

63

74


15. a. Since each position has two possible entries, thereare 210 possible sequences.

b. For any value of r, choose r positions in which toplace 1s. The remaining positions then have 0s.

This can be done in � � ways.

c. Since 0 � r � 10, by combining the first two parts

we obtain � � � � � � … � � � � 210.

d. L.S. � 1 � 10 � 45 � 120 � 210 � 252 � 210 � 120 � 45 � 10 � 1

� 1024� 210.

17. b. The expression represents the number of sequencesof 50 items of which 20 are alike of one kind, 15are alike of a second kind, and 15 are alike of athird kind. Since the number of such sequences iscreatable, the number is a positive integer.

18. The number of strings with no restriction is �58!3!!

� � 56.

If there are no consecutive 1s, there are six positionsrelative to the five 0s in which a single 1 can be

placed. There are � � � 20 ways of doing this. The

number of strings having at least two consecutive 1s is56 � 20 � 36.

19. The number of bit strings of length 8 is 28 � 256. We count the number of strings with no consecutive 1s.If there are eight 0s, there is 1 string.

If there are seven 0s and one 1, there are � �� 8 strings.

If there are six 0s and two 1s, there are � �� 21 strings.

If there are five 0s and three 1s, there are � �� 20 strings.

If there are four 0s and four 1s, there are � �� 5 strings.

If there are three or fewer 0s, all strings haveconsecutive 1s. The number of strings with at leastone pair of consecutive 1s is 256 � (1 � 8 � 21 � 20 � 5) � 201.

20. The number of sequences without restriction is

�4!

130!!3!

� � 4200.

The as and bs can be arranged in �47!3!!

� � 35 ways.

For each of these, there are eight positions in which

the three cs can be placed. This can be done in

� � � 56 ways. The number of sequences with at least

one pair of consecutive cs is 4200 � 35 � 56 �

2240. As a fraction, this is �24224000

� � �185�.

21. Any walk can be described by five Rs and three Us.

The number of sequences is �58!3!!

� � 56.

22. a. Since a random walk is represented by Es and Vs,we are creating sequences using two symbols. Thisis synonymous with binary sequences.

b. The number of paths ending at (20, 12) is

�20

3!212!� � � �.

c. If the paths pass through (10, 10) we calculate thenumber of paths to (10, 10) and the number from(10, 10) to (20, 12). The number of paths is

� �� .

d. The number passing through (8, 4) and (12, 8) is

� �� .

23. a. Choose three of 13 positions in which to place

the Cs. This can be done in � �ways. The

remaining 10 positions now are filled with six As inthe first six positions and the last four with Bs.

The number of arrangements is � � � 286.

b. Again, place the Cs in � � ways. The first empty

spot must then be filled with A. There are now ninespots in which to put the remaining five As. This

can be done in � � ways. The number of

arrangements is � �� 36 036.95

133

95

133

133

133

124

84

124

122

2010

3212

83

54

63

72

81

63

1010

101

100

10r


c. Position C in space 1 and then choose two other

spaces for the other two Cs in � � ways. Now put

an A in the first empty spot and choose five spaces

for the remaining As in � � ways. The number of

arrangement is � �� 8316.

24. a. The possible end positions are (12, 0), (12, 2),(12, 4), (12, 6), (12, 8), (12, 10), (12, 12).

b. Each move can be labelled RU, right and up, orRD, right and down. By matching RU with 0 andRD with 1, we represent each possible path as abinary sequence of 0s and 1s.

c. If the path ends at (12, 0), there must be six 0s and

six 1s, and the number of paths is � � � 924.

Exercise 11.5

1. With no restriction there are 26k words. If Z isexcluded, there are 25k words. The number of wordsthat contain Z is 26k � 25k.

2. If the subset is of length k, possible final integers inthe subset are k, k � 1, k � 2, k � 3, … , 100. Thenumber of final integers is 100 � (k � 1) � 101 � k,and this is the number of subsets.

3. For 1 � L � 4, if the largest element is L we have L � 1 integers from which to choose 4. This can be

done in � � ways.

4. In the set there are n � 1 odd numbers and n evenones. From these, o odd numbers can be chosen in

� � ways, and e even numbers can be

chosen in � � ways. The number of subsets is

� �� .ne

n � 1o

ne

n � 1o

L � 14

126

95

122

95

122


7. If the sequence begins with 1, there are � �sequences possible, some of which also end with 1.

If the sequence ends with 1, there are � �sequences possible, some of which start with 1. If the

sequence begins and ends with 1, there are � �sequences possible. The number of sequences possible

is 2� � � � �.

We can also consider three cases: the string beingswith 1 and ends in 0, it begins in 0 and ends in 1, andit begins and ends with 1. Then the number of strings

is 2� � � � �. Show that these results are

identical.

8. There are two ways of arranging 1 and n. If they areseparated by one term, the first of the two can be inany position from 1, 2, 3, … , n � 2. There are n � 2possibilities.The remaining elements can then fill the other n � 2positions. The number of sequences is 2(n � 2)(n � 2)!.

9. The number of sequences of length r containing 1 and

2 is � �.

The r elements can be arranged in r! ways and inexactly half of them 1 precedes 2. The number of

sequences is �r2!�� .

10. The total number of sequences of 2n elements with

two alike of n types is �((22n!))n!

� � �(2

2nn)!

�.

Of these, exactly half will have 1 before the first 2,

since the likelihood of a 1 before a 2 is �12

�. The

number of sequences is �(22nn+)1!

�.

998r � 2

998r � 2

n � 2k � 2

n � 2k � 1

n � 2k � 2

n � 1k � 1

n � 2k � 2

n � 1k � 1

n � kk � 1

Alternate solution:

Choose four positions from 2n in � � ways.

Place a 1 in the first of these. The remaining 1 andtwo 2s can be placed in the other three positions inthree ways. Now the remaining elements can be

assigned to the open positions in �(2n

2n�

– 24)!

� ways.

The number of sequences is 3� ��(2n2n

�– 2

4)!��

= �(2n

2n�

– 24)!

�

= �(22nn+)1!

�.

11. Arrange the Bs and Cs in �(b

b�

!c!c)!

� ways. There

are now b � c � 1 positions in which to place As.

This can be done in � � ways. The number

of arrangements is �(b

b�

!c!c)!

� � �.

Review Exercise

5. a. There are �5!

130!!2!

� � 2520 arrangements.

b. There are 3! � 6 arrangements.

c. Placing a spruce at either end leaves five cedars

and three pines. They can be arranged in

�58!3!!

� � 56 ways.

d. If the same species is at both ends, there are 56

arrangements if they are spruce, �58!2!!

� � 168 if they

are pine, and �3!

83!!2!� � 560 if they are cedar. The

total number of arrangements is 784.

e. The cedars and pines can be ordered in �58!3!!

� � 56

ways. There are now nine positions from which

to choose two for the spruce. The number of

arrangements is 56� � � 2016.

f. If the row starts with a spruce, there are �59!3!!

�

arrangements. If the row ends with a spruce, there

are �59!3!!

� arrangements. If the row starts and ends

with a spruce, there are �58!3!!

� arrangements. By the

principle of inclusion-exclusion, the number of

arrangements is 2 �59!3!!

� � �58!3!!

� � 952.

g. The pines and spruce can be arranged in �35!2!!

� � 10

ways. There are now six positions in which to plant

the cedars. This can be done in � � � 6 ways. The

number of arrangements is 60.

7. a. Case 1 includes the possibility that the sequenceends with a vowel and Case 2 includes thepossibility that the sequence begins with a vowel.Double counting of those sequences that start and end with a vowel is also possible. The number of such sequences is 2 � 4 � 3 � 1 � 24. The correct number of sequences is 2 � 120 � 24� 216.

b. There is duplication of counting. If a is selected asthe vowel, then selecting e as one of the three

chosen in � � means that a and e are both in the

sequence. But selecting e as the vowel allows forthe possibility that a is selected later. Hence thissolution includes sequences with both a and eincluded twice. Sequences including both a and e

number � �4! � 144. The correct number of

sequences is 480 � 144 � 336.

8. a. If the word begins with a, there are 4 � 3 � 2 � 1� 24 possibilities. If the word begins with b, thereare 24 possibilities. If the word begins with c, thesecond letter can be a, b, or d, and there are 3 �3 � 2 � 1 � 18 possibilities. There are 66 wordsbefore those beginning with ce. The first wordbeginning ce is ceabd, and the second is ceadb.Then there are 67 words before ceadb.

