linear algebra chapter 4- vectors

7/27/2019 Linear Algebra Chapter 4- VECTORS

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LINEAR ALGEBRA

W W L CHEN

c W W L Chen, 1982, 2008.This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990.

It is available free to all individuals, on the understanding that it is not to be used for financial gain,

and may be downloaded and/or photocopied, with or without permission from the author.

However, this document may not be kept on any information storage and retrieval system without permission

from the author, unless such system is not accessible to any individuals other than its owners.

Chapter 4

VECTORS

4.1. Introduction

A vector is an object which has magnitude and direction.

Example 4.1.1. We may be travelling north-east at 50 kph. In this case, the direction of the velocityis north-east and the magnitude of the velocity is 50 kph. We can describe our velocity in kph as

50

2,

502

,

where the first coordinate describes the speed with which we are moving east and the second coordinatedescribes the speed with which we are moving north.

Example 4.1.2. An object in the sky may be 100 metres away in the south-east direction 45 degreesupwards. In this case, the direction of its position is south-eastand 45 degrees upwards and the magnitudeof its distance is 100 metres. We can describe the position of the object in metres as

50,50, 100

2

,

where the first coordinate describes the distance east, the second coordinate describes the distance northand the third coordinate describes the distance up.

The purpose of this chapter is to study some relationship between algebra and geometry. We shallfirst study some algebra which is motivated by geometric considerations. We then use the algebra laterto better understand some problems in geometry.

Chapter 4 : Vectors page 1 of 24


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Linear Algebra c W W L Chen, 1982, 2008

4.2. Vectors in R2

A vector on the plane R2 can be described as an ordered pair u = (u1, u2), where u1, u2 R.

Definition. Two vectors u = (u1, u2) and v = (v1, v2) in R2 are said to be equal, denoted by u = v, if

u1 = v1 and u2 = v2.

Definition. For any two vectors u = (u1, u2) and v = (v1, v2) in R2, we define their sum to be

u + v = (u1, u2) + (v1, v2) = (u1 + v1, u2 + v2).

Geometrically, if we represent the two vectors u and v byAB and

BC respectively, then the sum

u + v is represented byAC as shown in the diagram below:

C

A Bu

u+vv

The next diagram demonstrates geometrically that u + v = v + u:

D C

A B

u

u

u+vv

v

PROPOSITION 4A. (VECTOR ADDITION)(a) For everyu, v

R

2, we have u + v

R

2.

(b) For everyu, v, w R2, we have u + (v + w) = (u + v) + w.(c) For everyu R2, we have u + 0 = u, where 0 = (0, 0) R2.(d) For everyu R2, there exists v R2 such that u + v = 0.(e) For everyu, v R2, we have u + v = v + u.

Proof. Write u = (u1, u2), v = (v1, v2) and w = (w1, w2), where u1, u2, v1, v2, w1, w2 R. To checkpart (a), simply note that u1 + v1, u2 + v2 R. To check part (b), note that

u + (v + w) = (u1, u2) + (v1 + w1, v2 + w2) = (u1 + (v1 + w1), u2 + (v2 + w2))

= ((u1 + v1) + w1, (u2 + v2) + w2) = (u1 + v1, u2 + v2) + (w1, w2)

= (u + v) + w.

Part (c) is trivial. Next, if v = (u1,u2), then u + v = 0, giving part (d). To check part (e), notethat u + v = (u1 + v1, u2 + v2) = (v1 + u1, v2 + u2) = v + u.



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Definition. For any vector u = (u1, u2) in R2 and any scalar c R, we define the scalar multiple to be

cu = c(u1, u2) = (cu1, cu2).

