applied mathematics 1a (eng) mathematics 132: vectors … · 5. d. lay: linear algebra and its...

APPLIED MATHEMATICS 1A (ENG)

Mathematics 132: Vectors and Matrices

University of KwaZulu-NatalPietermaritzburgc©C. Zaverdinos, 2012.

All rights reserved. No part of this book may be reproduced, in any form or by any means,without permission in writing from the author.

2

PrefaceThis course has grown out of lectures going back well over 30 years. Names which come tomind are J.Nevin, R.Watson, G.Parish and J. Raftery. I have been involved with teachinglinear algebra to engineers since the early 90s and some of my ideas have been incorporated inthe course. What is new in these notes is mainly my approach to the theoretical side of thesubject. Several of the numerical examples and exercises come from earlier notes, as do a fewtheoretical problems.

I would like to thank Dr. Paddy Ewer for showing me how to use the program LatexCad forthe diagrams and last, but not least, the Reverend George Parish for proof-reading an earlierversion and also for elucidating No.5 of Exercise 2, as well as Professors John and Johann vanden Berg for their encouragement in developing these notes.

About this course

The course is meant to be studied as a whole. Many examples and exercises in later Chaptersrefer the reader to earlier chapters. This is especially true of Chapter 3.

Chapter 1 motivates the idea of a vector through geometry and discusses lines and planes andtransformations related to such geometric objects. A vector can be thought of as a displacementin space and an ordered triple of numbers.

Chapter 2 generalizes the idea of a triple to an n-tuple and motivates linear algebra throughthe problem of finding solutions to simultaneous linear equations in n unknowns. The coeffi-cient matrix of such equations is known as a matrix. Simplification of such a matrix by rowoperations forms the major part of this chapter.

Chapter 3 considers matrices in detail and looks at them dynamically in the converse senseof Chapter 2: A matrix defines a transformation of points in n-dimensional space. Matrixmultiplication is introduced in terms of the composition of such transformations and some otherkey concepts such as linear independence, the rank and inverse of a matrix are discussed.Abstract vector spaces are never mentioned, but we remark that the the alternative proof ofthe basic theorem of linear independence in section 3.6 goes through word-for-word for suchspaces and also leads to the well-known Replacement Theorem of Grassmann.

Chapter 4 is about determinants and the cross product (also called the vector product).The theory of determinants predates that of matrices, going back to Leibnitz in the 17th Cen-tury. One of the founders of linear algebra, the 19th Century mathematician Arthur Cayley,once remarked that many things about determinants should really come after the study ofmatrices, and this is the modern approach adopted by us.The cross product is used extensively in mechanics, im particular dynamics, which is studiedin Mathematics 142. Algebraic properties of the cross product are derived from those of 3× 3determinants, while the exercises can serve as an introduction to some of its applications.

Note 1 Some exercises (particularly those marked with an asterisk *) are harder and, at thediscretion of the instructor, can be omitted or postponed to a later stage.

3

Bibliography

The following books can be consulted but they should be used with caution since differentauthors have a variety of starting points and use different notation. Many books also have atendency to become too abstract too early. Unless you are a mature reader, books can confuserather than help you.

1. A.E. Hirst: Vectors in 2 or 3 Dimensions (Arnold).This may be useful for our Chapter 1 since it makes 3-dimensional space its central theme.

2. R.B.J.T. Allenby: Linear Algebra (Arnold).This is a quite elementary text.

3. J.B. Fraleigh & R.A. Beauregard: Linear Algebra (Addison-Wesley).

4. K. Hardy: Linear Algebra for Engineers and Scientists (Prentice-Hall).

5. D. Lay: Linear Algebra and its Applications (3rd Edition, Pearson). This booktreats matrix multiplication in much the same way as we do, but its treatment of geometricaspects is less thorough.It has over 560 pages, becomes abstract and advanced after p.230, but will probably beuseful in later years of study.

6. H. Anton: Elementary Linear Algebra (6th Edition, John Wiley and Sons).

7. E.M. Landesman & M.R. Hestenes: Linear Algebra for Mathematics, Science andEngineering (Prentice-Hall). This is quite advanced.

I recommend especially (12) for Chapter 1 and (4) and (5) for later chapters of the notes.Because of its elementary nature, (2) is good for Chapter 2. If used with care the above bookscan be helpful.

C. ZaverdinosPietermaritzburg, February 6, 2012

Contents

1 Two and Three-Dimensional Analytic Geometry. Vectors 7

2 Matrices and the Solution of Simultaneous Linear Equations 51

3 Linear Transformations and Matrices 89

4 Determinants and the Cross Product 151

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6 CONTENTS

Chapter 1

Two and Three-DimensionalAnalytic Geometry. Vectors

1.1 Points in three-dimensional space

The study of the geometry of lines and planes in space provides a good introduction to Lin-ear Algebra. Geometry is a visual subject, but it also has an analytical aspect which is thestudy of geometry using algebra: how geometric problems can be expressed and solved al-gebraically. This geometry is also called Cartesian Geometry, after its founder, the 17thCentury philosopher and mathematician Rene Descartes.

From school you are already familiar with the Cartesian plane. The x−axis and y−axismeet at right-angles at the origin O. Every point A in the x− y plane is uniquely representedby an ordered pair (a, b) of real numbers, where a is the x−coordinate and b is the y−coordinateas in Figure 1.1. We write A = (a, b). Here M is the foot of the perpendicular from A to thex−axis and likewise N is the the foot of the perpendicular from A to the y−axis.Notice that M = (a, 0) and N = (0, b).

x− axis

y− axis

O

Figure 1.1

b

a

A = (a, b).........................

.......................................a

M

b

N

In order to represent a point A in space we add the z−axis which is perpendicular to both the

7

8CHAPTER 1. TWO AND THREE-DIMENSIONAL ANALYTIC GEOMETRY. VECTORS

x− and y−axes as in Figure 1.2. Here A = (a, b, c) is a typical point and a is the x−coordinate,b is the y−coordinate and c is z−coordinate of the point A. In the diagram P is the foot of theperpendicular from A to the y− z plane. Similarly, Q and R are the feet of the perpendicularsfrom A to the z − x and x− y planes respectively.

O

x− axis

y− axis

R

P

Q

L

M

z− axis

N

aa

b

b

b

c c

Figure 1.2

A = (a, b, c)

aa

1.1.1 The Corkscrew Rule and right-handed systems

You wish to open a bottle of wine using a corkscrew. You first turn the corkscrew so thatit enters the cork. As observed by the opener the turning is clockwise, but observed fromthe underside of the bottle the sense is reversed: it is anticlockwise. The direction in whicha corkscrew moves as it is turned is given by the Corkscrew Rule. In Figure 1.2, rotatingfrom the x−axis to the y−axis and applying this rule produces motion along the (positive)z−axis. Rotating from the y−axis to the z−axis and applying the same rule gives motionalong the (positive) x−axis. Finally, rotating from the z−axis to the x−axis and applying therule produces motion along the (positive) y−axis. The axes x, y, z (in that order) are said toform a right-hand system. This can be represented by the scheme

x z−−−−−−→y x−−−−−−→z y−−−−−−→x

Have a look at yourself in the mirror while opening a bottle of wine and observe that whatis clockwise for you is anticlockwise for the person in the mirror. Thus a right-hand system ofaxes observed in a mirror does not follow the corkscrew rule and is called a left-hand system.The usual convention is to use only right-hand systems.

1.1.2 Displacements and vectors

Consider two numbers a and b on the real line <. The difference c = b−a is the displacementfrom a to b. Given any two numbers from a, b and c, the third is uniquely determined by the

1.1. POINTS IN THREE-DIMENSIONAL SPACE 9

equation c = b − a. For example, let a = −2 and b = 3. The displacement from a to b isc = 3− (−2) = 5. The displacement from 6 to 11 is also 5.Let A = (a1, a2, a3), and B = (b1, b2, b3) be two points in space. The displacement from Ato B is denoted by AB and is given by

AB = (b1 − a1, b2 − a2, b3 − a3) (1.1)

1.1.3 Units and distance

The unit can be any agreed upon distance, like the meter or kilometer. Units (of length, time)are important in applications but for the most part we won’t specify them.The distance of A = (a, b, c) from the origin O is given by

|OA| =√

a2 + b2 + c2

To see this, see Figure 1.2 and use the theorem of Pythagoras:

|OA|2 = |OR|2 + |RA|2= |OL|2 + |LR|2 + |RA|2= a2 + b2 + c2

The distance from A to B is written as |AB| and is given by

|AB| =√

(b1 − a1)2 + (b2 − a2)

2 + (b3 − a3)2 (1.2)

and is also called the length or magnitude of AB.

Because the displacement AB has both direction and magnitude it is called a vector andpictured as an arrow going from A to B. The tail of the arrow is at A and its head is at Bas in Figure 1.3.As an example, let A = (−1, 2.3,−4.5) m and B = (2.1, 3.6,−2.5) m. ThenAB = (2.1− (−1) , 3.6− 2.3,−2.5− (−4.5)) = (3.1, 1.3, 2.0) m.

Two displacement vectors AB and CD are equal if they are equal as displacements. Henceif C = (−3.5, 1.1, 3.3) and D = (−.4, 2.4, 5.3) and A, B are as above then

CD = (−.4, 2.4, 5.3)− (−3.5, 1.1, 3.3) = (3.1, 1.3, 2.0) = AB.

The magnitude of this vector is |AB| = |CD| =√

(3.1)2 + (1.3)2 + (2.0)2 = 3.9115 m.Various vectors (displacement, velocity, acceleration, force) play an important role in mechan-ics and have their own units. For example, if positions are measured in meters and time ismeasured in seconds, a velocity vector will have units ms−1. So if a particle moves in such away that each second it displaces (−1, 2, 3) meters, we say its velocity is constantly (−1, 2, 3)ms−1.

Depending on the geometric interpretation, the tail (head) of displacement vectors may beat any point. Vectors are often written as u = (u1, u2, u3), v = (v1, v2, v3), a = (a1, a2, a3),b = (b1, b2, b3) etc.

We now have three ways of expressing the position of a point A in space. If A hascoordinates a = (a1, a2, a3), then

A = (a1, a2, a3) = a = OA (1.3)

Although equation (1.3) identifies a point with its position, geometrically speaking we liketo distinguish the point A itself from its position vector OA. When we say “a is the position


O

x− axis

:

A

BAB

z− axis

y− axis

Figure 1.3

a3

b3

b2

a2

b1a1

vector of point A” it is understood that the tail of a is at the origin and its head is at the pointA.

Our convention is that we use capital letters A,B, . . . to denote points in space.

Given the point C, there is a unique point D in space such that CD = AB (see Exercise 2,No.4). There is also a unique point D such that OD = AB and we may write D = AB. Inthat case the tail of D can only be O. The notation OD brings out the vectorial nature of thedisplacement from O to D.

1.1.4 More systematic notational convention

Instead of speaking of the x−, y− and z−axis, we also refer to these as the x1−, x2− andx3−axis respectively. Accordingly, given vectors a, b, c,... it will be convenient to assume(unless otherwise stated) that a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), .... Also byconvention, points A, P , Q,... have position vectors a, p, q,...., unless otherwise specified.

1.1.5 Addition of vectors

Let u = (u1, u2, u3) and v = (v1, v2, v3) be two vectors. Their sum u + v is defined by theequation

u + v = (u1 + v1, u2 + v2, u3 + v3)

The vector −v is defined as −v = (−v1,−v2,−v3) and the difference

u− v = u + (−v) = (u1 − v1, u2 − v2, u3 − v3)

The zero vector is 0 = (0, 0, 0).