42

53

65

92

b � c � 1a

b � c � 1a

3(2n)(2n � 1)(2n � 2)(2n � 3)��

4 · 3 · 2 · 1

2n4

2n4


b. Following adcbe only adceb, adebc, and adecbbegin with ad. There are three words beginning ad.If the word begins ae there are 3 � 2 � 1 � 6words. If the word begins with b or c there are 4 � 3 � 2 � 1 � 24 words. If the word beginswith da, the first word is dabce, the second isdabec, and the third is dacbe. The number of wordsbetween adcbe and dacbe is 3 � 6 � 2 � 24 � 2� 59.

c. Since there are 24 words beginning with a, 24beginning with b, 6 beginning ca, and 6 beginningcb, the 61st word is cdabe.

9. a. If A and B are included, we select r � 2 elements

from n � 2 in � � ways.

b. If A is included there are � � subsets,

and if B is included there is an equal number. If

both are included there are � �. The number of

subsets is 2� � � � �.

c. If neither A nor B is included there are � �subsets.

10. a. If the sequence contains A and B, there are

� � ways of choosing the remaining elements

and r! ways of ordering the subset. Of these, halfwill have A before B. The number of sequences is

�12

�� r!.

b. Choose three positions in � � ways and fill them

with A, B, and C in order. Now choose r � 3

elements from n � 3 in � � ways and order

them in (r � 3)! ways. The number of sequences is

� �� (r � 3)!

� � ��3r!!�.

c. This is exactly as the previous case, allowing for A and B to be interchanges. The number of

sequences is � ��r3!�.

Chapter 11 Test

1. P(10, 3) � � � � 10 · 9 · 3 � �10 ·

39!

· 8�

� 600.

2. a. There are four choices for the first digit, then 8, 7, 6,and 5 for the next four. There are 4 · 8 · 7 · 6 · 5 � 6720 numbers.

b. There are 4 · 3 · 7 · 6 · 5 � 2520 numbers.

c. Three even digits can be chosen in � � ways.

Two odd digits can be chosen in � � ways. The five

chosen digits can be permuted in 5! ways. There

are � �� 5! � 4800 numbers.

3. P(n, r) is the number of sequences of length r that canbe formed using n distinct symbols. Any subset of r

symbols can be arranged in r! ways. Then if � � is the

number of subsets of r distinct symbols chosen from r,

� �r! � P(n, r)

or � � � �P(n

r!, r)�.

4. a. Four people can be chosen from 12 in � �ways �� 495�.

b. If Bob and Mary are included, then two of theremaining 10 must be chosen. This can be done in

� � ways. �� 45�.

c. If Bob and Mary will not serve together, the numberof committees is determined by the number withoutrestriction less the number with both included. The

number of committees is � � � � � �450�.

5. There are five sets of four questions; from each ofwhich we are to select two. Two of four questions

with answer A can be chosen in � � � 6 ways. The

total number of sets, without ordering the sets, is 65 � 7776.

42

102

124

102

102

124

124

nr

nr

nr

52

43

52

43

103

n � 3r � 3

n � 3r � 3

n � 3r � 3

r3

n � 3r � 3

r3

n � 2r � 2

n � 2r � 2

n � 2r

n � 2r � 2

n � 1r � 1

n � 2r � 2

n � 1r � 1

n � 2r � 2


6. The number of orderings with no restriction is �27!3!!

�.

If the Ts are together, consider them as one letter,so there are now six letters with three alike. Hence

there are �63!!� orderings of these. The number of words

with the Ts separated by at least one letter is

�27!3!!

� � �63!!� � 300.

7. There are 2n sequences possible with no restriction.There are four first-last possibilities, 10, 01, 00, and 11. Three of the four are acceptable. Then the number of

sequences is �34

� 2n � 3 · 2n – 2.

8. There are 36 symbols. If there is no restriction, thenumber of passwords of length 6, 7, or 8 is

366 � 367 � 368

� 366 (1 � 36 � 362)� 1333 · 366. The number of passwords with no letter is 106 � 107 � 108 � 111 · 106. The number of passwords with no digit is 266 � 267 � 268 � 703 · 266. The number of pass-words with at least one letter and at least one digit is 1333 � 366 � 111 � 106 � 703 � 266.

9. The number of subsets is

� � � .

The number of subsets containing A is

� � � .

The fraction of sets containing A is �Nr�.

(N � 1)(N � 2) … (N � r � 1)��

(r � 1)!N � 1r � 1

N(N � 1)(N � 2) … (N � r � 1)��

r!Nr



Beginning Exercise

6. If a sequence is both arithmetic and geometric it canbe represented by a, a � d, a � 2d, … and a, ar, ar2,… where a � d � ar.

Then d � a (r � 1).Also a � 2d � ar2,

so a � 2d (r � 1) � ar2

2ar � a � ar2

r2 � 2r � 1 � 0 since a ≠ 0r � 1.

The only sequence possible has r � 1 and d � 0. Thisis the sequence a, a, a, a, … .

7. a. t1

� 3, t2

� 6, t3

= 10, t4

� 15, t5

� 21. It is not

arithmetic.

b. In an

� tn + 1

� tn

we obtain a1

� 3, a2

� 4,

a3

� 5, a4

� 6.

In general, an

� tn+1

� tn

= � � � � �= �

(n � 3)2(n � 2)� � �

(n � 2)2(n � 1)�

= �n �

22

�.

This is a linear function so the sequence isarithmetic.

8. a1

� log g1

� log a

a2

� log g2

� log a � log r

a3

� log g3

� log a � 2 log r...a

n� log g

n� log a � (n � 1) log r

an + 1

� log gn + 1

� log a � n log r

Then an + 1

� an

� log r, which is a constant.

Then {an} is an arithmetic sequence with first term

log a and common difference log r.

Exercise 12.1

3. p10

� 2000

p11

� 2000 (1.05) � 2000 � p10

(1.05) � 2000

p12

� (p10

(1.05) � 2000) 1.05 � 2000

� p10

(1.05)2 � p10

(1.05) � 2000

p18

� p10

(1.05)8 � p10

(1.05)7 � … � 2000

� 2000 ��(11..0055)9

�

�

11

�� $22 053.13

4. An arithmetic sequence with first term a, commondifference d, and having 20 terms is given by a, a � d,a � 2d, … , a � 17d, a � 18d, a � 19d.Grouping the first and last terms, second and secondlast terms, and so on, and summing, gives (a � a � 19d) � (a � d � a � 18d) �

(a � 2d � a � 17d) � …= 10 (2a � 19d)

= 20 ��2a �

219d��.

5. Let the arithmetic sequence be 1, 1 � d, 1 � 2d, … ,and the geometric sequence be 1, r, r2, … . Then 1 � d � r because the second terms are equal.Since the third terms are also equal1 � 2d � r2

� (1 � d)2.Then d � 0, so r � 1.Then each sequence is 1, 1, 1, 1, … .

6. a. If the new sequence is the square of the given then

gn

� (3.2n – 1)2

� 9.4n – 1.

This is a geometric sequence with a � 9 and r � 4.

b. T1

� 3, T2

� t3

� 3.22 � 3.4

T3

� t5

� 3.24 � 3.42

and Tn

� t2n – 1

� 3.22n – 2 � 3.4n – 1.

This is a geometric sequence with a � 3 and r � 4.

n � 2n

n � 3n � 1

Chapter 12: Sequences 175

Chapter 12 • Sequences

c. h1

� 3, h2

� t2

� t1

� 9 � 2 � 18,

h3

� t3

� t2

� (3.22)(3.2) � 72

hn – 1

� tn – 1

� tn – 2

� (3.2n – 2)(3.2n – 3)

� 9.22n – 5

hn

� tn

� tn – 1

� (3.2n – 1)(3.2n – 2)

� 9.22n – 3.

Then hn

� 4hn – 1

.

This sequence is not geometric since h2

≠ 4h1.

7. This is the sequence from question 6 c. Since g2

� 18

and gn

� 4gn – 1

, g1

� �92

�.

9. a. At the end of the first month the amount owning is$5000 (1.0075) � 100 � $4937.50.

c. Setting a spreadsheet as indicated,

A

1 = 5000 (1.0075) � 100

2 = A1

(1.0075) � 100

entry 62 has an amount $89.55. Paying off the debtrequires 63 months.