Example 4.2.1. Suppose that u = (2, 1). Then 2u = (4, 2). Geometrically, if we represent the twovectors u and 2u by OA and OB respectively, then we have the diagram below:

A

O

B

u

2u

PROPOSITION 4B. (SCALAR MULTIPLICATION)(a) For everyc R and u R2, we have cu R2.(b) For everyc R and u, v R2, we have c(u + v) = cu + cv.(c) For everya, b R and u R2, we have (a + b)u = au + bu.(d) For everya, b R and u R2, we have (ab)u = a(bu).(e) For everyu R2, we have 1u = u.

Proof. Write u = (u1, u2) and v = (v1, v2), where u1, u2, v1, v2 R. To check part (a), simply notethat cu1, cu2 R. To check part (b), note that

c(u + v) = c(u1 + v1, u2 + v2) = (c(u1 + v1), c(u2 + v2))

= (cu1 + cv1, cu2 + cv2) = (cu1, cu2) + (cv1, cv2) = cu + cv.

To check part (c), note that

(a + b)u = ((a + b)u1, (a + b)u2) = (au1 + bu1, au2 + bu2)

= (au1, au2) + (bu1, bu2) = au + bu.

To check part (d), note that

(ab)u = ((ab)u1, (ab)u2) = (a(bu1), a(bu2)) = a(bu1, bu2) = a(bu).

Finally, to check part (e), note that 1u = (1u1, 1u2) = (u1, u2) = u.

Definition. For any vector u = (u1, u2) in R

2, we define the norm of u to be the non-negative realnumber

u =

u21 + u22.

Remarks. (1) The norm of a vector is simply its magnitude or length. The definition follows from thefamous theorem of Pythagoras.

(2) Suppose that P(u1, u2) and Q(v1, v2) are two points on the plane R2. To calculate the distance

d(P, Q) between the two points, we can first find a vector from P to Q. This is given by (v1u1, v2u2).The distance d(P, Q) is then the norm of this vector, so that

d(P, Q) =

(v1 u1)2 + (v2 u2)2.



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(3) It is not difficult to see that for any vector u R2 and any scalar c R, we have cu = |c|u.

Definition. Any vector u R2 satisfying u = 1 is called a unit vector.

Example 4.2.2. The vector (3, 4) has norm 5.

Example 4.2.3. The distance between the points (6, 3) and (9, 7) is

(9 6)2 + (7 3)2 = 5.

Example 4.2.4. The vectors (1, 0) and (0,1) are unit vectors in R2.

Example 4.2.5. The unit vector in the direction of the vector (1, 1) is (1/

2, 1/

2).

Example 4.2.6. In fact, all unit vectors in R2 are of the form (cos , sin ), where R.

Quite often, we may want to find the angle between two vectors. The scalar product of the two vectorsthen comes in handy. We shall define the scalar product in two ways, one in terms of the angle betweenthe two vectors and the other not in terms of this angle, and show that the two definitions are in factequivalent.

Definition. Suppose that u = (u1, u2) and v = (v1, v2) are vectors in R2, and that [0, ] represents

the angle between them. We define the scalar product u v of u and v by

u v = uv cos if u = 0 and v = 0,

0 if u = 0 or v = 0.(1)

Alternatively, we write

u v = u1v1 + u2v2. (2)

The definitions (1) and (2) are clearly equivalent if u = 0 or v = 0. On the other hand, we have thefollowing result.

PROPOSITION 4C. Suppose that u = (u1, u2) and v = (v1, v2) are non-zero vectors inR2, and that [0, ] represents the angle between them. Then

uv cos = u1v1 + u2v2.

Proof. Geometrically, if we represent the two vectors u and v byOA and

OB respectively, then the

difference v

u is represented by

AB as shown in the diagram below:

| |

B

A

O

vu

u

v

By the Law of cosines, we have

AB2

= OA2

+ OB2 2OAOB cos ;



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in other words, we have

v

u

2 =

u

2 +

v

2

2

u

v

cos ,

so that

uv cos = 12 (u2 + v2 v u2)= 12 (u

21 + u

22 + v

21 + v

22 (v1 u1)2 (v2 u2)2)

= u1v1 + u2v2

as required.