1.1.6 Basic properties of addition of vectors

For all vectors u, v and w

1. u + (v + w) = (u + v) + w (associative law)

2. u + v = v + u (commutative law)

3. 0 + u = u (the vector 0 = (0, 0, 0) behaves like the number 0)

4. u + (−u) = 0

An important result for addition is the following:

1.1.7 Geometrical interpretation: The triangle law for addition ofvectors

AB + BC = AC

To see this, let the position vectors of A, B and C be a, b and c respectively. Then AB = b−aand BC = c− b. Hence by the properties of addition,

AB + BC = (b− a) + (c− b) = (b− b) + (c− a)= 0 + c− a = c− a = AC

Figure 1.4 illustrates this geometrical result in which it is understood that the head B ofAB is the tail of BC etc.The point D is chosen so that AD = BC. It follows that DC = AB (why?). We have ageometric interpretation of the above commutative law (known as the parallelogram law foraddition of vectors):

BC + AB = AD + DC = AC = AB + BC

A

µ-

6

C1

.........................................................6

B

.......................................

D

AB

BC

AC

O

AD

6

Figure 1.4

DC

:b

....................

AB = DCAD = BC

AC = AB + BC

1.1.8 Multiplication of vectors by scalars. Direction of vectors. Par-allel vectors

In contrast to a vector, a scalar is just a real number, i.e. an element of the real number system<. We use α, β, a, b, r, s, t, ... to denote scalars. Note that f (f underlined) denotes a vector


while f is a scalar.

Let u = (u1, u2, u3) be a vector and α ∈ < be a scalar. The product αu is defined by

αu = (αu1, αu2, αu3)

The vector αu is said to be a (scalar) multiple of u. We usually omit the qualification ‘scalar’.

1.1.9 Properties of the product αu

For all scalars α, β and vectors u and v,

1. α (u + v) = αu + αv

2. (α + β) u = αu + βu

3. α (βu) = (αβ)u

4. 1u = u

5. αu = 0 if, and only if, α = 0 or u = 0

6. |αu| = |α| |u| (the length of αu is |α| times the length of u)

The proofs of the first four are left to you. We prove the last:

|αu| = |(αu1, αu2, αu3)|=

√(αu1)

2 + (αu2)2 + (αu3)

2

=√

α2u21 + α2u2

2 + α2u23

=√

α2 (u21 + u2

2 + u23)

=√

α2

√u2

1 + u22 + u2

3

= |α|√

u21 + u2

2 + u23

= |α| |u|

1.1.10 Geometric interpretation of αu. Parallel vectors

Since |αu| = |α| |u|, multiplication of u by the scalar α multiplies the length of u by |α|.Let u be non-zero, so that it has a definite direction. If α > 0 the direction of αu is the sameas that of u, while if α < 0 the direction of αu is opposite to that of u.

Let u and v be non-zero vectors. We say they have the same direction if u = αv for somescalar α > 0 and opposite directions if u = αv for some α < 0. The vector u is parallel tov if u is a multiple of v, that is, u = αv for some scalar α. Necessarily α 6= 0 and we write u ‖ vto express this relation.Notice that we only speak of vectors being parallel if they are non-zero.

1.1.11 Properties of parallelism

For all non-zero u, v and w,

1. u ‖ u

2. u ‖ v implies v ‖u


3. u ‖ v and v ‖ w together imply u ‖ w.

These properties are geometrically evident, while analytic proofs are left as an exercise. Forexample, to see property (2) analytically, let u = αv. As α 6= 0, v = 1

αu and v ‖ u, as expected.

Example 1.1.1 Let u = AB, where A is the point (−1, 7,−4) and B is the point (−4, 11, 1),so that u = (−3, 4, 5). Let v = CD, where C = (2, 9,−11) and D = (−7, 21, 4), so CD =(−9, 12, 15) = 3u. Hence u and v have the same direction while u and −v have oppositedirections but are parallel.

1.1.12 The dot (scalar) product of two vectors

Let u = (u1, u2, u3) and v = (v1, v2, v3) be any two vectors. Their dot product u · v is definedas

u · v = u1v1 + u2v2 + u3v3 (1.4)

Note that the dot product of two vectors is a scalar and not a vector (that is why it is alsocalled the scalar product).

1.1.13 Properties of the dot product

For all scalars α, β and vectors u, v and w,

1. u · v = v · u2. α (u · v) = (αu) · v = u · (αv)

3. (αu + βv) · w = (αu · w) + (βv · w)

4. u · u = |u|2, so |u| = +√

u · u5. u · u > 0 and u · u = 0 if, and only if u = 0.

The proofs of these these properties are left as exercises (Exercise 2 No.8).

1.1.14 Geometric interpretation of the dot product

Let u = AB and v = AC be non-zero vectors. Suppose that AB and AC make an angle θbetween them at the vertex A, where 0 ≤ θ ≤ 180 in degrees, or 0 ≤ θ ≤ π in radians (recallthat 180 degrees = π radians). Then

u · v = |u| |v| cos θ (1.5)

To see this result, see Figure 1.5 and use AB + BC = AC, so that BC = AC −AB = v−uand

|BC|2 = (v − u) · (v − u)= v · v + (−u) · (−u)− v · u− u · v= |v|2 + |u|2 − 2u · v

Then by the cosine rule (which applies whether θ is acute or obtuse),

|BC|2 = |AB|2 + |AC|2 − 2 |AB| |AC| cos θ

|v|2 + |u|2 − 2u · v = |u|2 + |v|2 − 2 |u| |v| cos θ (from the previous equation)

Cancelling |v|2 + |u|2 from both sides leads to the required result (1.5).


>

-~A

B

Cθ

θ acute

K

-

A

B

C

θ

u

v

u

vj

θ obtuse

Figure 1.5

Example 1.1.2 Let A = (1, 0, 1), B = (2, 0, 3), C =(4,√

10, 2)

and D = (2, 2,−2). Find theangle θ between AB and AC. Decide if the angle φ between AB and AD is acute or obtuse.

Solution: AB = (2, 0, 3)− (1, 0, 1) = (1, 0, 2) and AC =(4,√

10, 2)− (1, 0, 1) =

(3,√

10, 1).

Hence

cos θ =AB ·AC

|AB| |AC| =(1, 0, 2) · (3,

√10, 1

)√

12 + 02 + 22

√32 +

(√10

)2+ 12

=12

and θ = 60 degrees or π3 radians.

Similarly, AD = (1, 2,−3) and

cos φ =AB ·AD

|AB| |AD| =(1, 0, 2) · (1, 2,−3)

√12 + 02 + 22

√12 + 22 + (−3)2

= −√

514

Since the dot product is negative the angle φ is obtuse.

1.1.15 Some further properties of the dot product

1. The non-zero vectors AB and AC are at right angles (are perpendicular AB ⊥ AC) if,and only if, (AB) · (AC) = 0.Such vectors are also called mutually orthogonal.

2. |u · v| ≤ |u| |v| for any vectors u and v (Cauchy-Schwartz inequality).

3. |u + v| ≤ |u|+ |v| for any vectors u and v (Cauchy’s inequality).

The first result follows from the fact that AB ⊥ AC if, and only if, cos θ = 0.The second result is left as an exercise (Exercise 2 No.8).To see the third result, use (2) and consider

|u + v|2 = (u + v) · (u + v)= u · u + v · v + 2u · v≤ u · u + v · v + 2 |u| |v|= |u|2 + |v|2 + 2 |u| |v|= (|u|+ |v|)2


As |u + v|2 ≤ (|u|+ |v|)2, it follows that |u + v| ≤ |u|+ |v|.

Example 1.1.3 Find a non-zero vector perpendicular to both u = (2, 1,−1) and v = (3, 3,−1).

Solution:

Let x = (x1, x2, x3) satisfy x ⊥ u and x ⊥ v. Then

x · u = 2x1 + 1x2 + (−1)x3 = 0x · v = 3x1 + 3x2 + (−1)x3 = 0

By subtraction −x1−2x2 = 0, or x1 = −2x2. Thus from the first equation, x3 = 2x1 +x2 =−3x2. We may let x2 be any non-zero number, e.g. x2 = 1. Then

x = (−2, 1,−3)

is perpendicular to u and to v.We will see later why it is always possible to find a vector x 6= 0 perpendicular to two non-zerovectors. This fact will also come into our study of planes (see 1.3.4).

1.1.16 Unit vectors. Components. Projections

A vector u of length 1 is called a unit vector. For example, u =(

35 , 0,− 4

5

)is a unit vector.

A vector c 6= 0 defines the vector

c =1|c|c

which is the unique unit vector having the same direction as c. To see that c has length1 we note

|c|2 = c · c =(

1|c|c

)·(

1|c|c

)=

1|c|2 c · c =

|c|2|c|2 = 1

The component of the vector f in the direction of (or along) the non-zero vector c isdefined as f · c.Suppose that f 6= 0 and that the angle between f and c is θ. Then

f · c =∣∣f

∣∣ cos θ (1.6)

To see equation (1.6), use equation (1.5):

f · c = f · 1|c|c =

f · c|c| =

∣∣f∣∣ |c| cos θ

|c| =∣∣f

∣∣ cos θ

(See Figure 1.6). Notice that if π2 < θ ≤ π then the component (1.6) is negative (as is only

natural).The projection of f on c is the vector

(f · c) c =

(f · c|c|2

)c (1.7)

Figure 1.7 shows the geometric interpretations of the projection of the non-zero vector f = PQon c and on −c. The projection PR is the same in both cases: PR is the projection (in thenatural sense) of the vector f on the line parallel to c passing through the tail P of f .

More generally, if d ‖ c, the projection of f on c is the same as the projection of f on d. (SeeExercise 2, No.13a.)


*

-PS = c

Figure 1.6

θS

f

|f | cos θ

|f | sin θ (f 6= 0)

Q

RP

¾PS = c

Á.............................

Á.............................-

f

--

f

PT = −c

Figure 1.7

P

θ

Q Q

R RP

π − θ

TS

The situation with components is different. As noted, in Figure 1.7 the component of f along−c is −|f | cos θ, . In general, if d and c are parallel with the same direction then the componentsof f along c and d are equal. If d and c have opposite directions then the component of falong d is the negative its component along c.

We return to projections in subsections 1.2.4 and 1.2.5.

Example 1.1.4 1. If we imagine that c is pointing due East and f is a displacement, then(f · c) c is the Easterly displacement, while

∣∣f∣∣ cos θ is the Easterly component and

∣∣f∣∣ sin θ

is the Northerly component f .

You may think of f as a vector representing a force that is dragging a box placed atits tail. Then

∣∣f∣∣ sin θ is the lifting effect of f while

∣∣f∣∣ cos θ is the horizontal dragging

effect of the force.

2. Let f = (11,−3, 4) and c = (5,−1,−2). The length of c is |c| =√

52 + (−1)2 + (−2)2 =√30.

The unit vector c in the direction of c is

c =1√30

(5,−1,−2)


The component of f along c is

f · c = (11,−3, 4) · 1√30

(5,−1,−2) =53

√30

and the projection of f on c can be found from equation (1.7):

(f · c) c =

53

√30

(1√30

(5,−1,−2))

=53

(5,−1,−2)

1.1.17 The special unit vectors i, j, k

The vectors i, j and k are defined as i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1). They haveunit length and are directed along the x−, y− and z−axis respectively. It is clear that theyare also mutually orthogonal and that any vector a = (a1, a2, a3) has a unique expression

a = a1i + a2j + a3k (1.8)

The scalars a1, a2 and a3 are the components of a in the directions i, j and k respectively. Seealso No.17 of Exercise 2 below.

1.1.18 Linear combinations

Let a and b be two vectors and λ and µ two scalars. The vector x = λa + µb is called a linearcombination of a and b with coefficients λ and µ. Similarly, for scalars r, s and t, the vectorra + sb + tc is a linear combination of a, b, and c with coefficients r, s and t.

Some concrete examples.

1. Equation (1.8) shows that every vector is a linear combination of i, j and k.

2. Let a = (1,−1, 2) and b = (−3, 2, 1).

4a + 5b = 4 (1,−1, 2) + 5 (−3, 2, 1) = (−11, 6, 13)

is a linear combination of a and b with coefficients 4 and 5.

3. If a = (−1, 2, 3), b = (5,−7, 1), c = (−4, 8,−1) then

12a +

23b +

(−5

6

)c =

(376

,−313

, 3)

is a linear combination of a, b, and c with coefficients 12 , 2

3 and − 56 .