10. The first seven rows are shown. Numbers satisfying a

n� 3n � 1 are circled.

k � 1: 1 2 3 4 5 6 7 8 9 10 r1

� 3

k � 2: 11 12 13 14 15 16 17 18 19 20 r2

� 3

k � 3: 21 22 23 24 25 26 27 28 29 30 r3

� 3

k � 4: 31 32 33 34 35 36 37 38 39 40 r4

� 4

k � 5: 41 42 43 44 45 46 47 48 49 50 r5

� 3

k � 6: 51 52 53 54 55 56 57 58 59 60 r6

� 3

k � 7: 61 62 63 64 65 66 67 68 69 70 r7

� 4

We conclude that r1

� 3, r3k

� 3, r3k – 1

� 3, r3k+1

� 4,for k � 1.

11. Since a1

� a and a1

� a2

� 5, then

a2

� 5 � a.

Since a2

� a3

� 5,

a3

� 5 � a2

� a.

Then the sequence is a, (5 � a), a, (5 � a), … .

The general term an

is an

� �a if n � 2k

5 � a if n � 2k � 1

.

12. a. The first four points in the sequence are (1,0),(2,2), (3,4), and (4,6).

b. Since Pn + 1

� Pn

� (1,2), the slope of the linesegment connecting two adjacent points is aconstant 2. Hence the sequence defines points on a straight line with slope 2 and passing throughpoint (1,0).

13. a. f(x) � 2x (1 � x)f (0.5) � 0.5 for all n � 2.

b. x1

� 0.3

x2

� f(x1) � 0.6(0.7) � 0.42

x3

� f(x2) � 0.84(0.58) � 0.4872

x4

� f(x3) � 0.9744(0.5128) � 0.499 672 32

x5

� 0.499 999 785 3.

As n increases xn

→ 0.5.

c. For any x such that 0 < x1

< 1, xn

→ 0.5.

d. For x1

� 0.5, x6

� 0.730 959 919 5,

x7

� 0.589 972 546 8,

x8

� 0.725 714 822 4,

x9

� 0.597 158 456 9

x10

� 0.721 680 702 7,

x11

� 0.602 572 998 2.

It appears that values oscillate from 0.72 to 0.60.

14. a. s1

� 2, s2

� 4, s3

� 7, s4

� 11.

c. Since sn

� sn – 1

� n, n � 2, we can write

sn

� sn – 1

� n.

Then s1

� 2

s2

� s1

� 2

s3

� s2

� 3

s4

� s3

� 4

...

sn

� sn – 1

� n.

Adding sn

� 2 � 2 � 3 � 4 �… � n

� (1 � 2 � 3 � 4 � … � n) � 1

� �n(n

2� 1)� � 1

� �n2 �

2n � 2�, n � 2.

176 Chapter 12: Sequences

15. a. fn

(x) � g[fn – 1

(x)]

f1

(x) � �1 �

xx

�

f2

(x) � g[f1(x)] � g��1 �

xx

��

�

� �1 �

x2x

�

f3

(x) � �1 �

x3x

�

It appears that fn

(x) � �1 �

xnx

�.

To show that it is, we show that if fn

(x) � �1 �

xnx

�

then fn + 1

(x) � �1 � (n

x� 1)x�

fn + 1

(x) �

�

��1 � (n

x� 1)x�.

We can now be sure that fn

(x) � �1 �

xnx

�.

b.

c. If terms are constant, then fn

(x0) � f

n + 2(x

0) for

n � 1, 2, 3, … .

Then �1 �

x0

nx0

� � �1 � (n

x0

� 1)x0

�

1 � nx0

� 1 � (n � 1)x0

x0

� 0.

For x0

� 0 we have the sequence 0, 0, 0, … .

d. If terms alternate in value, then f1

(x1) � f

3(x

1) and

f2

(x1) � f

4(x

1).

Then �1 �

xx

� � �1 �

x3x

�

and x � 0.But for x � 0 all terms are 0 and there is noalternating of values.

16. a. a1

� 1, a2

� 3, a3

� 9, a4

� 27.

b. For each triangle removed, three new triangles areformed from each triangle at the previous stage.

c. a1

� 30 � 1

a2

� 31 � 3.

We assume that an – 1

� 3n – 2 and use the statement

from part b to see that an

� 3an – 1

� 3n – 1.

d. If a side of the original triangle is 1, the length of a

side of the first set of smaller triangles is �12

�.

Then b1

� 1

b2

� �12

� b1

and in general bn

� �12

� bn – 1

.

e. Operating repeatedly, we obtain bn

� ��12

��n – 1.

f. Let the original area A1

be A.

Then A1

� A

A2

� �34

�A

A2

� �34

� A2

� ��34

��2A

and An

� ��34

��n – 1A.

As n becomes large, An

→ 0.

y

x

y = x

x = –1

4321

–1– 2– 3– 4

1 2 3 4–4 –3 –2 –1

y = 1

�1 �

xnx

�

��1 �

1(�

n �

nx1)x

�

�1 �

xnx

�

��1 � �

1 �

xnx

�

�1 �

xx

�

��1 � �

1 �

xx

�


17. Let the area of D1

be A(D1).

At stage 2 three new triangles are formed,

each �19

� A(D1) in area.

Hence A(D2) � A(D

1) � �

13

� A(D1) � A(D

1) �1 � �

13

��.

At stage 3 12 new triangles are formed, each with

area �19

� A(D2) � �

811� A(D

1).

Then A(D3) � A(D

1) �1 � �

13

� � �247��.

Then by extension,

A(Dn) � A(D

1) �1 � �

13

� � �247� � �

43

2

5� � … � �34.9

n

n

–

–

2

2��

� 1� �13

� �1 � �49

� � ��49

��2� …�.

As n → ∞, A(Dn) � A(D

1)�1 � �

� �53

� A(D1).

Let the perimeter of D1

be P(D1) � 3.

At stage 2 we eliminate the equivalent of one side butadd the equivalent of two sides.

Then P(D2) � P(D

1) � �

13

� P(D1) � P(D

1) �1 � �

13

�� 3��

43

��.

At stage 3 we eliminate 12 pieces and add

24 pieces, each �19

� units in length.

Then P(D3) � 4 � �

43

� � �136� � ��

43

��23.

At stage 4 there are 48 lines each �19

� unit in length

from which we delete �13

� of their length and replace by

96 lines each �217� unit in length.

Then P(D4) � ��

43

��23 � �

4287�

� ��43

��33.

By extension, P(Dn) � ��

43

��n – 1 3.

As n → ∞, P(Dn) → ∞ and the perimeter is infinitely

large.

Exercise 12.2

5. a. A � �1100� �

100

i � 1a

i.

c. �100

n � 1b

n� �

100

n � 1(a

n� A)

� �100

n � 1an � 100A

� 0.

6. a. �10

n � 1g

n� 30 � 3 –1 � 3 –2 � … � 3 –9

� 1 � �13

� � �312� � … � �

319�

�

� �32

� �1 � ��13

��10�.

b. �n

j � 1g

j� �

32

� �1 � ��13

��n�.

c. �8

j � 1g

2j – 1� g

1� g

3� g

5� g

7� g

9� g

11�

g13

� g15

� 30 � 3 –2 � 3 –4 � … � 3 –14

� 1 � �312� � �

314� � … � �

3114�

�

� �98

� �1 � ��13

��16�.

1�1 � ��312��8�

��

1 � ��312��

1 �1 � ��13

��10��

1 � �13

��13

�

�

1 � �49

�


7. �60

i � 1g

i� a � ar � ar2 � … � ar59

� a (1 � r � r2 � … � r59).

�120

i � 61g

i� ar60 � ar61 � ar62 � … � ar119

� ar60 (1 � r � r2 � … � r59).

Then � �r160�.

(Note: Do not use sum formula when it is unnecessary.)

8. Sn

� 9 � 99 � 999 � …� (10 � 1) � (100 � 1) � (1000 � 1) � …� (10 � 102 � 103 � … � 10n) �

(1 � 1 � 1 � … � 1)

� �10

1(100�

n �

11)

� � n

� �10 (10n �

91) � 9n�.

Every term in 1, 11, 111, … is the corresponding termin 9, 99, 999, … , divided by 9. Hence

sum is �10(10n �

811) � 9n�.

For k, kk, kkk, we obtain �10 (10n �

811) � 9n� k.

9. Since � � is the number of binary sequences of

length n with exactly k 1s, then

�n

k � 0� � � � � � � � � � … � � � represents the

number of strings of length n with 0, 1, 2, … , n 1s.This is all possible strings. But since every elementcan be either 0 or 1 there are 2n possible strings.

Then �n

k � 0� � � 2k.