Remarks. (1) We say that two non-zero vectors in R2 are orthogonal if the angle between them is /2.It follows immediately from the definition of the scalar product that two non-zero vectors u, v R2 areorthogonal if and only if u v = 0.

(2) We can calculate the scalar product of any two non-zero vectors u, v R2 by the formula (2) andthen use the formula (1) to calculate the angle between u and v.

Example 4.2.7. Suppose that u = (

3, 1) and v = (

3, 3). Then by the formula (2), we have

u v = 3 + 3 = 6.

Note now that

u

= 2 and

v

= 2

3.

It follows from the formula (1) that

cos =u vuv =

6

4

3=

3

2,

so that = /6.

Example 4.2.8. Suppose that u = (

3, 1) and v = (3, 3). Then by the formula (2), we haveu v = 0. It follows that u and v are orthogonal.

PROPOSITION 4D. (SCALAR PRODUCT) Suppose that u, v, w R2 and c R. Then(a) u v = v u;(b) u (v + w) = (u v) + (u w);(c) c(u v) = (cu) v = u (cv);(d) u u 0; and(e) u u = 0 if and only if u = 0.

Proof. Write u = (u1, u2), v = (v1, v2) and w = (w1, w2), where u1, u2, v1, v2, w1, w2 R. Part (a) istrivial. To check part (b), note that

u (v + w) = u1(v1 + w1) + u2(v2 + w2) = (u1v1 + u2v2) + (u1w1 + u2w2) = u v + u w.

Part (c) is rather simple. To check parts (d) and (e), note that u u = u21 + u22 0, and that equalityholds precisely when u1 = u2 = 0.



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Consider the diagram below:

P

R A

Q

O

av

w

u

(3)

Here we represent the two vectors a and u byOA and

OP respectively. If we project the vector u on to

the line OA, then the image of the projection is the vector w, represented byOQ. On the other hand,

if we project the vector u on to a line perpendicular to the line OA, then the image of the projection is

the vector v, represented by OR.Definition. In the notation of the diagram (3), the vector w is called the orthogonal projection of thevector u on the vector a, and denoted by w = proj

au.

PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u, a R2. Then

proja

u =u aa2 a.

Remark. Note that the component of u orthogonal to a, represented byOR in the diagram (3), is

u proja

u = u u a

a

2

a.

Proof of Proposition 4E. Note that w = ka for some k R. It clearly suffices to prove thatk =

u aa2 .

It is easy to see that the vectors u w and a are orthogonal. It follows that the scalar product(u w) a = 0. In other words, (u ka) a = 0. Hence

k =u aa a =

u aa2

as required.

To end this section, we shall apply our knowledge gained so far to find a formula that gives theperpendicular distance of a point (x0, y0) from a line ax + by + c = 0. Consider the diagram below:

P

n = (a, b)

Q

O

D

u

ax+by+c=0



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Suppose that (x1, y1) is any arbitrary point O on the line ax + by + c = 0. For any other point (x, y) onthe line ax + by + c = 0, the vector (x x1, y y1) is parallel to the line. On the other hand,

(a, b) (x x1, y y1) = (ax + by) (ax1 + by1) = c + c = 0,so that the vector n = (a, b), in the direction

OQ, is perpendicular to the line ax + by + c = 0.

Suppose next that the point (x0, y0) is represented by the point P in the diagram. Then the vector

u = (x0 x1, y0 y1) is represented by OP, and OQ represents the orthogonal projection projnu of uon the vector n. Clearly the perpendicular distance D of the point (x0, y0) from the line ax + by + c = 0satisfies

D = projn

u =u nn2 n

= |(x0 x1, y0 y1) (a, b)|a2 + b2 =|ax0 + by0 ax1 by1|

a2 + b2=|ax0 + by0 + c|

a2 + b2.

We have proved the following result.

PROPOSITION 4F. The perpendicular distance D of a point (x0, y0) from a line ax + by + c = 0 isgiven by

D =|ax0 + by0 + c|

a2 + b2.