Exercise 2 Partial Solutions

1. In Figure 1.2 express ON , MP , OM , LP , NR, NM and OP as vectors involving a,b and c. Write down these vectors as linear combinations of i, j and k and find theirlengths in terms of a, b and c.

Solution: ON = (0, 0, c) = MP (length |c|), OM = (0, b, 0) (length |b|),, OP = (0, b, c)(length |OP | =

√b2 + c2), LP = OP − OL = (0, b, c) − (a, 0, 0) = (−a, b, c) (length

|LP | = √a2 + b2 + c2) NR = OR−ON = (a, b, 0)− (0, 0, c) = (a, b,−c) (length |NR| =√

a2 + b2 + c2) NM = OM − ON = (0, b, 0) − (0, 0, c) = (0, b,−c). (length |NM | =√b2 + c2)

2. Interpret the equation (r − (2,−1, 5)) · (r − (2,−1, 5)) = 49 geometrically.

Solution: If r = (x, y, z) this reads (x− 2)2 + (y + 1)2 + (z − 5)2 = 49, so then equationrepresents the surface of a sphere centered at (2,−1, 5) and having radius 7.


3. Let u = (1, 3,−2), v = (2, 0, 1) and w = (3,−3, 4).

(a) Find the linear combinations u + 3v − 2w, (−5) u + 2v + 3w and 34u− 1

2v + 14w.

Solution: u + 3v − 2w = (1, 3,−2) + 3 (2, 0, 1)− 2 (3,−3, 4) = (1, 9,−7)(−5)u + 2v + 3w = −5 (1, 3,−2) + 2 (2, 0, 1) + 3 (3,−3, 4) = (8,−24, 24)34u− 1

2v + 14w = 3

4 (1, 3,−2)− 12 (2, 0, 1) + 1

4 (3,−3, 4) =(

12 , 3

2 ,−1)

(b) If the tail of u is the point A = (−2,−11, 13), what is the head of u? If the head ofw is B = (2,−9, 10), what is the tail of w?

Solution: If the head of u is B, then AB = OB − OA = u, so B = OB =OA+u = (−2,−11, 13)+(1, 3,−2) = (−1,−8, 11). (Note that A is the same as OA,so B −A = OB −OA).If the tail of w is A, then AB = OB − OA = w and so A = OA = OB − w =(2,−9, 10)− (3,−3, 4) = (−1,−6, 6).

(c) Find coefficients α, β such that w = αu + βv.

Solution: Want (3,−3, 4) = α (1, 3,−2) + β (2, 0, 1). These needs 3 = α + 2β,−3 = 3α, 4 = −2α + β. So α = −1. So 2β = 3− (−1) = 4, β = 2 and (if a solutionexists) 4 = −2 (−1) + β, yes! β = 2

(d) Is 3 (3,−1, 3) a linear combination of u, v and w?This means you must look for scalars p, q, r such that pu + qv + rw = 3 (3,−1, 3),and this involves three equations in three unknowns.

Solution: pu+qv+rw = p (1, 3,−2)+q (2, 0, 1)+r (3,−3, 4) = (p + 2q + 3r, 3p− 3r,−2p + q + 4r).Hence p + 2q + 3r = 9, 3p − 3r = −3, −2p + q + 4r = 9. There is no solution tothese three simultaneous equations, so the answer is NO.

4. (Generalization of No.3b). Let u be a given vector. Show that if the tail A of u is giventhen the head B of u is uniquely determined. State and prove a similar result if the headof u is given. Draw a picture.

Solution: If the head B of u is given then the tail A of u is uniquely determined.For, AB = OB −OA = u, so A = OB − u.

5. See Figure 1.8. Imagine that in each case the axes are part of a room in which you arestanding and looking at the corner indicated. Decide which of the systems of mutuallyorthogonal axes are left-handed and which right-handed. If you are outside the room,what would your answer be?

Solution: The first and third are right-handed systems, the other two are left-handed. Ifyou were looking at the room from the outside, exactly the opposite would apply.

6. Let A0,...,A9 be ten points in space and put am = Am−1Am for m = 1, · · · , 9 and a10 = A9A0.Find the sum a1 + · · ·+ a10 .

Solution: The answer is the zero vector 0, as the ten-sided vector-figure is closed.

7. Complete the proofs in statements 1.1.6, 1.1.9 and 1.1.11 above.

Hint 3 These properties reflect similar properties of numbers. For example, the commu-tative law of addition of vectors is proved by

u + v = (u1 + v1, u2 + v2, u3 + v3)= (v1 + u1, v2 + u2, v3 + u3)= v + u


yz zx x

xxyy

z yz

Figure 1.8

The proof of the first item of 1.1.9 is

α(u + v) = α(u1 + v1, u2 + v2, u3 + v3)= (α(u1 + v1), α(u2 + v2), α(u3 + v3))= (αu1 + αv1, αu2 + αv2, αu3 + αv3)= αu + αv

8. Complete the proofs of subsections 1.1.13 and 1.1.15.

Hint 4 For the Cauchy-Schwartz inequality it may be assumed that the vectors are non-zero. For a geometric proof use equation (1.5). For a purely algebraic proof see the nextquestion.

9. (a) Show that a dot product (αr + βs) · (γu + δv) of linear combinations multiplies outjust like in ordinary arithmetic.

Solution: (αr + βs) · (γu + δv) = (αr + βs) · γu + (αr + βs) · δv= αr · γu + βs · γu + αr · δv + βs · δv= (αγ) (r · u) + (βγ) (s · u) + (αδ) (r · v) + (βδ) (s · v).

(b) Show that |αu + βv|2 = α2 |u|2 + β2 |v|2 + 2αβu · v.Solution: If α = γ and β = δ and r = u and s = v, we get

(αα) (u · u) + (βα) (v · u) + (αβ) (u · v) + (ββ) (s · v) = α2 |u|2 + β2 |v|2 + 2αβu · v,since u · v = v · u.

(c) By putting α = |v| and β = − |u| in 9b prove the Cauchy-Schwartz inequality.

Solution:0 ≤ |αu + βv|2 = |v|2 |u|2 + (− |u|)2 |v|2 + 2 |v| (− |u|)u · v = 2 |v|2 |u|2− 2 |v| |u|u · v.Since we mat assume that the vectors are non-zero, cancelling |v| |u|, gives the result.

10. Show that if the vectors u and v in equation (1.5) are unit vectors, then cos θ = u · v.Solution: Obvious.

11. Show that the non-zero vectors u and v are parallel if, and only if, |u · v| = |u| |v|.Solution: The vectors are parallel if, and only if, the angle between them is 0 or πradians.


12. Let a = (3,−1, 2) and b = (−7, 3,−1). Find

(a) The unit vectors a and b.

Solution: a = 1√9+1+4

(3,−1, 2) = 1√14

(3,−1, 2), b = 1√49+9+1

(−7, 3,−1) =1√59

(−7, 3,−1) .

(b) The component of a along b and the component of b along a.

Solution: The component of a along b is a · b = (3,−1, 2) · 1√59

(−7, 3,−1) = − 26√59

.The component of b along a is b · a = (−7, 3,−1) · 1√

14(3,−1, 2) = − 26√

14

(c) The projection of a on b and the projection of b on a.

Solution: The projection of a on b is

(a · b

)b =

a · b|b|2 b = −26

59b = −26

59(−7, 3,−1)

The projection of b on a is

b · a|a|2 a = −26

14(3,−1, 2) = −13

7(3,−1, 2)

13. (a) If γ 6= 0, show that the projections of f on c and f on γc are equal.

Solution: The projection of f on γc is

f · γc

|γc|2 γc =γ2f · cγ2 |c|2 c =

f · c|c|2 c

(b) Show that if f 6= 0 then the projection of f on itself is f .

Solution: The projection of f on f is

f · f∣∣f ∣∣2 f = f

(c) Let g be the projection of f on c 6= 0. Show that the component of f along c is g · c.Solution: g =

(f · c) c, so that

g · c =[(

f · c) c] · c =

(f · c) (c · c) = f · c

(We have used property 2 of subsection 1.1.13).

14. A man walks from a point A in the N-E direction for 15 km to a point B. Find thecomponent of the displacement AB in the following directions

(a) East

(b) South

(c) N30◦E

Solution: Taking the unit vectors i and j due East and due North respectively,AB = 15√

2(1, 1). Hence the respective answers are 15√

2(1, 1) · (1, 0) = 15

2

√2 km,

15√2

(1, 1) · (0,−1) = − 152

√2 km and 15√

2(1, 1) · ( 1

2 ,√

32 ) = 15

2

√2

(12

√3 + 1

2

)= 14. 489

km.


15. A wind is blowing at 10 km/h in the direction N30◦W. What is its component in thedirection S15◦W?

Solution: Solution: Using the same unit vectors as before, the velocity of the wind isw = 10

(− 1

2 ,√

32

). A unit vector in the direction S15◦W is (cos (−105) , sin (−105)). The

required component is 10(− 1

2 ,√

32

)· (cos (−105) , (−105)) = −7. 071 1 km/h.

16. Let u and v be two vectors. The set of all linear combinations of u and v is called thespan of the vectors and is denoted by sp (u, v). (This is discussed more fully in item 2 ofsubsection 3.1.2 of Chapter 3).

(a) Show that if u = (−1, 2, 3) and v = (2, 4,−5), then sp (u, v) consists of all vectors ofthe form (−α + 2β, 2α + 4β, 3α− 5β), where α and β are arbitrary scalars.

(b) How do you think the span of three vectors u, v, w should be defined?Describe sp ((−2, 3, 1), (5,−4, 2), (−1, 1,−3)).

(c) * Let u, v, a and b be four vectors. Show that

sp (u, v) ⊆ sp (a, b)

if, and only if each of the vectors u and v are linear combinations of the vectors aand b.

SolutionLet u and v be linear combinations of a and b, say u = pa + qb and v = ra + sb forscalars p, q, r, s. Then for scalars α and β,

αu + βv = α (pa + qb) + β (ra + sb) = (αp + βr) a + (αq + βs) b

This shows that sp (u, v) ⊆ sp (a, b). Since the converse is obvious, the result follows.

17. * Let l, m and n be three mutually orthogonal unit vectors. Placing the tails of thesevectors at the origin, it is visually clear that we may use these vectors as a frame ofreference in place of i, j and k. (An analytic proof of this will follow from work done inChapter 3 - See Exercise 78, No.9). Assuming this result, any vector a can be expressedas a linear combination

a = γ1l + γ2m + γ3n

Show that necessarily the coefficients γ1, γ2, γ3 are the components (in the above sense)of a in the directions l, m and n respectively. In fact, γ1l, γ2m, γ3n are the projectionsof the vector a on l, m and n respectively. Conclude that

|a|2 = γ21 + γ2

2 + γ23

Show that l = 1√6

(1,−1, 2), m = 1√3

(−1, 1, 1) and n = 1√2(1, 1, 0) is such a system of

three mutually orthogonal unit vectors and find the coefficients γ1, . . . , γ3, if a = (−3, 2, 1).

Hint 5 Take the dot product of both sides of the above equation with each of l, m and n.The coefficients for the specific example are the components a · l = γ1 etc. A longerprocedure to find the coefficients for the specific example is to use the method of 3d above.

Solution: Since l ·m = l · n = 0, it follows that l · a = γ1. Etc.For the specific example, we find the components γ1 = a · l = (−3, 2, 1) · 1√

6(1,−1, 2) =

− 12

√6, γ2 = a·m = (−3, 2, 1)· 1√

3(−1, 1, 1) = 2

√3, and γ3 = a·n = (−3, 2, 1)· 1√

2(1, 1, 0) =

− 12

√2.


18. What familiar result from Euclidean Geometry does |b− a| ≤ |a|+ |b| represent?

Solution: “The sum of two sides of a triangle is larger than the third”. If |b− a| =|a|+ |b|, what is happening? Then (b− a) · (b− a) = (|a|+ |b|)2 and a · b = − |a| |b|: Thevertices of the triangle are on a straight line - the triangle is degenerate.