10. �n

i � 1(�1)i t

i� �t

1� t

2� t

3� t

4� t

5� t

6� …

� �a � (a � 2d) � (a � 2d) � (a � 3d) � (a � 4d) � (a � 6d) � … .

For n even this gives d � d � d � … � d � �n2

� d.

For n odd this gives � a � d � d � d � … � d

� � a � �(n �

21)

� d.

�n

i � 1� (�1)iS

i� � S

1� S

2� S

3� S

4� S

5� S

6� …

� � a � ar � ar2 + ar3 � …

� �(�a)[

11

�

�(r�r)n]

�.

11. a. For the arithmetic sequence an

� 1 � 3(n � 1)� 3n � 2.

Then �10

i � 1a

i� �

10

i � 1(3i � 2)

� 3 �10

i � 1i � 20

� 3 �10

2· 11� � 20

� 145.For the geometric sequence

�10

i � 1g

i� �

10

i � 12i

� 2 � 22 � 23 � … � 210

� �2 (210

1� 1)� � 2046.

b. We have t1

� a2, t

2� a

4, t

3� a

8, and

tk

� a2k � 3.2k � 2.

Then �10

i � 1ti

� �10

i � 1(3.2i � 2)

� 3 �10

i � 12i � 20

� 6118.

nk

nn

n2

n1

n0

nk

�60

i � 1g

i

�

�120

i � 61g

i


13. Note that �i(i �

11)

� � �1i� � �

i �

11

�.

Then �n

i � 1�i(i �

11)

� � �11

� � �12

� � �12

� � �13

� � �13

� � �14

�

� … � �n�

11

� � �1n

� � �1n

� � �n �

11

�

= 1 � �n �

11

�

= �n �

n1

�.

14. �n

i � 1(2i � 3)2 � 4 �

n

i � 1i2 � 12 �

n

i � 1i � 9n

� 4 �n(n � 1)

6(2n � 1)�� 12 �

n(n2� 1)� � 9n

� .

15. �n

j � 1jt

j� t

1� 2t

2� 3t

3� 4t

4� … � nt

n.

Now S1

� �n

i � 1ti� t

1+ t

2+ t

3� … � t

n

S2

� �n

i � 2ti� t

2� t

3� … � t

n

S3

� �n

i � 3ti� t

3� … � t

n

...

Sn

� �n

i � n

ti� t

n.

Then �n

j � 1S

j � t

1� 2t

2� 3t

3� … � nt

n.

The equality holds.

n (4n2 �12n � 11)��

3


16. For gj� ar j – 1, we have

S � �n

j � 1jg

j� g

1� 2g

2� 3g

3� … � ng

n

S � a � 2ar � 3ar2 � 4ar3 � …

� narn – 1.

Now rS � ar � 2ar2 � 3ar3 � …

� (n � 1)arn – 1 � narn.

Then (1 � r)S � a � ar � ar2 � ar3 � …

� arn – 1 � narn

= �a (

rrn

�

�

11)

� � narn

Then S � �rn�

arn

1� � �

a(r(r

�

n �

1)12)

�.

17. For the sequence defined by tn

� n4, define a new

sequence with general term dn

� tn + 1

� tn.

Then �n

j � 1d

j� t

n + 1 � t

1� (n � 1)4 � 1

� n4 � 4n3 � 6n2 � 4n.

Also dj

� ( j � 1)4 � j4

� 4 j3 � 6 j2 � 4 j � 1.

Thus 4 �n

j � 1j3 � 6 �

n

j � 1j2 � 4 �

n

j � 1j � �

n

j � 11

� n4 � 4n3 � 6n2 � 4n

or 4 �n

j � 1j3 � n4 � 4n3 � 6n2 � 4n

� 6 �n

j � 1j2 � 4 �

n

j � 1j � �

n

j � 11

� n4 � 4n3 � 6n2 � 4n

� 6 �n(n�1)

6(2n�1)� � 4 �

n(n2�1)� � n

� n2 (n � 1)2.

Then �n

j � 1j3 � ��n(n

2�1)��2

.

18. f (x) � �n

i � 1(a

i� b

ix)2

= � �n

i � 1b

i�x2 � 2 � �n

i � 1a

ib

i�x � �n

i � 1a

i2

� 0 since f (x) is the sum of squares.

This is a quadratic inequality and hence D � 0.

Then 4 � �n

i � 1a

ib

i�2

� 4 �n

i � 1a

i2 �

n

i � 1b

i2 � 0

or � �n

i � 1a

ib

i�2

� �n

i � 1a

i2 �

n

i � 1b

i2.

For equality, expand using n � 2 and n � 3 to see

that the result is (a1

b2

� a2

b1)2 � 0 for n � 2 and

(a1

b2

� a2

b1)2 � (a

1b

3� a

3b

1)2 � (a

2b

3� a

3b

2)2

� 0 for n � 3.

In general, one obtains

�n

i, j � 1a

ia

jb

ib

j� �

n

i, j � 1a

i2 b

j2 ,

i ≠ j i ≠ j

which gives �n

i, j � 1(a

ib

j� a

jb

i)2 � 0.

i ≠ j

From this ai� kb

ifor all i. That is, if we consider

vectors A � (a1, a

2, … , a

n) and B � (b

1, b

2, … , b

n),

A � kB.For n � 3, we have, on the left side,

� �3

i � 1a

ib

i�2

� �a1

b1

� a2

b2

� a3

b3�

2� �A · B�

2,

which is the square of the dot product of the vectors.

On the right side,

�3

i � 1a

i2 �

3

i � 1b

i2 � (a

12 � a

22 a

32)(b

12 � b

22 � b

32)

� �A�2 �B�2.

We conclude that A · B � �A� �B�.

Exercise 12.3

2. c. Assume that Sn–1

� �(n�

21)

� �2a � (n�2)d�.

Then Sn

� Sn–1

� tn

= �n�

21

� �2a � (n�2)d� � a � (n�1)d.

= (n�1)a � �n�

21

� (n�2)d � a � (n�1)d

= na � �n�

21

� [(n�2)d � 2d]

= �n2

� �2a � (n�1)d�.

Hence if the formula is correct for n � k�1, then itis also correct for n � k.

4. b. If n � 1, �n(3n

2�1)� � 1. The statement is true for

n � 1.Assume that the statement is true for n � k � 1.Then 1 � 4 � 7 � … � [3(k�1) �2]

� �(k�1)(

23k�4)� is true.

Then 1 � 4 � 7 � … � [3(k�1) �2] � (3k�2)

� �(k�1)(

23k�4)� � (3k�2)

=

= �k (3k

2�1)�.

This is the given statement. Hence if the statementis true for n � k�1 it is also true for n � k.Since the statement is true for n � 1 and when truefor n � k�1 it is also true for n � k, then it is truefor all values of n � 1.

d. If n � 1, the left side is 2 and the right side is

�1(2

3)(3)� � 2.

The statement is true for n � 1.Assume that the statement is true for n � k.Then (1)(2) � (2)(3) � … � (k)(k�1)

� �k(k�1

3)(k�2)�.

3k2 � 7k � 4 � 6 k � 4��

2


Then (1)(2) � (2)(3) � … � (k)(k�1) � (k�1)(k�2)

� �k(k�1

3)(k�2)� � (k�1)(k�2)

� �(k�1)(k�

32)(k�3)�.

For n � k � 1 the given statement has right side

�(k�1)(k�

32)(k�3)�.

Hence if the statement is true for n � k it is also true for n � k � 1.Since the statement is true for n � 1 and if true for n � k it is also true for n � k � 1, then thestatement is true for all values of n � 1 bymathematical induction.

5. If n � 1 we have t1

� �1 � 3

6� 2� � 1, which is

an integer.

Assume that tk

� �k3 � 3

6k2 � 2k� is an integer.

Consider tk + 1

� tk

�

� �k3 � 3

6k2 � 2k�

� �3(k�1

6)(k�2)�.

Since (k�1) and (k�2) are consecutive integers,

3(k�1)(k�2) is a multiple of 6, then tk + 1

� tk

is an

integer and since tk

is an integer, tk + 1

is an integer.

Since the statement is true for n � 1 and if true for

n � k it is also true for n � k�1, then the statement is

true for all values of n � 1.

(Can students prove that tn

is an integer by examining

the expression?)

6. For n � 1 the expression gives a1

� 0, so it is true for

n � 1. (It is sometimes useful to use a second value

as a confidence check. Here a2

� 0 � 12 � 1 by the

recursion and a2

� �2(1

6)(3)� � 1 by the expression.)