Example 4.2.9. The perpendicular distance D of the point (5, 7) from the line 2x 3y + 5 = 0 is givenby

D =|10 21 + 5|

4 + 9=

613

.

4.3. Vectors in R3

In this section, we consider the same problems as in Section 4.2, but in 3-space R3. Any reader whofeels confident may skip this section.

A vector on the plane R3 can be described as an ordered triple u = (u1, u2, u3), where u1, u2, u3 R.

Definition. Two vectors u = (u1, u2, u3) and v = (v1, v2, v3) in R3 are said to be equal, denoted by

u = v, if u1 = v1, u2 = v2 and u3 = v3.

Definition. For any two vectors u = (u1, u2, u3) and v = (v1, v2, v3) in R3, we define their sum to be

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

Definition. For any vector u = (u1, u2, u3) in R3 and any scalar c R, we define the scalar multiple

to be

cu = c(u1, u2, u3) = (cu1, cu2, cu3).

The following two results are the analogues of Propositions 4A and 4B. The proofs are essentiallysimilar.



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PROPOSITION 4A. (VECTOR ADDITION)(a) For everyu, v R3, we have u + v R3.(b) For everyu, v, w R3, we have u + (v + w) = (u + v) + w.(c) For everyu R

3

, we have u + 0 = u, where 0 = (0, 0, 0) R3

.(d) For everyu R3, there exists v R3 such that u + v = 0.(e) For everyu, v R3, we have u + v = v + u.

PROPOSITION 4B. (SCALAR MULTIPLICATION)(a) For everyc R and u R3, we have cu R3.(b) For everyc R and u, v R3, we have c(u + v) = cu + cv.(c) For everya, b R and u R3, we have (a + b)u = au + bu.(d) For everya, b R and u R3, we have (ab)u = a(bu).(e) For everyu R3, we have 1u = u.

Definition. For any vector u = (u1, u2, u3) in R3, we define the norm of u to be the non-negative real

number

u =

u21 + u22 + u

23.

Remarks. (1) Suppose that P(u1, u2, u3) and Q(v1, v2, v3) are two points in R3. To calculate the

distance d(P, Q) between the two points, we can first find a vector from P to Q. This is given by(v1 u1, v2 u2, v3 u3). The distance d(P, Q) is then the norm of this vector, so that

d(P, Q) =

(v1 u1)2 + (v2 u2)2 + (v3 u3)2.

(2) It is not difficult to see that for any vector u R3 and any scalar c R, we have

cu = |c|u.

Definition. Any vector u R3 satisfying u = 1 is called a unit vector.

Example 4.3.1. The vector (3, 4, 12) has norm 13.

Example 4.3.2. The distance between the points (6, 3, 12) and (9, 7, 0) is 13.

Example 4.3.3. The vectors (1, 0, 0) and (0,1, 0) are unit vectors in R3.

Example 4.3.4. The unit vector in the direction of the vector (1, 0, 1) is (1/

2, 0, 1/

2).

The theory of scalar products can be extended to R3 is the natural way.

Definition. Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are vectors in R3, and that [0, ]represents the angle between them. We define the scalar product u v of u and v by

u v = uv cos if u = 0 and v = 0,

0 if u = 0 or v = 0.(4)

Alternatively, we write

u v = u1v1 + u2v2 + u3v3. (5)

The definitions (4) and (5) are clearly equivalent if u = 0 or v = 0. On the other hand, we have thefollowing analogue of Proposition 4C. The proof is similar.



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PROPOSITION 4C. Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are non-zero vectors inR3,

and that [0, ] represents the angle between them. Then

uv cos = u1v1 + u2v2 + u3v3.

Remarks. (1) We say that two non-zero vectors in R3 are orthogonal if the angle between them is /2.It follows immediately from the definition of the scalar productthat two non-zero vectors u, v R3 areorthogonal if and only if u v = 0.