19. Let u and v have the same length: |u| = |v|. Show that (u− v) · (u + v) = 0. In otherwords, if u− v and u + v are not zero they are at right angles.

Solution: (u− v) · (u + v) = (u− v) · u + (u− v) · v = u · u − v · u + u · v − v · v =−v · u + u · v = 0.

20. How are our general results affected (and simplified) when one of the coordinates is re-stricted to be zero for all vectors under consideration?

1.2 The straight line

Two distinct points determine a line. Points that lie on one and the same line are calledcollinear.

Let P , Q and R be three distinct points. Then it is geometrically obvious that if R lies on the(infinite) line through P and Q then PQ is parallel to PR. We will see that our definition of astraight line conforms to this criterion. See Exercise 6, No.2.

1.2.1 Generic (parametric) representations of a line `

Let A and B be two distinct points. Then as t ∈ < varies, the point R with position vector

r = r (t) = OA + tAB (1.9)

varies over the entire straight line ` through A and B.As in Figure 1.9, from O jump onto the line at A and then go parallel to AB to R. The pointR = A is obviously on ` and corresponds with t = 0. If R 6= A we have AR = tAB for somet 6= 0, so in general

r = OR = OA + AR = OA + tAB

When t = 1 then we are at R = B.As r = r (t) represents a general point on the line `, the equation (1.9) is called a generic or

parametric equation of the line through A and B. The variable t is called the parameter.

1.2.2 Some general observations

In what follows a = OA, b = OB, etc.

1. As t varies over the set < of real numbers the point r = OR = a + t (b− a) of equation(1.9) varies continuously over the line `. In this representation of the line ` we considerb− a = AB as the direction of `. In Figure 1.9 as t increases R moves to the right, as tdecreases R moves to the left.

2. As remarked, t = 0 corresponds to R = A and t = 1 to R = B. If t is between 0 and 1,say t = 1

2 we get

OR = a +12

(b− a) =12

(a + b)

Thus 12 (a + b) is the position vector of the point midway between A and B.

1.2. THE STRAIGHT LINE 23

..

¸

*

O

A

B

`

R

tABµ

OA

r = OR = OA + tAB

Figure 1.9

3. More generally, a point

r = r(t) = a + t (b− a) = (1− t) a + tb

where 0 ≤ t ≤ 1 lies between A and B and the set AB of these points is called the linesegment joining A and B.If A 6= B and 0 < t < 1, the point r(t) is said to lie strictly between A and B.The line segment AB is a set of points and it must be distinguished from the vector ABand also from the distance |AB|.

4. Let u 6= 0 define a direction and suppose a is the position vector of the point A. Then ageneric equation of the line ` through A with direction u is

r (t) = a + tu (1.10)

A typical point a + tu on the line is called a generic point. As t varies over the realnumbers in the generic equation (1.10) we obtain a set of points which we identify with` and (1.10) is its analytic definition.

5. A generic equation is not unique.

(a) If C 6= D are two points on the line given by equation (1.9), then since these pointsalso determine the line, another generic equation (now with parameter s) is

r (s) = OC + sCD

The parameters t and s are of course related. See Example 1.2.1 of subsection 1.2.3and No.1b of Exercise 6.

(b) When do two generic equations like (1.10) define the same line? Let u and v benon-zero vectors and suppose that a + tu is a generic equation of line `1 and b + sva generic equation of line `2.Then `1 = `2 if, and only if, u ‖ v and a− b is a multiple of u. (See No.4 of Exercise6).


1.2.3 Some illustrative Examples

Example 1.2.1 Find two generic equations for the line ` passing through A = (1, 2, 3) andB = (−1, 1, 4).

Solution: One generic equation is

r = (1, 2, 3) + t ((−1, 1, 4)− (1, 2, 3))= (1, 2, 3) + t (−2,−1, 1) = (1− 2t, 2− t, 3 + t)

Another is

r = (−1, 1, 4) + s ((1, 2, 3)− (−1, 1, 4))= (−1, 1, 4) + s (2, 1,−1) = (−1 + 2s, 1 + s, 4− s)

In the second representation we have used r = OB + sBA, with parameter s. The relation-ship between s and t is s = 1− t, as can be easily seen.

Example 1.2.2 Find the position vector of the point R lying on the line ` of Example 1.2.1that is a distance 1

3 |AB| from A but on the side of A opposite to that of B.

Solution: To find R put t = − 13 in OR = OA + tAB to get

OR = OA +(−1

3

)AB

= (1, 2, 3) +(−1

3

)(−2,−1, 1)

=(

53,73,83

)

Example 1.2.3 Find the foot Q of the perpendicular from the point P = (2,−1, 4) to the line` of Example 1.2.1. Hence find the shortest distance between P and `.

Solution:

Let Q = (1− 2t, 2− t, 3 + t) be the required foot of the perpendicular from P = (2,−1, 4) to theline `. Then PQ = (−1− 2t, 3− t,−1 + t) must be perpendicular to the direction (−2,−1, 1)of `:

(−1− 2t, 3− t,−1 + t) · (−2,−1, 1) = 0

This gives t = 13 and so Q = q =

(13 , 5

3 , 103

). The shortest distance of P from ` is thus∣∣PQ

∣∣ =∣∣(− 5

3 , 83 ,− 2

3

)∣∣ = 13

√93. (See Figure 1.10).

1.2.4 A general result for the foot of a perpendicular to a line

There is a better way to visualize the position vector q of the foot Q of the perpendicular fromfrom P to the line ` with generic equation r = a+ tu. From Figure 1.11 we see that AQ = q−ais the projection of AP = p− a on the vector u. Using equation (1.7) with f = p− a and c = uwe obtain

q = a +

(1|u|2

(p− a

) · u)

u (1.11)

Note that (as in example 1.2.3) the shortest distance from P to ` is |PQ|.


Q

P = (2,−1, 4)

Figure 1.10

r = (1− 2t, 2− t, 3 + t)

A

Q

P

*

Á

u

AQ = q − a

Figure 1.11

AP = p− a

line `

Example 1.2.4 Let us apply the formula (1.11) to the above example 1.2.3.

Solution:

For the line (1, 2, 3) + t (−2,−1, 1) we have |u|2 = 6 and the foot of the perpendicular fromP = (2,−1, 4) to the line has position vector

q = (1, 2, 3) +(

16

((2,−1, 4)− (1, 2, 3)) · (−2,−1, 1))

(−2,−1, 1)

=(

13,53,103

)

This is the same result as before.

In Exercise 6, No.9 you are asked to derive the general formula 1.11 using the the first methodused to solve this problem.

Example 1.2.5 Find the foot Q of the perpendicular from a general point P = (p1, p2, p3) tothe line r = a + tu if a = (1, 2, 3) and u = (−2,−1, 1).


Solution:

Using equation (1.11) - effectively equation (1.7) - we obtain

q − a =((p1, p2, p3)− (1, 2, 3)) · (−2,−1, 1)

|(−2,−1, 1)|2 (−2,−1, 1) =−2p1 − p2 + p3 + 1

6(−2,−1, 1)

=(−1

3+

23p1 +

13p2 − 1

3p3,−1

6+

13p1 +

16p2 − 1

6p3,

16− 1

3p1 − 1

6p2 +

16p3

)

and so

q =(

23

+23p1 +

13p2 − 1

3p3,

116

+13p1 +

16p2 − 1

6p3,

196− 1

3p1 − 1

6p2 +

16p3

)

1.2.5 Projections on a line through the origin

When a line ` passes through the origin, the foot Q of the perpendicular from P to ` is calledthe projection of P on `.The projection of p on the line tu is just the projection of p on the vector u and is the resultof putting a = 0 in equation (1.11):

q =

(p · u|u|2

)u (1.12)

Example 1.2.6 Find the projection of P = (p1, p2, p3) on the line t (−2,−1, 1).

Solution:

We can find the required projection by letting a = 0 in example 1.2.5 or directly from (1.12):

q =(p1, p2, p3) · (−2,−1, 1)

|(−2,−1, 1)|2 (−2,−1, 1) =−2p1 − p2 + p3

6(−2,−1, 1)

=(

23p1 +

13p2 − 1

3p3,

13p1 +

16p2 − 1

6p3,−1

3p1 − 1

6p2 +

16p3

)(1.13)

We will return to this example in Chapter 3 (See Example 3.3.1).

Example 1.2.7 A straight line lies in the x − y plane and passes through the origin. Theline makes an angle of 30◦ with the positive x−axis. Find the foot Q of the projection fromP = (p1, p2) on the line.

Solution:

Let u =(cos π

6 , sin π6

)=

(√3

2 , 12

). The parametric equation of the line is r = t

(√3

2 , 12

)and

the foot Q of the perpendicular from P to the line is the projection of p = (p1, p2) on the unitvector u:

q =

(√3p1

2+

p2

2

)(√3

2,12

)=

(14

(3p1 +

√3p2

),14

√3p1 +

14p2

)(1.14)

We shall return to this example in subsection 1.4 below and again in Chapter 3 Example 3.2.1.


7 Á

-6

-

A

B C

D

1..............................................

XAB = DCAD = BC

AX = XC

Figure 12

A well-known result on parallelograms

Example 1.2.8 Show that the diagonals of a parallelogram bisect each other.

Solution:

In Figure 1.12 ABCD is a parallelogram with AB = DC and BC = AD. It is understood thatthe parallelogram is non-degenerate in that no three of A, B, C, D are collinear.

Let X be the midpoint of AC. ThenAB + BX = AX = XC = XD + DC = XD + AB, so BX = XD.

1.2.6 Intersecting, parallel and skew lines

Let `1 and `2 given by generic equations a1 + su1 and a2 + tu2 respectively be two straightlines. They are parallel if u1 ‖ u2.Exactly one of the following holds (see Exercise 6, No.5* for a rigorous proof):

1. The lines are identical.

2. They are distinct and parallel.

3. They are not parallel and meet in a (unique) point.

4. They are not parallel and do not intersect (they are by definition skew).

`1

P

Q

`2 Figure 13


A familiar example of skew lines is provided by a road passing under a railway bridge. Atrain and car can only crash at a level crossing!In the case of skew lines there will be points P on `1 and Q on `2 such that PQ is perpendicularto the directions of both lines and the distance

∣∣PQ∣∣ is minimum. See Figure 1.13.

Example 1.2.9 Let `1 be the line through A = (0, 0, 0) and B = (1, 2,−1) and suppose `2 isthe line through C = (1, 1, 1) and D = (2, 4, 3). Are the lines skew? In any case, find theshortest distance between them.

Solution: Generic equations for `1 and `2 are t(1, 2,−1) and (1, 1, 1)+s(1, 3, 2) respectively.Since (1, 2,−1) 6‖ (1, 3, 2), the lines are not parallel (and so cannot possibly be identical). Theywill meet if for some t and s

t(1, 2,−1) = (1, 1, 1) + s(1, 3, 2)

This means that the three equations t = 1+s, 2t = 1+3s and −t = 1+2s must hold. From thefirst and last equation, 1+s = − (1 + 2s) and thus s = − 2

3 , t = 13 . But then 2t = 2

3 6= 1+3(− 2

3

)and the second equation fails. It follows that `1 and `2 are skew.Let P = t(1, 2,−1) and Q = (1, 1, 1) + s(1, 3, 2) be the points where PQ is minimum. ThenPQ = (1, 1, 1)+s(1, 3, 2)− t(1, 2,−1) is perpendicular to both directions (1, 2,−1) and (1, 3, 2).Hence:

((1, 1, 1) + s(1, 3, 2)− t(1, 2,−1)) · (1, 2,−1) = 2 + 5s− 6t = 0((1, 1, 1) + s(1, 3, 2)− t(1, 2,−1)) · (1, 3, 2) = 6 + 14s− 5t = 0

Solving these two equations gives s = − 2659 and t = − 2

59 . Hence,

PQ = (1, 1, 1) + s(1, 3, 2)− t(1, 2,−1) =(

3559

,−1559

,559

)

The shortest distance between the lines is∣∣( 35

59 ,− 1559 , 5

59

)∣∣ = 5√59

.