(k�1)3 � 3(k�1)2 � 2(k�1)��

6


Assume that the statement an

� �n(n – 1)

6(2n – 1)�

is true for n � k � 1, so that

ak – 1

� is true.

By the recursion, ak

� ak – 1

� (k – 1)2

= � �6(k�

61)2

�

=

= �(k�1) k

6(2k�1)�.

Then if the statement is true for n � k � 1 it is alsotrue for n � k.Since the statement is true for n � 1, and if true for n � k � 1 it is also true for n � k, then bymathematical induction it is true for all values of n � 1.

7. a. For n � 1, en

� 0, so the statement is true for

n � 1.

Assume that en

is an even integer for n � k; that is,

assume that ek

� k2 � k is even.

Then ek + 1

� (k � 1)2 � (k � 1)

� k2 � k

� (k2 � k) � 2k.

Now k2 � k is even by the assumption and 2k is

even for all values of k. Hence ek + 1

is even.

Since en

is even if n � 1, and if it is even for n � k

then it is even for n � k � 1, by the principal of

mathematical induction it is always even.

b. en

� n(n � 1).

Then n and n � 1 are consecutive integers so one

or the other is even, and the product is even.

(k�1)(2k2 � 7k � 6 � 6k �6)��

6

(k�1)(k�2)(2k�3)��

6

(k�1)(k�2)(2k�3)��

6

9. The sum of the cubes of three consecutive integers can

be represented by Sn

� n3 � (n � 1)3 � (n � 2)3 for

n � 1.

For n � 1, S1

� 1 � 23 � 33 � 36, which is divisible

by 9.

Assume that for n � k � 1, Sk – 1

� (k�1)3 � k3

� (k � 1)3 is divisible by 9.

Then Sk

� k 3 � (k � 1)3 � (k � 2)3

� Sk – 1

� (k � 2)3 � (k � 1)3

� Sk – 1

� 9(k2 � k � 1).

Now Sk – 1

is divisible by 9 and so is 9(k2 � k � 1).

Then Sk

is divisible by 9 if Sk – 1

is divisible by 9.

Since Sk

is divisible by 9 if n � 1 and if it is divisible

by 9 for n � k � 1 then it is also for n � k, Sn

is

divisible by 9 for all values of n � 1.

10. a. We have t1

� 2 but if tn

� n! then t1

� 1. Hencestep 1 fails.However, assuming that if n � k, t

k� k! is true,

then by the recursion,

tk + 1

� (k � 1) tk

� (k � 1) k!

� (k � 1)!Hence if the statement is true for n � k it is alsotrue for n � k � 1.Since step 1 does not hold, we see the importanceof both steps in the induction process.

11. Using matrix multiplication

� � � � � � �,

then A2 � � � � � � � �.

The statement is true if n � 2.Assume that the statement is true for n � k � 1.

That is, assume Ak – 1 � �.

Then Ak � Ak–1A

� � �� .

The statement is true for n � k if true for n � k�1.Since the statement is true for n � 2 and if true for n � k�1 then it is true for n � k, then it is true for allvalues of n � 2.

12. The proof is in Section 12.4.

13. We give step 2 only.

Assume that for n � k the statement

vk

� 5.2k–1 �2 � k is true. By the recursion,

vk + 1

� 2vk

� k � 1

= 5.2k � 4 � 2k � k � 1

= 5.2k � 2 � (k � 1).

If the statement is true for n � k then it is true for

n � k � 1.

16. a. Clearly M1

� 1 since one move suffices.

For two pegs, move the smaller disk to peg 2, then

the larger disk to peg 3, and then the smaller disk

on top of it. M2

� 3.

b. Assume that if n � k � 1 then it requires

Mk – 1

� 2k – 1 � 1 moves to move the disks

from peg 1 to peg 2 in the correct order. Then

for Mk

we have the following:

With Mk – 1

moves we move the top k�1 disks from

peg 1 to peg 2. It requires one move to put the last

disk on peg 3 and then Mk – 1

moves to put the first

(k � 1) disks on top of it on peg 3.

Hence Mk

� 2(2k – 1 � 1) � 1

� 2k � 1.If the statement is true for n � k � 1 then it is alsotrue for n � k.Since the statement is true for n � 1 and if true forn � k � 1 it is true for n � k, then the statement istrue for all values of n � 1 by mathematicalinduction.

17. This problem was considered in Chapter 1 onexamining what it means to prove a statement. Itillustrates how ideas connect. Here,

t40

� 41 � 40 � 402

� 41 � 40(41)

Clearly 41�t40

, so it is not prime.

18. This problem was posed in section 12.1. Notice hownicely mathematical induction cleans up the approachtaken there.

01

1ka

01

1a

01

1(k�1)a

01

1(k�1)a

01

12a

01

1a

01

1a

aq � bscq � ds

ap � brcp � dr

qs

pr

bd

ac


19. For proof 1, the use of mathematical induction gives

a straightforward approach. Notice also that

tn

� 2n3 � 3n2 � n � n(2n2 � 3n � 1)

� n(n � 1)(2n � 1).

Now n and n � 1 are consecutive integers, so one is

divisible by 2. Also note that n can always be written

as 3k, 3k � 1, or 3k � 2, where k � 0. If n � 3k it is

divisible by 3, so tn

is divisible by 6. If n � 3k � 1

then 2n � 1 � 6k � 3, so tn

is divisible by 6. Lastly,

if n � 3k � 2 then n � 1 � 3k � 3, so tn

is divisible

by 6. Then tn

is always divisible by 6.

20. a. We have 2k – 1 � 2k – 1 � 0 and

2k � 1 � 2k – 1 � (2k – 1 � 1).

There are 2k – 1 � 1 � 1 � 2k – 1 terms in the

sequence.

b. �2k – 1

i � 2k – 1

f(i) � �2k

1– 1� � �

2k – 11

� 1� � �

2k – 11

� 2�

� … � �2k �

11

�

> �2k �

11

� � �2k �

11

� � �2k �

11

�

� … � �2k �

11

�

� �22k

k

�

�1

1�

> �12

� since 2�2k – 1� � 2k > 2k � 1.

c. The sum in part b gives an expression for 2k – 1

terms starting with the term �2k

1– 1� in the sum.

Hence we block off the sum S(2k) � �11

� � �12

� � �13

�

� �14

� � … � �21k� and note that, from part b, the first

term is 1 > �12

�, the sum of �12

� � �13

� > �12

�, the sum

�14

� � �15

� � �16

� � �17

� > �12

�, and so on.

But there are k such groupings, so the sum

�11

� � �12

� � �13

� � �14

� � … � �21k� > �

2k

�.

21. For g(n) � �n12� we have t(n) � �

n

i � 1g(i)

� �11

� � �212� � �

312� � … � �

n12�

For n � 1, t(1) � 1 and for n � 2,

t(2) � 1 � �14

� < 2 � �12

�.

The statement is true for n � 1 and for n � 2.

Suppose that the statement is true for n � k � 1;

that is, assume that t(k � 1) < 2 � �n �

11

�.

Now t(k) � t(k � 1) � �k12�

� 2 � �k �

11

� � �k12�

� 2 � �kk

2

2�

(kk�

�

1)1

�

� 2 � �k2

k(

2

k�

�

k1)

�

� 2 � �1k

�.

If the statement is true for n � k�1 then it is also truefor n � k.Since the statement is true for n � 1 and if true for n � k�1 then it is true for n � k, then bymathematical induction it is true for all n � 1.

Exercise 12.4

7. The expansion is true for all values of x. Hence it istrue specifically for x � 1, and since (1 � 1)(2)n � 0,the sum of the coefficients is 0.

8. The expression is the expansion of ��13

� � �23

��6� 1.


13. a. The kth term is � �(a – 1)24 – k (3a2)k

or � � 3k a3k – 24.

Then for a –15 we require 3k � 24 � �15 and k � 3. The fourth term contains a –15.

b. For a10 we require 3k � 24 � 10 and k � �334�.

Since this is not an integer, there is no termcontaining a10.

14. The general term is tk

� � �(�1)k xk.

For k � 2, � � � 15.

n (n � 1) � 30n � 6 or �5

Then n � 6.

15. a. (z2 � 2z5)10 � �10

k � 0� �(�1)k zk z20 + 3k.

The general term is (�1)k �2k � � z20 + 3k.

b. The middle term is t6

with k � 5.

t6

� �25 � � z35

� �8064 z35.

c. For z41, 20 � 3k � 41k � 7.