(2) We can calculate the scalar product of any two non-zero vectors u, v R3 by the formula (5) andthen use the formula (4) to calculate the angle between u and v.

Example 4.3.5. Suppose that u = (2, 0, 0) and v = (1, 1,

2). Then by the formula (5), we haveu v = 2. Note now that u = 2 and v = 2. It follows from the formula (4) that

cos = u vuv = 24 = 12 ,

so that = /3.

Example 4.3.6. Suppose that u = (2, 3, 5) and v = (1, 1,1). Then by the formula (5), we haveu v = 0. It follows that u and v are orthogonal.

The following result is the analogue of Proposition 4D. The proof is similar.

PROPOSITION 4D. (SCALAR PRODUCT) Suppose that u, v, w R3 and c R. Then(a) u v = v u;(b) u

(v + w) = (u

v) + (u

w);

(c) c(u v) = (cu) v = u (cv);(d) u u 0; and(e) u u = 0 if and only if u = 0.

Suppose now that a and u are two vectors in R3. Then since two vectors are always coplanar, we candraw the following diagram which represents the plane they lie on:

P

R A

Q

O

av

w

u

(6)

Note that this diagram is essentially the same as the diagram (3), the only difference being that whilethe diagram (3) shows the whole ofR2, the diagram (6) only shows part ofR3. As before, we represent

the two vectors a and u byOA and

OP respectively. If we project the vector u on to the line OA, then

the image of the projection is the vector w, represented byOQ. On the other hand, if we project the

vector u on to a line perpendicular to the line OA, then the image of the projection is the vector v,represented by

OR.

Definition. In the notation of the diagram (6), the vector w is called the orthogonal projection of thevector u on the vector a, and denoted by w = proj

au.



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The following result is the analogue of Proposition 4E. The proof is similar.

PROPOSITION 4E. (ORTHOGONAL PROJECTION) Suppose that u, a R3. Then

proja

u =u aa2 a.

Remark. Note that the component of u orthogonal to a, represented byOR in the diagram (6), is

u proja

u = u u aa2 a.

4.4. Vector Products

In this section, we shall discuss a product of vectors unique to R3. The idea of vector products has wideapplications in geometry, physics and engineering, and is motivated by the wish to find a vector that isperpendicular to two given vectors.

We shall use the right hand rule. In other words, if we hold the thumb on the right hand upwardsand close the remaining four fingers, then the fingers point from the x-direction towards the y-direction,while the thumb points towards the z-direction. Alternatively, if we imagine Columbus had never livedand that the earth were flat, then taking the x-direction as east and the y-direction as north, then thez-direction is upwards!

We shall frequently use the three vectors i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1) in R3.

Definition. Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are two vectors in R3. Then the vector

product u v is defined by the determinant

u v = det i j ku1 u2 u3

v1 v2 v3

.

Remarks. (1) Note that

i j = (j i) = k,j k = (k j) = i,k i = (i k) = j.

(2) Using cofactor expansion by row 1, we have

u v = det

u2 u3v2 v3

i det

u1 u3v1 v3

j + det

u1 u2v1 v2

k

=

det

u2 u3v2 v3

, det

u1 u3v1 v3

, det

u1 u2v1 v2

= (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).

We shall first of all show that the vector product u v is orthogonal to both u and v.



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PROPOSITION 4G. Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are two vectors inR3. Then

(a) u (u v) = 0; and(b) v (u v) = 0.

Proof. Note first of all that

u (u v) = (u1, u2, u3)

det

u2 u3v2 v3

,det

u1 u3v1 v3

, det

u1 u2v1 v2

= u1 det

u2 u3v2 v3

u2 det

u1 u3v1 v3

+ u3 det

u1 u2v1 v2

= det

u1 u2 u3u1 u2 u3

v1 v2 v3

,

in view of cofactor expansion by row 1. On the other hand, clearly

det

u1 u2 u3u1 u2 u3v1 v2 v3

= 0.