1. let A = (1, 2,−3), B = (−1, 1, 2), C = (5, 4,−13), D = (−5,−1, 12) and E = (−1,−2, 3).Answer the following questions.

(a) Find a generic equation with parameter t of the straight line ` through A and B inthe form of equation (1.9).

Solution: Generic equation is (1, 2,−3)+ t ((−1, 1, 2)− (1, 2,−3)) = (1− 2t, 2− t,−3 + 5t).(Direction with this parametrization is (−2,−1, 5)).

(b) Is E on `? Show that C and D are on ` and find a corresponding generic equationof the line with parameter s. Find the relationship between s and t.

Solution: AC = (5, 4,−13)− (1, 2,−3) = (4, 2,−10) = (−2) (−2,−1, 5)AD = (−5,−1, 12) − (1, 2,−3) = (−6,−3, 15) = 3 (−2,−1, 5). Hence C and D areon `.Another generic equation for ` is (5, 4,−13)+s ((−5,−1, 12)− (5, 4,−13)) = (5− 10s, 4− 5s,−13 + 25s)The relationship between the two parameters is 1 − 2t = 5 − 10s or 5s − t = 2.AE = (−1,−2, 3)− (1, 2,−3) = (−2,−4, 6) / (−2,−1, 5). So E is not on `.

(c) Find two points on ` at a distance√

24 from E.

Solution. E = (−1,−2, 3). Let (1− 2t, 2− t,−3 + 5t) be such a point. (1− 2t, 2− t,−3 + 5t)−(−1,−2, 3) = (2− 2t, 4− t,−6 + 5t).We want |(2− 2t, 4− t,−6 + 5t)| =

√24 or (2− 2t)2 + (4− t)2 + (−6 + 5t)2 = 24,

Solution is : t = 815 , t = 2. The two points are: With t = 2, (1− 2t, 2− t,−3 + 5t) =

(−3, 0, 7).and with t = 8

15 , (1− 2t, 2− t,−3 + 5t) =(− 1

15 , 2215 ,− 1

3

).


(d) Find an equation describing the line segment joining A and B. In particular find themidpoint of the segment and also the three points strictly between A and B dividingthe segment into four equal parts.

Solution: (1− 2t, 2− t,−3 + 5t) (0 ≤ t ≤ 1) describes segment AB. t = 12 defines

midpoint (1− 2t, 2− t,−3 + 5t) =(0, 3

2 ,− 12

). The required three points dividing the

segment into 4 equal parts are the midpoint and the points corresponding to t = 14

and t = 34 , i.e.

(12 , 7

4 ,− 74

)and

(− 12 , 5

4 , 34

)respectively.

2. Let P , Q and R be three distinct points on the line of equation (1.10). Show that they arecollinear, i.e., PQ ‖ PR.

Hint 7 Let P = a + t1u, Q = a + t2u and R = a + t3u.

Solution: PQ = (t2 − t1)u and PR = (t3 − t1)u. Since t1, t2, t3 are distinct, it followsthat PQ = t2−t1

t3−t1PR, and the result follows.

3. Let the generic equation r(t) = a + tu represent a line `, where u 6= 0, as in the previousquestion. Suppose that P with position-vector p is any point. Show that P lies on `

(a) if, and only if, p− a is a multiple of u.(b) if, and only if, the equation

a + tu = p

has a unique solution for t.

(Compare this with No.12 below for another criterion).

4. * Let u and v be non-zero vectors and suppose that a+ tu and b+sv are generic equationsof lines `1 and `2 respectively.Show that necessary and sufficient conditions for the generic equations to represent thesame set of points (i.e. `1 = `2) are

(i) u ‖ v and(ii) a− b is a multiple of v.

Using this result show that two lines sharing two distinct points are identical.

Solution:Suppose that `1 = `2. Then a = b + tv for some t (showing (ii) holds). Furthermore,a + u = b + t1v for some t1. Subtracting, u = (t1 − t) v, so (i) holds.

Conversely, let the conditions (i) and (ii) hold. Then u = βv for some scalar β anda− b = γv for a scalar γ. So, given t, we can solve

a− b + tu = (γ + tβ) v = sv

(uniquely) for s. This shows that all points on `1 are on `2. Since the conditions (i) and(ii) are symmetrical (that is, v ‖ u and b − a is a multiple of u), a symmetric argumentshows that all points on `2 are on `1. Thus `1 = `2.Leta+ tu = b+ sv and a+ t1u = b+ s1v where t 6= t1 and s 6= s1. Then (t− t1)u = (s− s1)v.It follows that (i) and (ii) hold and that the lines are identical.

5. * Use the results of the previous exercise to show that the four conditions in subsection1.2.6 are in fact mutually exclusive and exhaustive.

Solution:Either (A) u ‖ v or (B) not. If case (A) holds, suppose the lines have a point in common,so that a + tu = b + sv has a solution for s and t. Then it follows that a− b is a multipleof v and so the lines are identical.If (B) is the case the lines either meet or do not meet.


6. Using equation (1.11), or otherwise, find the point Q of the line ` with parametric equation(1− 2t, 2− t,−3 + 5t) which is closest to

(a) the origin,

Solution: Q is(− 4

15 , 4130 , 1

6

)

(b) the point P = (5, 8, 4),

Solution: Q =(− 2

5 , 1310 , 1

2

)

(c) the point P = (−3, 2, 1).

Solution: Q =(− 13

15 , 1615 , 5

3

)

7. Find the projections of the following points P on the line (−2t,−t, 5t).

(a) P = (4, 6, 7)

Solution: The projection is

(4, 6, 7) · (−2,−1, 5)|(−2,−1, +5)|2 (−2,−1, 5) =

710

(−2,−1, 5)

(b) P = (−1,−2, 3).

Solution: The projection is

(−1,−2, 3) · (−2,−1, 5)|(−2,−1, 5)|2 (−2,−1, 5) =

1930

(−2,−1, 5)

8. Find the projection of the point P with position vector p = (p1, p2, p3) on the line (−2t,−t, 5t).

Solution. Direction of line is (−2,−1, 5). Projection is

(p1, p2, p2) · (−2,−1, 5)|(−2,−1, 5)|2 (−2,−1, 5) =

−2p1 − p2 + 5p3

30(−2,−1, 5)

9. * Derive the formula (1.11) that finds the foot Q of the perpendicular from P = (p1, p2, p3)onto the line a + tu using the technique of example 1.2.3.

Hint 8 Show that the solution to(a + tu− p

) · u = 0 is

t = −(a− p

) · u|u|2

Solution: Substitute this value of t in a + tu to get the required formula

a +

(−

(a− p

) · u|u|2

)u = a +

((p− a

) · u|u|2

)u

10. Show that the closest the line ` with parametric equation a + tu comes to the point P is

d =1|u|

√(|p− a||u|)2 − ((p− a

) · u)2 (1.15)


Solution: From equation (1.11) we get the square d2 of the required distance as follows.Let a− p = s, then

d2 =∣∣q − p

∣∣2 =

∣∣∣∣∣a− p−((

a− p) · u

|u|2)

u

∣∣∣∣∣

2

=

(s−

(s · u|u|2

)u

)·(

s−(

s · u|u|2

)u

)

= |s|2 − 2

(s · u|u|2

)s · u +

(s · u|u|2

)2

|u|2

= |s|2 − (s · u)2

|u|2

=1|u|2

(∣∣a− p∣∣2 |u|2 − ((

a− p) · u)2

)

11. Show that |r (t)|2 in equation (1.10) is a quadratic in t. How does your knowledge ofquadratics tie up with the formula (1.11) and equation (1.15)?

Solution: |r (t)|2 = (a + tu) · (a + tu) = |a|2 + 2ta · u + t2 |u|2. The minimum of this iswhen

t = −a · u|u|2

In other words, r (t) comes closest to the origin for this value of t. But r (t) comes closestto the point p just when r (t)− p comes closest to the origin. This gives

t = −(a− p

) · u|u|2

exactly as before.

12. Deduce from equation (1.15) that ` passes through the point P if, and only if,(p − a) · u = ±|p − a||u|. Note that geometrically this is clear in view of the formula forthe angle between p− a and u if p− a 6= 0.

Solution: ` passes through the the point P if, and only if, the shortest distance d fromP to ` is 0 which is the case if, and only if, |a− p|2|u|2 − ((

a− p) · u)2 = 0 if, and only

if, (a− p) · u = ±|a− p||u|.13. * If t is time in seconds and u is the displacement in meters per second (the constant

velocity), then equation (1.10) can be interpreted as the position of (say) an aircraft inmeters at time t.

(a) In a situation like this we would be interested in how close the aircraft comes to acertain point P for t ≥ 0. Find a condition for (1.15) to hold at some time t ≥ 0.If this condition fails, what is the closest the aircraft gets to P and when does thisoccur?

Solution. Equation (1.15) holds for some t ≥ 0 just when(a− p

) · u ≤ 0. If thiscondition fails, what is the closest the aircraft gets to P is when t = 0 and the closestis

∣∣a− p∣∣.

(b) How can your findings be used to find the closest distance two aircraft travellingwith constant velocities get to each other, assuming that as time t ≥ 0 increases thedistance between the aircraft decreases?


Hint 9 Let a + tu and b + tv describe the positions of the two aircraft at time t.Consider the line b−a+ t (v − u). You want its closest distance from the origin. Byassumption, why is (b− a) · (v − u) < 0?

Solution: (b− a) · (v − u) < 0 is the condition found in (b) since the aircraft getcloser and closer as t increases.

14. A rhombus is a (non-degenerate) parallelogram all of whose sides have the same length.Show that the diagonals of a rhombus intersect at right angles.

Hint 10 In Example 1.2.8 (Figure 1.12) use |AB| = |BC| as well as AC = AB + BCand AB + BD = AD, so BD = AD − AB = BC − AB. Now use the result of Exercise2 No.19.

Solution: Follows straight from the hint: (AB + BC) · (BC −AB) = 0

15. *(Generalization of previous exercise). Let a + tu and b + sv represent two straight linesand suppose that A = a + t1u, B = b + s1v, C = a + t2u, and D = b + s2v are points onthese lines with A 6= C and B 6= D.Show that |AB|2 + |CD|2 = |BC|2 + |DA|2 if, and only if, the lines are perpendicular.

Hint 11 Expand |AB|2 = (b + s1v − (a + t1u)) · (b + s1v − (a + t1u)) etc. and show that|AB|2 + |CD|2 − |BC|2 − |DA|2 = 2u · v(t1 − t2)(s2 − s1).

16. Let A, B and C be three points not on a line and so forming the distinct vertices of atriangle. The median from A is the line joining A to the midpoint of the opposite sideBC. Show that all three medians intersect in a single point M . This point is called thecentroid of the triangle and is the centre of mass of three equal masses placed at thevertices of the triangle.

Hint 12 Let a, b, and c, be the position vectors of A, B and C respectively. Then themidpoint of BC has position vector 1

2 (b + c). A generic equation of the line through Aand the midpoint of BC is

a + t

(12(b + c)− a

)

Put t = 23 and simplify the result. You should conclude that the centroid lies two thirds of

the way from any vertex along its median toward the opposite side.

Solution:

a +23

(12(b + c)− a

)=

13

(a + b + c)

The symmetry of this equation proves that the medians intersect at the point M =13 (a + b + c). The distance of A from M is

23

∣∣∣∣(

12(b + c)− a

)∣∣∣∣

17. (Cartesian equations for a straight line). Let equation (1.10) define a straight line, whereu = (u1, u2, u3) and each ui 6= 0. Show that r = (x, y, z) is on the line if, and only if,

x− a1

u1=

y − a2

u2=

z − a3

u3

What happens if one or two of the ui = 0? Find Cartesian equations for the line ofQuestion 1a above. Find generic and Cartesian equations for the x−, y− and z−axis.