The coefficient is (�1)7 27 � � � �15 360.

d. For z36, 20 � 3k � 363k � 16.

There is no integer value for k, so z36 does not appear.

e. No.

107

105

10k

10k

n2

nk

24k

24k


16. (1 � x2)(1 � x)2n � (1 � x2)�1 � 2nx � � �x2 � …�� 1 � 2nx � �� 1� x2 � … .

Then � � � 1 � 189

2n2 � n – 190 � 0

n � 10 or n � ��129�.

Since n > 0 and is an integer n � 10.

19. Let (1 � ax)n � 1 � nax � � � a2x2 � …

� 1 � 21x � 189x2 � … .

Then na � 21 and �n(n

2� 1)� a2 � 189.

Then (na)2 � na(a) � 378a � 3n � 7.

The function is (1 � 3x)7.

20. (1 � x)3 (1 � x)n � (1 � 3x � 3x2 � x3) �

�… � � �xr – 3 � � �xr – 2 � � �xr – 1 � � �xr � …�� … � �� 3 � � � 3 � � � � �� xr � …

The coefficient of xr is as given.

Now, in (1 � x)n + 3 the coefficient of xr is � �.

These must be equal, and the equality follows.

21. S(x) is a geometric series with a � 1, r � (1 � x), and n terms. Usethis and the sum of the series.

n � 3r

nr � 3

nr � 2

nr � 1

nr

nr

nr � 1

nr � 2

nr � 3

n2

2n2

2n2

2n2

22. Since r and s are the roots of x2 � 2x�1 � 0, we obtain

r � �1 � �2 and s � �1��2.

Then rn � sn � ��1 � �2�n� ��1 � �2�n

.

Determining tk + 1

in each expansion gives tk + 1

in the expression.

tk + 1

� � �(�1)n – k ��2�k� � � (�1)n – k ��2�k

� (�1)n – k � � ��2�k �1 � (�1)k�.

If k is even, ��2�kis an integer and t

k + 1is an integer.

If k is odd, ��2�kis not an integer but 1 � (�1)k � 0,

so tk + 1

� 0.

Then rn � sn is the sum of a number of integers and is itself

an integer.

25. � � � � � � �(n �

nk!)! k!� �

� n! � � �� n! ��(n �

nk�

�

11)! k!

�� .

26. Set x � 1 in (1 � x)10.

27. Setting x � 1, (1 � x)12 � (1 � x)12 � 212.

Also (1 � x)12 � (1 � x)12

� 1 � � �x � � �x2 � � �x3 � … � � �x12

� 1 � � �x � � �x2 � � �x3 � … � � �x12

� 2 �1 � � �x2 � � �x4 � … � � �x12�.1212

124

122

1212

123

122

121

1212

123

122

121

n � 1k

k��(n � k � 1)! (k � 1)! k

n � k � 1��(n � k � 1)(n � k)! k!

n!��(n � k � 1)! (k � 1)!

nk � 1

nk

nk

nk

nk


Setting x � 1 in this expression, we obtain

2 �1 � � � � � � � … � � �� 212

from which � � � � � � … � � � � 211 � 1.

30. The sum of all entries in the kth row is (1 � 1)k – 1.Then the sum of all entries to and including the nth

row is

S � 20 � 21 � 22 � … � 2n – 1

� 2n � 1.

31. �2x � �1x

�� n

k � 0� �(2x)n – k (x) – k.

For x3 we have n � 2k � 3

and � � 2n – k � 672 � 25 � 21.

n(n � 1)2n – k – 1 � 25 � 21.

Now n and n � 1 are consecutive integers. Given theright side, there are only two possible values for n,n � 7, in which case n(n�1) � 7 � 6 and 26 – k � 24

so k � 2; or n � 8, in which case n(n � 1) � 8 � 7 and 27 – k � 22 � 3 and there is no possible value for k.Then n � 7.

32. a. In (1 � x)m + n the coefficient of xk is � �.

In (1 � x)m (1 � x)n we write the expansion as

�� x � � �x2 � … � � �xk – 1

� � �xk � … � � �xm�� x � � �x2 � … � � �xk – 1

� � �xk � … � � �xn�.

In this the coefficient of k, where 0 � k � m,0 � k � n, is

� �� …

� � �� .

The equality follows.

n0

mk

nk – 2

m2

nk – 1

m1

nk

m0

nn

nk

nk – 1

n2

n1

n0

mm

mk

mk – 1

m2

m1

m0

m � nk

nk

nk

1212

24

122

1212

124

122

b. If we choose i blue objects from m, where 0 � i � k, and k � i red objects from n, we obtain

the expression �k

i � 0� �� for the

number of possible subsets.

33. a. If the sequence ends in k 1s they must be precededby a 0. Then there are n 0s and r – k 1s to arrange

before this 0. This can be done in � � ways, and this is the number of possible sequences,where 0 � k � r.

b. Every one of the above sequences is distinct sincethe number of 1s at the end changes as k changes.Hence the total number of sequences is

�r

k � 0� �.

But the number of sequences possible for n � 1 0s

and r 1s is � �.

Then �r

k � 0� � � � � � � �

� … � � � � � �.

34. This is a generalization of question 22.

35. This problem illustrates that, while the symbol � �counts the number of subsets possible in a situation,the symbol, in a broader sense, generates a number.Hence it is reasonable to assume that negative valuesof n can be used. Using the definition given,

� � �

= (�1)k

= (�1)k � �.

36. By division, the first five terms in each case are

(1 � x) –1 � 1 � x � x2 � x3 � x4 � …

(1 � x) –1 � 1 � x � x2 � x3 � x4 � …

(1 � x) –2 � 1 � 2x � 3x2 � 4x3 � 5x4 � …

(1 � x) –2 � 1 � 2x � 3x2 � 4x3 � 5x4 � … .

37. S1(x) � 1 � x � x2 � x3 � x4 � … � xn – 1

� xn � …

xS1(x) � x � x2 � x3 � x4 � … � xn – 1 � xn

� xn + 1 � …

(1 � x)S1(x) � 1

S1(x) � (1 � x) –1.

38. S2(x) � 1 � 2x � 3x2 � 4x3 � … � nxn – 1

� (n � 1)xn � …

xS2(x) � x � 2x2 � 3x3 � … � (n � 1)xn – 1

� nxn � …

(1 � x) S2(x) � 1 � x � x2 � x3 � … � xn – 1

� xn � …

= (1 � x) –1

Then S2(x)� (1 � x) – 2.

39. For n � 2,

(1�x) –2 � 1 � � �(�x) � � �(�x)2

� � �(�x)3 � …

� 1 � (�1)� �(�x) � (�1)2� �(�x)2

� (�1)3 � �(�x)3 � (�1)4� �(�x)5 � …

� 1 � 2x � 3x2 � 4x3 � 5x4 � … .

40. c. (1�x) – 3 � �∞

k � 0(�1)k � �(�x)k

� �∞

k � 0� �xk

� 1 � � �x � � �x2 � � �x3 � …

� 1 � 3x � 6x2 � 10x3 � …

� �(k � 2)

2(k � 1)� xk � … .

53

42

31

2 � kk

3 � k � 1k

54

43

32

21

–23

–22

–21

n � k � 1k

n(n � 1)(n � 2)(n � 3) … (n � k � 1)��

k!

�n(�n�1)(�n�2)(�n�3) … [�n � (k�1]��

k!�nk

nk

n � r � 1r

n � rr

n � 11

n0

n � r – kr – k

n � r � 1r

n � r – kr – k

n � r – kr – k

m � nk

nk – i

mi


d. (1 � x) –4 � �∞

k � 0(�1)k � �(x)k

� �∞

k � 0(�1)k � �xk

� 1 � � �x � � �x2 � � �x3 � …

� 1 � 4x � 10x2 � 20x3 � …

� (�1)k � �xk � … .

e. (1 � 2x) – 4 � �∞

k � 0(�1)k � �( –2x)k

� �∞

k � 0� �2k xk

� 1 � � �2x � � �22 x2

� � �23 x3 � …

� 1 � 8x � 40x2 � 160x3 � …

� � �2k xk � … .

f. (1 � 3x) –5 � �∞

k � 0(�1)k � �(3x)k

� �∞

k � 0(�1)k � �3k xk

� 1 � � �3x � � �32 x2

� � �33 x3 � …

� 1 � 15x � 135x2 � 945x3 � … .