This proves part (a). The proof of part (b) is similar.

Example 4.4.1. Suppose that u = (1,1, 2) and v = (3, 0, 2). Then

u v = det i j k1 1 2

3 0 2

=

det

1 20 2

, det

1 23 2

, det

1 13 0

= (2, 4, 3).

Note that (1,1, 2) (2, 4, 3) = 0 and (3, 0, 2) (2, 4, 3) = 0.

PROPOSITION 4H. (VECTOR PRODUCT) Suppose that u, v, w R3 and c R. Then(a) u v = (v u);(b) u

(v + w) = (u

v) + (u

w);

(c) (u + v) w = (u w) + (v w);(d) c(u v) = (cu) v = u (cv);(e) u 0 = 0; and(f) u u = 0.

Proof. Write u = (u1, u2, u3), v = (v1, v2, v3) and w = (w1, w2, w3). To check part (a), note that

det

i j ku1 u2 u3

v1 v2 v3

= det

i j kv1 v2 v3

u1 u2 u3

.

To check part (b), note that

det i j ku1 u2 u3

v1 + w1 v2 + w2 v3 + w3

= det i j ku1 u2 u3v1 v2 v3

+ det i j ku1 u2 u3w1 w2 w3

.

Part (c) is similar. To check part (d), note that

c det

i j ku1 u2 u3

v1 v2 v3

= det

i j kcu1 cu2 cu3

v1 v2 v3

= det

i j ku1 u2 u3

cv1 cv2 cv3

.

To check parts (e) and (f), note that

u 0 = det

i j k

u1 u2 u3

0 0 0

= 0 and u u = det

i j k

u1 u2 u3

u1 u2 u3

= 0

as required.



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Next, we shall discuss an application of vector product to the evaluaton of the area of a parallelogram.To do this, we shall first establish the following result.

PROPOSITION 4J. Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are non-zero vectors in R3

,and that [0, ] represents the angle between them. Then(a) u v2 = u2v2 (u v)2; and(b) u v = uv sin .

Proof. Note that

u v2 = (u2v3 u3v2)2 + (u3v1 u1v3)2 + (u1v2 u2v1)2 (7)

and

u2v2 (u v)2 = (u21 + u22 + u23)(v21 + v22 + v23 ) (u1v1 + u2v2 + u3v3)2. (8)

Part (a) follows on expanding the right hand sides of (7) and (8) and checking that they are equal. Toprove part (b), recall that

u v = uv cos .

Combining with part (a), we obtain

u v2 = u2v2 u2v2 cos2 = u2v2 sin2 .

Part (b) follows.

Consider now a parallelogram with vertices O,A,B,C. Suppose that u and v are represented byOA

andOC respectively. If we imagine the side OA to represent the base of the parallelogram, so that

the base has length u, then the height of the the parallelogram is given by v sin , as shown in thediagram below:

C B

O A

v sin v

u

u

It follows from Proposition 4J that the area of the parallelogram is given by u v. We have provedthe following result.

PROPOSITION 4K. Suppose that u, v R3. Then the parallelogram with u and v as two of its sideshas area u v.

We conclude this section by making a remark on the vector product u v of two vectors in R3.Recall that the vector product is perpendicular to both u and v. Furthermore, it can be shown that thedirection of u

v satisfies the right hand rule, in the sense that if we hold the thumb on the right hand

outwards and close the remaining four fingers, then the thumb points towards the u v-direction whenthe fingers point from the u-direction towards the v-direction. Also, we showed in Proposition 4J that



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the magnitude of uv depends only on the norm of u and v and the angle between the two vectors. Itfollows that the vector product is unchanged as long as we keep a right hand coordinate system. This isan important consideration in physics and engineering, where we may use different coordinate systems

on the same problem.