1.3. PLANES 33

Solution: We have(x, y, z) = (a1, a2, a3) + t(u1, u2, u3)

This says x−a1u1

= y−a2u2

= z−a3u3

= t, and the result follows. If say u1 = 0, then theCartesian equations are x = a1 and y−a2

u2= z−a3

u3. If u1 = u2 = 0, equations are x = a1,

y = a2.For the line (1− 2t, 2− t,−3 + 5t) = (1, 2,−3) + t (−2,−1, 5) of 1a the Cartesian equa-tions are

x− 1−2

=y − 2−1

=z + 3

5Generic equations for the x−, y− and z−axis are r(1, 0, 0), s(0, 1, 0), and t(0, 0, 1); Carte-sian equations are y = z = 0, x = z = 0 and x = y = 0 respectively.

18. Let `1 be the line through A = (−1,−2, 6) and B = (3, 2,−2) and `2 the line throughC = (1, 3, 1) and D = (1, 1, 5). Solve the following problems.

(a) Show that `1 and `2 are skew.Solution: Try to solve for s and t:

(−1,−2, 6) + s (4, 4,−8) = (1, 3, 1) + t (0,−2, 4)

Gives −2+4s = 0, −5+4s+2t = 0, 5− 8s− 4t = 0. Hence s = 12 , −5+2+2t = 0,

t = 32 . But 5 − 8

(12

) − 4(

32

)= 5 6= 0 and lines are skew as they are clearly not

parallel.

(b) Find points P on `1 and Q on `2 such that PQ is perpendicular to the directions ofboth lines. Hence find the shortest distance between `1 and `2.

Solution: We may suppose that P = (−1,−2, 6) + s (4, 4,−8) and Q = (1, 3, 1) +t (0,−2, 4).

QP = (−1,−2, 6) + s (4, 4,−8)− ((1, 3, 1) + t (0,−2, 4)))= (−2 + 4s,−5 + 4s + 2t, 5− 8s− 4t)

We must have QP perpendicular to the directions of both `1 and `2. Hence(−2 + 4s,−5 + 4s + 2t, 5− 8s− 4t)·(1, 1,−2) = −17+24s+10t = 0 and (−2 + 4s,−5 + 4s + 2t, 5− 8s− 4t)·(0,−1, 2) = 15 − 20s − 10t = 0. This gives 4s = 2, s = 1

2 and t = 12 . We get P =

(−1,−2, 6)+ 12 (4, 4,−8) = (1, 0, 2) and Q = (1, 3, 1)+ 1

2 (0,−2, 4) = (1, 2, 3). Shortestdistance of lines apart is

∣∣PQ∣∣ = |(1, 2, 3)− (−1,−2, 6)| = |(2, 4,−3)| = √

29.

19. Let `1 be the line through A = (2,−1, 4) and B = (3, 0, 5) and `2 the line throughC = (−1, 0, 1) and D = (−2, 1, 0). Show that `1 and `2 meet and find the point ofintersection.

Solution: Solve for s and t: (2,−1, 4)+s ((3, 0, 5)− (2,−1, 4)) = (−1, 0, 1)+t ((−2, 1, 0)− (−1, 0, 1)).We have (3 + s + t,−1 + s− t, 3 + s + t) = (0, 0, 0). The solution is given by s − t = 1,s + t = −3 so s = −1, t = −2. Point of intersection is (1,−2, 3).

20. How are our general results affected (and simplified) when one of the coordinates (say thez− coordinate) is restricted to be zero? In particular what does equation (1.11) reduce to?

1.3 Planes

1.3.1 Generic Equation for a plane

Our first object is to find a generic equation for the plane passing through three non-collinearpoints A = (a1, a2, a3), B = (b1, b2, b3) and C = (c1, c2, c3).


7

A

qÁ

Á

q

q

B

C

R

P

Q

tAC = AQ = PR

sAB = AP = QR

}

6

0Figure 14

If four points lie on a plane, they are called coplanar. So if R is a point, our first questionis “when are A, B, C and R coplanar?” The answer is suggested by Figure 1.14:Suppose that R lies in the same plane as A, B and C. Let the line through R parallel to ACmeet the line through A and B at the point P . Then AP = sAB for some scalar s. Similarly,let the line through R parallel to AB meet the line through A and C at the point Q, so thatAQ = tAC for some scalar t. The figure APRQ forms a parallelogram and

OR = OA + AR = OA + (AP + PR) = OA +(AP + AQ

)

= OA + sAB + tAC (1.16)

This equation describes the position vector r = OR of a general point on the plane throughA, B and C and is called a generic equation of the plane through A, B and C.

Example 1.3.1 Let A = (1,−1, 2), B = (1, 2, 3) and C = (−1, 0, 1) be three given points.They are not collinear since the vectors AB = (0, 3, 1) and AC = (−2, 1,−1) are not parallel.Hence a generic equation of the plane through A, B and C is

r = (1,−1, 2) + s(0, 3, 1) + t(−2, 1,−1)

In equation (1.16) the vectors u = AB and v = AC are non-zero and non-parallel and arethought of as parallel to the plane described. This suggests the following definition:

1.3.2 Analytic definition of a plane

Let u and v be two non-zero, non-parallel vectors and A = a a given point. Then the set Π ofpoints R = r satisfying the generic (parametric) equation

r = r(s, t) = a + su + tv (1.17)

describes a plane. The parameters are s and t and we call a + su + tv a generic point of theplane.

1.3. PLANES 35

1.3.3 Equality of planes

As with lines, a generic equation for a plane is not unique. Let

a1 + s1u1 + t1v1 and a2 + s2u2 + t2v2

Represent planes Π1 and Π2 respectively. Then Π1 = Π2 if, and only if,

(i)sp (u1, v1) = sp (u2, v2)

and

(ii) a1 − a2 is in this common span.For the meaning of ‘span’ see No.16 of Exercise 2 and compare No.4 of Exercise 6.

The proof of the above statement is left as an exercise. See No.13 of Exercise 15 below.

1.3.4 Cartesian Equation of a plane

A vector n = (n1, n2, n3) 6= 0 is called a normal of the plane defined by equation (1.17) if itis perpendicular to every line lying in the plane. So for a variable point R on the plane wemust have n · AR = 0. Thus as in Figure 1.15, all points on the plane with position vectorr = (x1, x2, x3) must satisfy

n · (r − a) = 0

6

:

A

n

R = (x1, x2, x3)

plane (r − a) · n = 0

AR

Figure 15

This is a Cartesian equation of the plane.More fully this reads

n1x1 + n2x2 + n3x3 − (n1a1 + n2a2 + n3a3) (1.18)= n1x1 + n2x2 + n3x3 + c = 0

where c = − (n1a1 + n2a2 + n3a3)A Cartesian equation is unique up to non-zero constant multiples of (n1, n2, n3, c) since

evidently multiplying n1, n2, n3 and c by β 6= 0 cannot change a solution (x1, x2, x3).


1.3.5 Finding a normal n

If n is a normal to the plane given by equation (1.17), then in particular n ⊥ u and n ⊥ v. Inother words, simultaneously

n1u1 + n2u2 + n3u3 = 0 (1.19)n1v1 + n2v2 + n3v3 = 0

Conversely, if equation (1.19) holds then n must be a normal (see Exercise 15, No.5 below).Because equation (1.19) involves two equations for three unknowns, a solution n 6= 0 can alwaysbe found. See Chapter 2, 2.2.15 for a proof of this.

Example 1.3.2 Find a normal n 6= 0 to the plane of Example 1.3.1 and hence its Cartesianequation.

Solution:

We solve

(n1, n2, n3) · (0, 3, 1) = 3n2 + n3 = 0(n1, n2, n3) · (−2, 1,−1) = −2n1 + n2 − n3 = 0

simultaneously for n1, n2 and n3. (We did much the same thing in Example 1.1.3). Fromthe first equation, n3 = −3n2. Substituting this into the second equation, we get n1 =12 (n2 − n3) = 2n2. Thus, provided n2 6= 0, the vector

n = (2n2, n2,−3n2) = n2 (2, 1,−3)

is perpendicular (normal) to the plane. We may take n2 = 1 and n = (2, 1,−3). WithA = (1,−1, 2) we get a Cartesian equation for the plane:

(2, 1,−3) · ((x1, x2, x3)− (1,−1, 2)) = 2x1 + x2 − 3x3 + 5 = 0

Alternate way to find the cartesian equation of a plane

Consider again example 1.3.1 with generic equation r = (1,−1, 2) + s(0, 3, 1) + t(−2, 1,−1).Writing x = 1− 2t, y = −1 + 3s + t and z = 2 + s− t, we can eliminate s and t as follows.

Substituting t = 1−x2 into y = −1 + 3s + t and z = 2 + s − t gives y = −1 + 3s + 1−x

2 andz = 2 + s− 1−x

2 . Hence

y − 3z = −1 + 3s + 1−x2 − 3

(2 + s− 1−x

2

)= −2x− 5 and the Cartesian equation of the plane

is 2x + y − 3z + 5 = 0.

1.3.6 Some remarks on planes

Remark 13 The following statements are geometrically obvious, but analytic proofs can begiven after we have studied the subject more deeply, in particular after we ave done Chapter 3.See No.10 in Exercise 78.

1. Provided n = (n1, n2, n3) 6= 0, an equation n1x1 + n2x2 + n3x3 + c = 0 always representsa plane with normal n. (If n = 0 the equation represents nothing if c 6= 0 and all of spaceif c = 0).

2. By definition, two distinct planes are parallel if their normals are parallel. Such planesare a constant positive distance apart and have no points in common.

1.3. PLANES 37

3. Two non-parallel planes meet in a line.

4. By definition, a line is parallel to a plane if it is perpendicular to its normal (drawa picture to see that this is correct). A line not parallel to a plane meets the plane inexactly one point.

5. A line and a point not on it determine a unique plane.

6. There are an infinity of planes containing a given line.

Example 1.3.3 Find a generic equation for the line of intersection of the planes

x1 + x2 + x3 + 1 = 02x1 + x2 − x3 + 2 = 0

Solution:

Note that (1, 1, 1) 6‖ (2, 1,−1) so the planes are not parallel. Eliminating x1 gives x2 + 3x3 = 0or x2 = −3x3. Substituting x2 = −3x3 into the first (or second) equation gives x1 = 2x3 − 1.Hence, with t = x3 the parameter,

r = (x1, x2, x3) = (2t− 1,−3t, t)

is the line of intersection of the two planes.

Example 1.3.4 Find a generic equation of the plane with Cartesian equation

x1 − 2x2 + 5x3 = 10 (1.20)

Solution:

By far the easiest solution isr = (10 + 2x2 − 5x3, x2, x3)

Here the independent parameters are x2 and x3 and they can vary freely over <. We couldof course let s = x2 and t = x3, but this is only an artificial device and changes nothing.

A more long-winded approach is to find three non-collinear points A, B and C on the planeand then proceed as in Example (1.3.1).

Letting x2 = x3 = 0 gives x1 = 10 and A = (10, 0, 0) as one point. Putting x1 = x2 = 0gives x3 = 2 and B = (0, 0, 2). With x1 = x3 = 0 we have x2 = −5 and C = (0,−5, 0). Thepoints A, B and C do not lie on a line and another generic equation of the plane is

r = OA + sAB + tAC

= (10− 10s− 10t,−5t, 2s)

Remark 14 Instead of A, B and C we could have chosen any three non-collinear points sat-isfying equation (1.20). The parametric equation will be different but will represent the sameplane.

Example 1.3.5 Find two non-parallel vectors u and v both perpendicular to n = (1,−2, 5).

Solution:

From the previous example two such vectors are u = AB and v = AC

Example 1.3.6 Find the point of intersection of the line through (1,−1, 0) and (0, 1, 2) andthe plane

3x1 + 2x2 + x3 + 1 = 0


Solution:

For some t the point

r = (1,−1, 0) + t ((0, 1, 2)− (1,−1, 0))= (1,−1, 0) + t (−1, 2, 2) = (1− t,−1 + 2t, 2t)

must be on the given plane, i.e.