73

62

51

4 � kk

5 � k � 1k

k � 3k

63

52

41

3 � kk

4 � k � 1k

3 � kk

63

52

41

3 � kk

4 � k � 1k


41. The assumption that (1 � x) –n � �∞

k � 0� � xk

is true for n � 1, as demonstrated in question 37.Now assume that the assumption is true for n � t;

that is, (1 � x) –t � �t

k � 0� � xk.

Now (1 � x) –(t + 1)

� (1 � x) –t (1 � x) –1

� ��∞

k � 0� �xk� �1 � x � x2 � x3 � … � xk � …�

� �t

k � 0� �

k

i � 0� �� xk.

Now �k

i � 0� �

� �k

i � 0� �

� � � � � � � � � � …

� � �� from question 33.

� � �.

Then (1 � x) –(t + 1) � �∞

k � 0� � xk.

This is precisely what the statement leads us to expect.Hence if the statement is true for n � t, it is is also truefor n � t � 1.Since the statement is true for n � 1 and if true for n � t it is also true for n � t � 1, then it is true for all n � 1 by the induction hypothesis.

t � kk

t � kk

(t � 1) � k � 1k

(t � 1) � kk

(t � 1) � 22

(t � 1) � 11

t � 10

(t � 1) � ii

t � i � 1i

t � i � 1i

t � k � 1k

t � k � 1k

n � k � 1k

Review Exercise

4. a. If {ak} � a, a � d, a � 2d, … , a � (n � 1)d, then

An

� �n2

� �[2a � (n

n� 1)d]�� a � (n � 1) �

d2

�.

This is an arithmetic sequence with difference halfthat of the given sequence. Also

Gn

� �na(a �d)(a � 2d) … (a � (n � 1)d).

Then

�G

G

n

n

– 1

� � n �� n

a � (n � 1)d.

Since this is dependent on n, Gn

is neither anarithmetic nor geometric sequence.

b. If {ak} � a, ar, ar2, … , arn – 1, then

An

�

� �an((rr

n

�

�

11))

�.

This gives neither an arithmetic nor a geometricsequence.

Gn

� �na � ar � ar2 � …� arn– 1

� nan r

�(n –

21�)n��

� ar�n –

21

�

� a(�r)n – 1

This defines a geometric sequence with common

ratio �r.

6. a. un

� (1 � 0.012 � 0.008 � 0.005)un – 1

� 100 000

= 0.999 un – 1

� 100 000, n � 1 and

u0

� 30 000 000.

b. Since 0.001 u0

� 30 000, which is less than thenumber of immigrants, the population will grow.

c. If there are 30 000 immigrants the population is stable.

7. Since the moves involve the vector (2, �1), after k moves the point is at the point (2k, �k). We requirethat 4k2 � k2 � 400, k an integer.

k > 8.944.The point escapes the circle on the 9th move.

8. a. If hn

is the height of Dn, then h

1� 1.

Now h2

� 1 � �13

� � h1

� �13

�

h3

= 1 � �13

� � ��13

��2� h

2� ��

13

��2.

By extension, hn

� hn – 1

� ��13

��n – 1.

If An

is the area of Dn, then A

1� 1.

Now A2

� 1 � ��13

��2� A

1� ��

13

��2

A3

� 1 � ��13

��2� ��

13

��4� A

2� ��

13

��4.

By extension An

� An – 1

� ��13

��2n – 2.

b. hn

� 1 � �13

� � ��13

��2� … � ��

13

��n – 1

� �32

� �1 � ��13

��n�As n → ∞ , h

n� �

32

�

An

� 1 � ��13

��2� ��

13

��4� … � ��

13

��2n – 2

� �98

� �1 � ��13

��n�.

As n → ∞ , An

� �98

�.

a � ar � ar2 � … � arn – 1

��n

a(a � d)(a � 2d) … [a � (n � 1)d]��a(a � d)(a � 2d) … [a � (n � 2)d]


9. a. di� t

i� t

i – 1

� (i � 1)i � (i)(i � 1)

� 2i.

b. d2

� t2

� t1

d3

� t3

� t2

d4

� t4

� t3

...

dn

� tn

� tn – 1

�n

i � 2d

i� t

n� t

1.

c. �n

i � 2d

i� 2[2 � 3 � 4 � … � n] � (n � 1)n � 2.1.

Then 2 (1 � 2 � 3 � … � n) � n(n � 1)

and �n

i � 1i � �

n(n2� 1)�.

10. For n = 1 we have �1 �

13

� � �2 �

11

�, which is true.

Hence the statement is true for n � 1.

Assume that the statement is true for n � k; that is,

assume that

�1 �

13

� � �3 �

15

� � … � �(2k � 1)

1(2k � 1)��

2kk� 1�

is true.

Now, if n � k, we have

�1 �

13

� � �3 �

15

� � … � �(2k � 1)

1(2k � 1)�

� �(2k � 1)

1(2k � 3)�

� �2k

k� 1� � �

(2k � 1)1(2k � 3)�

��(2

2kk�

2 �

1)3(2kk�

�

13)

�

� �2kk�

�

13

�

� �2(k

k�

�

1)1� 1

�.

Hence the statement, if true for n � k, is true for n � k � 1.Since the statement is true for n � 1 and if true for n � k is also true for n � k � 1, then by mathematicalinduction the statement is true for all n � 1.


Note that while induction provides a proof, it is easierto note that breaking each term into partial fractionsallows a short, direct proof.

�(2k � 1)

1(2k � 1)��

2na� 1� � �

2nb� 1�

� � .

Then �1 �

13

� � �3 �

15

� � … � �(2n � 1)

1(2n � 1)�

� � � � � � � …

� � � �

� �

� �2n

n� 1�.

11. We give the main step only.

a. Assume that 3 does not divide xn – 1

; that is, xn – 1

is

not a multiple of 3.

Now xn

� 3xn – 1

� 1.

For any value of xn – 1

, 3xn – 1

is divisible by 3.

Hence 3xn – 1

is never divisible by 3.

If the statement is true for xn – 1

, then it is true for xn.

b. x2

� 4, so the statement is true for k � 1.

Assume that x2(k – 1)

is divisible by 4.

Then x2k

� 3x2k – 1

� 1

� 3(3x2k – 2

� 1) � 1

� 9x2(k – 1)

� 4.

Since x2(k – 1)

is divisible by 4, then x2k

is also.

�12

�

�2n � 1

�12

�

�1

�12

�

�2n � 1

�12

�

�2n � 1

�12

�

�2n � 1

�12

�

�2n � 3

�12

�

�7

�12

�

�5

�12

�

�5

�12

�

�3

�12

�

�3

�12

�

�1

�12

�

�2n � 1

�12

�

�2n � 1

12. We give the main step only.

Assume that f(k � 1) � 2k – 1 � (k � 1)k is positive

for k > 5.

Then f(k) � 2k � k(k � 1)

= 2.2k – 1 � (k � 1)k � 2k

= 2k – 1 � (k � 1)k � 2k – 1 � 2k

= f(k � 1) � 2k – 1 � 2k.

Since f(k � 1) > 0, then clearly f(k) is also.If the statement is true for n � k � 1 then it is alsofor n � k.

13. For n � 1, the statement is f1

� f3

� 1. Since f1

� 1

and f3

� 2, the statement is true.

For n � 2, the statement is f1

� f2

� f4

– 1. Since

f1

= f2

� 1 and f4

� 3, the statement is true.

Assume the statement is true for n � k; that is,

assume that f1

� f2

� … � fk

� fk + 2

� 1.

We wish to show that for n � k � 1 the statement

f1

� f2

� … � fk

� fk + 1

= fk + 3

� 1 is true.

Using the assumption,

f1

� f2

� … � fk

� fk + 1

� (f1

� f2

� … � fk) �

fk � 1

� fk + 2

� 1 � fk + 1

� (fk + 1

� fk + 2

) � 1

� fk + 3

� 1.

If the statement is true for n � k then it is also truefor n � k � 1.Since the statement is true for n � 1 and 2, and if truefor n � k it is also true for n � k � 1, by the methodof mathematical induction it is true for all n � 1.

14. Here we are given no formula, so we must determineone that can be proven true by mathematicalinduction.

(1 � 2 � 3 � 4 � … � n)2 � ��n(n2� 1)��2

for n � 1, 13 � ��1 �

22

��2is true.


We now give the main step only.

Assume that for n � k,

13 � 23 � … � k3 � ��k(k2� 1)��2

is true. Then

13 � 23 � … � k3 � (k � 1)3 � ��k(k2� 1)��2

� (k � 1)3

� (k � 1)2 ��k2 �

24

2k � 4��

� ��(k � 1)2(k � 2)��2

.

If the statement is true for n � k, then it is also true for n � k � 1.

15. Here is the main step.Assume that for n � k the statement is true; that is,pk � 1 � (p � 1)f(p) where f(p) is a polynomial of degreek � 1.Then if n � k � 1 we have

pk + 1 � 1 � pk + 1 � pk � pk � 1

� pk(p � 1) � (p � 1) f(p)

� (p � 1) [pk � f(p)]Now pk � f(p) is a polynomial of degree k, since f(p) is apolynomial of degree k � 1 and we now have pk added.Hence if the statement is true for n � k then it is also true for n � k � 1.

19. 1 � � �2 � � �22 � … � � �26 � (1 � 2)6.

20.(1 � x)2n � (1 � x)2n

� �2n

k � 0� �x � �

2n

k � 0(�1)k � �xk

� 2 �n

k � 0� �x2k, since terms with k odd disappear.

Setting x � 1 we obtain

22n � 0 � 2 �n

k � 0� �

or � � � � � � � � � … � � � � 22n – 1.2n2n

2n4

2n2

2n0

2n2k

2n2k

2nk

2nk

66

62

61

Chapter 12 Test

1. Every integer from 1 to 99 inclusive is to be squared.These squares are each to be multiplied by 3, creating99 new integers, which are to be summed.

2. Since gi� ari – 1, g

i2 � a2r2(i – 1).

Then �n

i � 1g

i2 � a2 � a2r2 � a2r4 � a2r6 � … � a2r2n

� a2 � a2r2 � a2(r2)2 � a2(r2)3 � … � a2(r2)n

� �a2 [

r(r2

2

�

)n �

11]

�

� �a2

r(r2

2

�

n �

11)

�.

4. a. t0

� 4000

t1

� 4000 (1.01) � 200 � 3840.

b. For any given month n � 1, we have tn – 1

as the

amount owing.

Then tn

� 1.01 tn – 1

� 200, t1

� 4000.

4. For n � 1, x1

� 1, and 21 � 1 � 1. The statement

is true for n � 1.

Assume that the statement is true for n � k � 1;

that is, assume that xk – 1

� 2k – 1 � (k � 1) is true.

We wish to prove that xk

� 2k � k is also true by

using the recursion. We have

xk

� 2xk – 1

� k � 2

� 2 [2k – 1 � (k � 1)] � k � 2

� 2k � 2k � 2 � k � 2

� 2k � k.

If the statement is true for n � k � 1 then it is alsotrue for n � k.Since the statement is true for n � 1, and if true for n � k � 1 is also true for n � k, then by the principleof mathematical induction it is true for all n � 1.

5. This can be done by mathematical induction. We givean alternate proof.

We know that 12 � 22 � 32 � … � n2

� �n(n � 1)

6(2n � 1)�.

Then 12 � 22 � 32 � … � (2n)2

� .

Now 22 � 42 � 62 � … � (2n)2

� 4(12 � 22 � 33 � … � n2)

� 4 �n(n � 1)

6(2n � 1)�.

Then 12 � 32 � 52 � … � (2n � 1)2

� [12 � 22 � 32 � 42 � … � (2n)2]� [22 � 42 � 62 � … � (2n)2]

� ��4n(n � 1

6)(2n � 1)�

�

��n(2n � 1

3)(2n � 1)�

� �4n3

3� n�.

6. �x � �1x

��12� �

12

k � 0� � x12 – k ��

1x

��k

� �12

k � 0x12 – 2k.

a. For x4 we require 12 � 2k � 4k � 4.

The coefficient of x4 is � � � 495.

b. For x0 we require k � 6

The coefficient of x0 is � � � 924.

7. 1 � 2� � � 4� � � … � 2n� � � �n

k � 0� �2k

� (1 � 2)n

� 3n.

nk

nn

n2

n1

126

124

12k

2n(2n � 1) [4n � 1 � 2n � 2]��

6

2n(2n � 1)(4n � 1)��

6

2n (2n � 1)(4n � 1)��

6


8. Let the sequence be a, a � d, a � 2d, … , a � (n � 1)d.

Then a1

� a3

� … � a17

� a � (a � 2d) � (a + 4d) + … � (a � 32d)

� �92

� (2a � 16d)

� 9a � 72d.

Also a2

� a4

� a6

� … � a18

� (a � d) � (a � 3d) � (a � 5d) � … � (a � 17d)

� 9a � 81d.

Now 9a � 72d � 27

9a � 81d � 9

so 9d � �18

d � �2.

Then 9a � 171, so a � 19.

Then �36

i � 1a

i� �

326� [2(19) � 35(�2)]

� �576.

Note that by comparing the two given statements one canobtain 9d � �18 immediately.

9. Solution 1.

� � is the number of subsets of size k chosen from

n distinct elements.Identify one element as a key. Then the subset eitherincludes this element or it doesn’t.If it does, we choose k � 1 elements from the remainingn � 1 elements, which can be done in

� � ways.

If the key element is not included, then the subset of kelements must be chosen from n � 1 elements, which

can be done in � � ways.

Then � � � � � � � �.n � 1

kn � 1k � 1

nk

n � 1k

n � 1k � 1

nk



6. If �10

i � 1a

i� 20, then �

120� (2a � 9d) � 20

or 2a � 9d � 4.

If �20

i � 1a

i� 60, then �

220� (2a � 19d) � 60

or 2a � 19d � 6.

Solving, d � �15

� and a � �1110�.

Then �60

i � 1a

i� �

1110� � �

1130� � …

� �620� ��

151� � 59��

15

�� 420.

9. a. We have x1

� 4, xn

� 3xn – 1

� 2

x2

� 3(4) � 2 � 10

x3

� 3(10) � 2 � 28

x4

� 3(28) � 2 � 82.

b. In each case a number greater by 1 than a multiple

of 3 is multiplied by 3 and the result is lessened

by 2. Hence it appears that xn

� 3n � 1, n � 1.

Proof by induction.

For n � 1, x1

� 4. The statement is true for n � 1.

Assume that the statement is true for n � k; that is,

assume that xk

� 3k � 1. We wish to prove that

xk + 1

� 3k + 1 � 1.

Using the recursion,

xk + 1

� 3xk

� 2

� 3(3k � 1) � 2

� 3k + 1 � 1.If the statement is true for n � k, then it is true for n � k � 1.Since the statement is true for n � 1, and if truefor n � k is also true for n � k � 1, then bymathematical induction it is true for all n � 1.

10. a. This argument is incorrect because it includes somecommittees more than once. Suppose, for example,that A and B are Grade 11 students and x is a Grade12 student. In selecting one Grade 11 student wecan choose A; in now selecting two students fromthe remaining 10, we can choose B and X. However,in selecting one Grade 11 student we could chooseB, then choose A and X from the remaining 10. This selection also includes all committees of threeGrade 11 students three times.

b. Each committee consists of three Grade 11 students,two Grade 11 students, and one Grade 12 student, orone Grade 11 student and two Grade 12 students.The number of committees is

� � � � �� 155.

Another approach is to calculate the number ofcommittees with no restriction and subtract thenumber of committees made up only of Grade 12

students. This gives � � � � � � 155.

11. The only repeated letter, other than s, is e. Considersequences of length 7 in two cases; two es are bothincluded or they are not.If two es are included, we choose three of the

remaining five letters in � � ways. There are six

positionings of the ss combination with the first one inposition 1, 2, 3, … , or 6. The remaining five positions

are filled by five letters, of which two are es, in �52!!�

ways.

The number of sequences is 6� � �52!!�.

If there is no repeat of es, then choose five letters from

six available, and the number of sequences is 6� �5!

For eight-letter sequences, if there are two es, the

number of sequences is 7� � �62!!�.

If there is only one e included, the number ofsequences is 7.6!.The total number of sequences is

6� � �52!!� � 6� � 5! � 7� � �

62!!� � 7(6!)

� 6 � 10 � 60 � 6 � 6 � 120 � 7 � 5 � 360� 7 � 720

� 25 560.

12. If 1 � r � 4 the number of sequences is 0. For r � 5,

the number of sequences is 5!� �, since from the

r elements we select five and then permute them. Hence the number of sequences is 0 for 1 � r � 4

and 5!� � for 5 � r � n. r5

r5

54

65

53

54

65

53

53

53

113

52

61

51

62

63


solutions manual of geometry and discrete mathematics

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