3 (1− t) + 2 (−1 + 2t) + (2t) + 1 = 0

Thus t = − 23 and the point of intersection is

(53 ,− 7

3 ,− 43

).

1.3.7 Perpendicular from a Point to a Plane

If Π is a plane and P a point, the problem in this section is to find the foot Q of the perpendicularfrom P to the plane. See Figure 1.16.

O

P

Á

Q

plane Π

Figure 1.16 Perpendicular Q from P to plane Π

If the plane passes through the origin, the foot Q is called the projection of P on the plane.

Example 1.3.7 Find the foot Q of the perpendicular from the point P = (4,−1, 3) to the planex + 2y + z = −4. Hence find the shortest distance from P to the given plane.

Solution:

The vector (1, 2, 1) is normal to the plane so the line ` defined by (4,−1, 3) + t (1, 2, 1) =(4 + t,−1 + 2t, 3 + t) passes through P and has direction normal to the plane. Geometry tellsus that the line ` hits the plane at the required foot Q of the perpendicular. Then

∣∣PQ∣∣ will

be the shortest distance of P from the plane.To find t, substitute (4 + t,−1 + 2t, 3 + t) into the Cartesian equation of the plane

(4 + t) + 2 (−1 + 2t) + (3 + t) = −4

1.3. PLANES 39

and solve for t. This gives t = − 32 and the position vector of the point Q therefore has

position vector

q =(

4 +(−3

2

),−1 + 2

(−3

2

), 3 +−3

2

)=

(52,−4,

32

)

The required shortest distance of P from the plane is

∣∣PQ∣∣ =

∣∣∣∣(

52,−4,

32

)− (4,−1, 3)

∣∣∣∣

=∣∣∣∣(−3

2,−3,−3

2

)∣∣∣∣ =

√272

Example 1.3.8 Find the projection of the point P = (4,−1, 3) onto the plane x+2y+z = −0.

Solution:

The plane passes through the origin (that is why we call the foot Q of the perpendicular theprojection). As before, substitute (4 + t,−1 + 2t, 3 + t) into the Cartesian equation x+2y+z =−0:

(4 + t) + 2 (−1 + 2t) + (3 + t) = 0

This gives t = − 56 and the projection of P onto the plane x + 2y + z = −0.is the point Q =(

4− 56 ,−1 + 2

(− 56

), 3− 5

6

)=

(196 ,− 8

3 , 136

).

1.3.8 A general way to see the foot Q of the perpendicular

..................

..................

..................

..................

..................

............................................................................

............

............

............

............

............

............

...Q

P

Xw

6n

plane Π

Figure 17: Foot Q of pependicular from P to Π

Figure 17 represents a plane Π with Cartesian equation n1x1 + n2x2 + n3x3 + c = 0 andnormal n = (n1, n2, n3). The point X = (x1, x2, x3) is any point on Π and Q is the foot of theperpendicular from P = (p1, p2, p3) to the plane.

The thing to notice is that PQ is the projection of PX = (x1 − p1, x2 − p2, x3 − p3) =(x1 − p1, x2 − p2, x3 − p3) on the normal n (i.e. on the line through P and Q):


PQ =

(1|n|2 PX · n

)n

=(

1n2

1 + n22 + n2

3

(x1 − p1, x2 − p2, x3 − p3) · (n1, n2, n3))

(n1, n2, n3)

= −(

c + n · p|n|2

)n

Hence Q has position-vector OQ = OP + PQ:

q = p−(

c + n · p|n|2

)n (1.21)

The distance of P from the plane Π is

∣∣PQ∣∣ =

∣∣n · p + c∣∣

|n| (1.22)

Notice that these formulas are independent of the choice of X. See Exercise 15, No.(8) foran alternative proof of equation (1.21).

Example 1.3.9 Find the foot Q of the perpendicular from a general point P = (p1, p2, p3) to

1. the above plane x + 2y + z = −4, and

2. the parallel plane x + 2y + z = 0 which passes through the origin.

Solution

1. The foot has position vector

q = (p1, p2, p3)−(

4 + (1, 2, 1) · (p1, p2, p3)|(1, 2, 1)|2

)(1, 2, 1)

= (p1, p2, p3)− 4 + p1 + 2p2 + p3

6(1, 2, 1)

=(

5p1

6− p2

3− p3

6− 2

3,−p1

3+

p2

3− p33

−43,−p1

6− p2

3+

5p3

6− 2

3

)(1.23)

2. The projection of P on x + 2y + z = 0 is

q = (p1, p2, p3)−(

(1, 2, 1) · (p1, p2, p3)|(1, 2, 1)|2

)(1, 2, 1)

= (p1, p2, p3)− p1 + 2p2 + p3

6(1, 2, 1)

=(

5p1

6− p2

3− p3

6,−p1

3+

p2

3− p3

3,−p1

6− p2

3+

5p3

6

)(1.24)


1. Show that A = (1, 2, 1), B = (0, 1, 3) and C = (−1, 2, 5) are not collinear and so define aunique plane Π. Find:

1.3. PLANES 41

(a) A generic equation for Π

(b) A normal to Π and hence a Cartesian equation for Π

(c) The foot Q of the perpendicular from P = (1, 2, 3) to Π and so the shortest distancefrom P to the plane Π. (Use two methods, the formula (1.21) and the method ofExample 1.3.7).

(d) In two ways the foot Q of the perpendicular from P = (p1, p2, p3) to Π. Answer:

q = (p1, p2, p3)− −3 + 2p1 + p3

5(2, 0, 1) (1.25)

.

(e) The intersection of the line ` with parametric form (1 + t, 3 + 2t, 4− t) and Π, aswell as the acute angle between the line and the normal to Π.

(f) A plane containing the point (4, 5,−3) and the line (1 + t, 3 + 2t, 4− t).

Solutions to Question 1.

A = (1, 2, 1), B = (0, 1, 3) and C = (−1, 2, 5) AB = (0, 1, 3) − (1, 2, 1) = (−1,−1, 2)AC = (−1, 2, 5)− (1, 2, 1) = (−2, 0, 4).As AB ∦ AC the points A, B, C are not collinear.

1aGeneric equation of plane is r = (1, 2, 1) + s (−1,−1, 2) + t (−2, 0, 4)

1bSpot normal: n = (2, 0, 1) equation of plane is 2x + z + c = 0.

Substitute A = (1, 2, 1) in equation: 2 (1) + 1 + c = 0 c = −3. Cartesian equation ofplane is 2x + z − 3 = 0.

1cUsing equation (1.21) the foot is at

(1 + 2

(− 25

), 2, 3 +

(− 25

))=

(15 , 2, 13

5

). The shortest

distance from P to the plane Π is therefore∣∣(1, 2, 3)− (

15 , 2, 13

5

)∣∣ =∣∣( 4

5 , 0, 25

)∣∣ = 15

√20 =

2√5.

Alternatively, let the foot be at q = (1, 2, 3) + t (2, 0, 1) = (1 + 2t, 2, 3 + t). We need2 (1 + 2t) + (3 + t)− 3 = 0, Solution is t = − 2

5 , and q is as before.

1dThe alternative way to do is to substitute (p1, p2, p3) + t (2, 0, 1) = (p1 + 2t, p2, p3 + t)in2x + z − 3 = 0: 2 (p1 + 2t) + (p3 + t)− 3 = 0. The solution for t is t = − 2

5p1 − 15p3 + 3

5 ,giving the the same result.

1eIntersection of line (1 + t, 3 + 2t, 4− t) and plane 2x+z−3 = 0: 2 (1 + t)+(4− t)−3 = 0,t = −3 required point is (1 + (−3) , 3 + 2 (−3) , 4− (−3)) = (−2,−3, 7).

Acute angle between line and normal to plane is θ = arccos(

(2,0,1)·(1,2,−1)√5√

6

)= 1.387 2

radians or 180π (1.387 2) = 79.481◦.

1fA plane containing the point (4, 5,−3) and the line (1 + t, 3 + 2t, 4− t).

Three points on plane are (4, 5,−3), (1, 3, 4), (0, 1, 5) and a generic equation of the plane isr = (4, 5,−3)+s ((1, 3, 4)− (4, 5,−3))+t ((0, 1, 5)− (4, 5,−3)) = (4− 3s− 4t, 5− 2s− 4t,−3 + 7s + 8t)

2. Find a generic equation for the plane 2x + 3y − 4z + 5 = 0.

Solution: r =(x, y, 1

2x + 34y + 5

4

)


3. Show that the plane r = r(s, t) = a + su + tv passes through the origin if, and only if a isa linear combination of u and v.

Solution: One way is clear, so let a = αu+βv for scalars α and β. Then r(−α,−β) = 0.

4. Find generic and Cartesian equations for the x− y and y − z planes. medskipSolution: Generic equation of x − y plane: r = (x, y, 0) of y − z plane r = (0, y, z)Cartesian equations are respectively z = 0 and x = 0.

5. Prove that if n satisfies equation (1.19) then it is normal to the plane given by the genericequation (1.17). In other words, show that that n will be perpendicular to every vectorparallel to the plane and so to each line lying in the plane.

Solution: Let r1 = a + s1u + t1v and r2 = a + s2u + t2v represent two points R1 and R2

on the plane. Then,

n ·R1R2 = n · ((s2 − s1)u + (t2 − t1)v) = (s2 − s1)n · u + (t2 − t1)n · v = 0

6. Use equation (1.21) to deduce the shortest distance formula equation (1.22).

Solution:∣∣∣p−

(p−

(c+n·p|n|2

)n)∣∣∣ = |n·p+c|

|n|

7. Follow the method of Example 1.3.7 to do Example 1.3.9.

Solution:Consider the line (p1, p2, p3)+t (1, 2, 1) = (p1 + t, p2 + 2t, p3 + t). As above, we find wherethis line hits the given plane. The equation determining this point Q is

(p1 + t) + 2 (p2 + 2t) + (p3 + t) = −4

This gives t = − 4+p1+2p2+p36 . Hence Q is

q = (p1, p2, p3)− 4 + p1 + 2p2 + p3

6(1, 2, 1)

For the projection we must substitute (p1, p2, p3)+ t (1, 2, 1) = (p1 + t, p2 + 2t, p3 + t) intothe equation x + 2y + z = 0 of the plane:

(p1 + t) + 2 (p2 + 2t) + (p3 + t) = 0

This gives t = −p1+2p2+p36 and hence

q = (p1, p2, p3)− p1 + 2p2 + p3

6(1, 2, 1)

8. Follow the method of Example 1.3.7 to show that the foot Q of the perpendicular from thepoint P to the plane Π with Cartesian equation n · r + c = 0 is given by equation (1.21)Consider also the case c = 0. (Then Q is the projection of P on the plane).

Hint 16 The method of of Example 1.3.7 leads to the equation

n · (p + tn)

+ c = 0

for t.

Solution: p + tn = (p1 + tn1, p2 + tn2, p3 + tn3) is the line through P perpendicular tothe plane. (p1 + tn1)n1 +n2 (p2 + tn2)+n3 (p3 + tn3)+ c = 0 is the condition for hitting

1.3. PLANES 43

the plane. Solution is t = −p1n1+n2p2+n3p3+c|n|2 . Hence the foot Q is found by substituting

this value of t in p + tn:

q = (p1 + tn1, p2 + tn2, p3 + tn3)

=

(p1 −

p · n + c

|n|2 n1, p2 −p · n + c

|n|2 n2, p3 −p · n + c

|n|2 n3

)

= p−(

c + n · p|n|2

)n

9. Write the equation of a plane n · p + c = 0 in the form m · p + d = 0 where m is a unitvector.

Solution: 1|n|n · p + 1

|n|c = 0 , so m = n and d = 1|n|c

10. Let n · r + c1 = 0 and n · r + c2 = 0 represent planes Π1 and Π2 respectively. Showalgebraically that Π1 and Π2 are either identical or have no point in common. In the lattercase write down a formula for the shortest distance between Π1 and Π2 if P = (p1, p2, p3)is a given point on the first plane.

Solution: If P = (p1, p2, p3) is a given point on the first plane, the distance between them

is |n·p+c2||n| .

11. Find a generic equation of the line of intersection of the planes 2x + 3y + z + 1 = 0 andx− 4y + 5z − 2 = 0.

Solution: We have 2x + 3y + z + 1− 2 (x− 4y + 5z − 2) = 11y − 9z + 5 = 0. Take z asparameter, y = 9

11z − 511 and so x − 4

(911z − 5

11

)+ 5z − 2 = 0, and x = − 19

11z + 211 . A

generic equation of line is

r =(−19

11z +

211

,911

z − 511

, z

)

12. (Hirst, p.36). In 12a - 12c find the intersection of the three given planes. In each casegive a brief geometric description.Note: If you have difficulty in solving three equations for three unknowns, you maypostpone this question to Exercise 33 No.(2) in Chapter 2.

(a)

3x− 2y + 2z = 54x + y − z = 3x + y + z = 6

(b)

x + 2y − z = 32x + 4y − 2z = 5

x− y + z = 1

(c)

x + 2y − z = 2x− y + z = 1

3x + 3y − z = 5


Solution to 12a:let (i) 3x− 2y + 2z = 5 (ii) 4x + y − z = 3 (iii) x + y + z = 6

(i)+2(ii) 11x = 11 so x = 1 (ii)+(iii) 5x + 2y = 9 gives y = 2 and z = 3. The planesintersect in unique point (1, 2, 3)

Solution to 12b:

(i) x + 2y − z = 3 (ii) 2x + 4y − 2z = 5 (iii) x− y + z = 1

2(i) gives 2x + 4y − 2z = 6. Since this contradicts (ii), the first two equations representnon-intersecting parallel planes. The three planes have no point in common.

Solution to 12c:(i) x + 2y − z = 2 (ii) x− y + z = 1 (iii) 3x + 3y − z = 52(i)+(ii) 3x + 3y − z = 5 which is (iii). So we only have (i) and (ii).

(i)-(ii) x + 2y − z − (x− y + z) = 3y − 2z = 1. Take y for parameter. z = 32y − 1

2and x + 2y − (

32y − 1

2

)= 2, x = − 1

2y + 32 . The planes intersect in the line with generic

equation

r =(−1

2y +

32, y,

32y − 1

2

)

13. * Prove the statement of subsection 1.3.3.

Hint 17 Imitate the procedure of No.4 of Exercise 6.

Solution:In the notation of 1.3.3 suppose that Π1 = Π2. Since a1 ∈ Π2, a1−a2 = αu2+βv2 for somescalars α and β. Hence condition (ii) holds. Again, for the same reason, a1 − a2 + u1 =su2 + tv2 for some scalars s and t. Subtracting gives u1 = (s− α)u2 + (t− β) v2. Henceu1 ∈ sp (u2, v2). Similarly, v1 ∈ sp (u2, v2), so that sp (u1, v1) ⊆ sp (u2, v2). As Π1 = Π2

a symmetric argument shows sp (u2, v2) ⊆ sp (u1, v1), so that sp (u1, v1) = sp (u2, v2).This shows that (i) holds.

Conversely, let (i) and (ii) hold. Then for certain scalars α, β, γ, δ, ε and η we have:a1− a2 = αu2 +βv2, and u1 = γu2 + δv2 and v1 = εu2 + ηv2. So for any scalars s and t,

a1 + su1 + tv1 = a2 + (α + sγ + tε)u2 + (β + sδ + tη) v2

This shows Π1 ⊆ Π2. By symmetry, Π2 ⊆ Π1, and so Π1 = Π2

14. (See Figure 1.18) Let n · r + c = 0 and m · r + d = 0 represent non-parallel planes Π1 andΠ2 respectively. Show that the Cartesian equation of a plane Π that contains the line ofintersection of Π1 and Π2 has the form

λ (n · r + c) + µ (m · r + d) (1.26)= (λn + µm) · r + λc + µd = 0

where λ and µ are two scalars not both of which are zero.

Hint 18 First note that because n 6= 0 and m 6= 0 are not parallel, the linear combinationλn+µm is also non-zero if one or both of λ, µ are non-zero. In that case equation(1.26) represents a plane Π. Note that if (e.g.) λ 6= 0 and µ = 0 then Π is just Π1.Now check that a point on Π1 and Π2 is also on Π.

Solution: Clearly if n · r + c = 0 and m · r + d = 0 then (λn + µm) · r + λc + µd = 0

1.4. REFLECTIONS IN A LINE OR A PLANE 45

...........................................

..........

.............................

..........................

Π1

Π2

Π

..

...........................................

Figure 1.17

15. Find the form of a Cartesian equation of the plane Π that contains the intersection of theplanes

2x1 + 3x2 + x3 + 1 = 0 and x1 − 4x2 + 5x3 − 2 = 0

If the plane Π also contains the point (1,−1, 2), find its Cartesian equation.

Solution: The two planes are 2x1 + 3x2 + x3 + 1 = 0 and x1 − 4x2 + 5x3 − 2 = 0.They are not parallel.Directly: (2λ + µ)x1 + (3λ− 4µ) x2 + (λ + 5µ) x3 + λ − 2µ = 0 is the form of a planecontaining the line of intersection. Neither plane contains the point (1,−1, 2) , so we maylet µ = 1 and (2λ + 1) (1) + (3λ− 4) (−1) + (λ + 5) (2) + λ− 2 = 0, giving λ = − 13

2 .Required Cartesian equation is

(2

(− 132

)+ 1

)x1 +

(3

(− 132

)− 4)x2 +

((− 132

)+ 5

)x3 +(− 13

2

)− 2 = −12x1 − 472 x2 − 3

2x3 − 172 = 0 or −24x1 − 47x2 − 3x3 − 17 = 0.

1.4 Reflections in a line or a plane

A line or plane can serve as a mirror so that a point P is reflected in a point P ′ in such away that P and P ′ are symmetric with respect to the line or plane, as in Figure Figure 1.18 .From the figure the point P ′ is given by

p′ = OP ′ = OP + 2PQ = p + 2(q − p

)= 2q − p (1.27)

where Q is the foot of the perpendicular from P to the line or plane and we have written p′

for the position vector of P ′.

Example 1.4.1 Consider example 1.2.7 of a line in the x-y plane. Using equation (1.14) andequation (1.27), the reflection of p = (p1, p2) in the line is

p′ = 2q − p = 2(

14

(3p1 +

√3p2

),14

√3p1 +

14p2

)− (p1, p2) (1.28)

=(

12p1 +

12

√3p2,

12

√3p1 − 1

2p2

)


P

P′=

...............

®

P

Reflection in a planeReflection in a line

P′

QQ

`

Π

Π

`

Figure 18

Example 1.4.2 Consider (1) of Example 1.3.9 where we found the foot of the perpendicularfrom p to the plane x + 2y + z = −4 given by equation (1.23):

q = (p1, p2, p3)− 4 + p1 + 2p2 + p3

6(1, 2, 1)

From this it follows that the reflection P ′ of P in the plane is the point with position

p′ = 2q − p = p− 4 + p1 + 2p2 + p3

3(1, 2, 1)


1. Using equation (1.11) - which finds the foot Q of the perpendicular from P to the line `with parametric equation r = a + tu - find a formula for the reflection P ′ of P in the line`.

Solution: q = a +(

1|u|2

(p− a

) · u)

u is foot of perpendicular from P to line. Reflectionformula is

p′ = 2q − p = 2

(a +

(1|u|2

(p− a

) · u))

u− p

2. Using equation (1.21) - which finds the foot Q of the perpendicular from P to the planewith Cartesian equation n · r + c = 0 - find a formula for the reflection P ′ of P in thegiven plane.

Solution: q = p−(

c+n·p|n|2

)n is foot of perpendicular from P to plane.Reflection formula

is

p′ = 2q − p = 2

(p−

(c + n · p|n|2

)n

)− p = p− 2

(c + n · p|n|2

)n

1.4. REFLECTIONS IN A LINE OR A PLANE 47

3. Specialize your formulae when a = 0 and c = 0.

Solution: For Question 1 the specialized formula is

p′ =2|u|2

(p · u)

u− p

For Question 2 it is

p′ = p− 2n · p|n|2 n

4. Using the answers to Questions (2) and (3), or otherwise, find formulae for reflectionsin the planes x + 2y + z = 0, 3x− y − 2z = 1 and 3x− y − 2z = 0.

Solution: For both planes n = (3,−1,−2). The formula for reflection in the first planeis

p′ = p−(−1 + (3,−1,−2) · p

7

)(3,−1,−2)

= p−(−1 + 3p1 − p2 − 2p3

7

)(3,−1,−2)

For the second it is

p′ = p−(

(3,−1,−2) · p7

)(3,−1,−2)

= = p−(

3p1 − p2 − 2p3

7

)(3,−1,−2)

5. In the x− y plane find a formula for reflection in the line passing through the origin andmaking an angle of 60◦ with the positive x−axis.

Solution: Parametric equation of line is t(

12 ,√

32

). Foot Q of perpendicular from P =

p = (p1, p2) to line is q =((p1, p2) ·

(12 ,√

32

))p =

(12p1 + 1

2p2

√3)p. Formula for reflec-

tion isp′ = 2q − p =

(p1 + p2

√3− 1

)p

6. * In the x − y plane find a formula for reflection in the line passing through the originand making an angle of θ radians with the positive x−axis.

Hint 20 All these questions use equation (1.27)


1.5 Summary of Chapter 1

1.5.1 Points in Space, Vectors

(See section 1.1).A point P = (p1, p2, p3) in space is identified with its position vector OP = p. The displacementfrom A to B is AB = b− a.

Vectors u = (u1, u2, u3) are underlined.Scalars α,β ,...,s, t,... are real numbers.The product of a vector a by a scalar γ is γa = (γa1, γa2, γa3). Non-zero vectors a and b

are parallel if b = sa for a non-zero scalar s.Vectors a and b are added:

a + b = (a1 + b1, a2 + b2, a3 + b3)

Addition satisfies the triangle law AB + BC = AC.The dot product of vectors a and b is

a · b = a1b1 + a2b2 + a3b3

The length or magnitude of a is|a| = √

a · aThe angle θ between a and b is given by a · b = |a| |b| cos θ.The unit vector in the direction of a 6= 0 is

a =1|a|a

The component of a in the direction b 6= 0 and the projection of a on b are respectively

a · b and(a · b

)b

1.5.2 Straight lines

(See section 1.2).

A parametric equation of the straight line ` through a with direction u 6= 0 is

r = a + tu (t ∈ <)

The foot Q of the perpendicular from P to ` is found by solving(a + tu− p

) · u = 0

for t and then substituting this value back into a + tu.Lines `1 and `1 are parallel if they have parallel directions and are skew if they are not

parallel and do not intersect.

1.5.3 Planes

(See section 1.3).

A generic equation of the plane containing two non-parallel vectors u and v and passingthrough the point A is

r = a + su + tv (s, t ∈ <)

1.5. SUMMARY OF CHAPTER 1 49

A vector n 6= 0 is normal to the plane if it is perpendicular to every line lying in the plane.A Cartesian equation of the plane has the form

(r − a) · n + c = (x1 − a1) n1 + (x2 − a2)n2 + (x3 − a3)n3 + c = 0

where c is a constant.The foot Q of the perpendicular from a point P to the plane is given by equation (1.21):

q = p−(

c + n · p|n|2

)n

It can also be found by solvingn · (p + tn

)+ c = 0

for t and then substituting this into the line p + tn

1.5.4 Projections and reflections

(See equation 1.7, subsection ??, example 1.3.9 and subsection 1.4).

When a line or plane goes through the origin the foot Q of the perpendicular from P to theline or plane is called the projection of P on the plane.

The reflection P ′ of a point P in a line or plane is given by

p′ = 2q − p

where Q is the foot of the perpendicular from P to the line or plane.

applied mathematics 1a (eng) mathematics 132: vectors … · 5. d. lay: linear algebra and its...

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