4.5. Scalar Triple Products

Suppose that u, v, w R3 do not all lie on the same plane. Consider the parallelepiped with u, v, w asthree of its edges. We are interested in calculating the volume of this parallelepiped. Suppose that u, vand w are represented by

OA,

OB and

OC respectively. Consider the diagram below:

P A

C

O B

vw

u

w

v

| |

| |

| |

By Proposition 4K, the base of this parallelepiped, with O , B , C as three of the vertices, has area vw.Next, note that if OP is perpendicular to the base of the parallelepiped, then

OP is in the direction of

v w. If P A is perpendicular to OP, then the height of the parallelepiped is equal to the norm of theorthogonal projection of u on v w. In other words, the parallelepiped has height

projvwu = u (v w)v w2 (v w) = |u

(v

w)|v w .

Hence the volume of the parallelepiped is given by

V = |u (v w)|.We have proved the following result.

PROPOSITION 4L. Suppose that u, v, w R3. Then the parallelepiped with u, v and w as three ofits edges has volume |u (v w)|.

Definition. Suppose that u, v, w R3. Then u (v w) is called the scalar triple product of u, vand w.

Remarks. (1) It follows immediately from Proposition 4L that three vectors in R3 are coplanar if andonly if their scalar triple product is zero.

(2) Note that

u (v w) = (u1, u2, u3)

det

v2 v3w2 w3

,det

v1 v3w1 w3

, det

v1 v2w1 w2

= u1 det

v2 v3w2 w3

u2 det

v1 v3w1 w3

+ u3 det

v1 v2w1 w2

= det

u1 u2 u3v1 v2 v3

w1 w2 w3

, (9)

in view of cofactor expansion by row 1.



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(3) It follows from identity (9) that

u (v w) = v (w u) = w (u v).

Note that each of the determinants can be obtained from the other two by twice interchanging two rows.

Example 4.5.1. Suppose that u = (1, 0, 1), v = (2, 1, 3) and w = (0, 1, 1). Then

u (v w) = det 1 0 12 1 3

0 1 1

= 0,

so that u, v and w are coplanar.

Example 4.5.2. The volume of the parallelepiped with u = (1, 0, 1), v = (2, 1, 4) and w = (0, 1, 1) as

three of its edges is given by

|u (v w)| =det

1 0 12 1 4

0 1 1

= | 1| = 1.

4.6. Application to Geometry in R3

In this section, we shall study lines and planes in R3 by using our results on vectors in R3.

Consider first of all a plane in R3. Suppose that (x1, y1, z1) R3 is a given point on this plane.Suppose further that n = (a,b,c) is a vector perpendicular to this plane. Then for any arbitrary point(x,y,z) R3 on this plane, the vector

(x,y,z) (x1, y1, z1) = (x x1, y y1, z z1)

joins one point on the plane to another point on the plane, and so must be parallel to the plane andhence perpendicular to n = (a,b,c). It follows that the scalar product

(a,b,c) (x x1, y y1, z z1) = 0,

and so

a(x x1) + b(y y1) + c(z z1) = 0. (10)

If we write d = ax1 + by1 + cz1, then (10) can be rewritten in the form

ax + by + cz + d = 0. (11)

Equation (10) is usually called the point-normal form of the equation of a plane, while equation (11) isusually known as the general form of the equation of a plane.

Example 4.6.1. Consider the plane through the point (2,5, 7) and perpendicular to the vector(3, 5,

4). Here (a,b,c) = (3, 5,

4) and (x1, y1, z1) = (2,

5, 7). The equation of the plane is given

in point-normal form by 3(x2)+5(y + 5)4(z7) = 0, and in general form by 3x + 5y 4z +37 = 0.Here d = 6 25 28 = 37.



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Example 4.6.2. Consider the plane through the points (1, 1, 1), (2, 2, 0) and (4,6, 2). Then the vectors

(2, 2, 0)

(1, 1, 1) = (1, 1,

1) and (4,

6, 2)

(1, 1, 1) = (3,

7, 1)

join the point (1, 1, 1) to the points (2, 2, 0) and (4,6, 2) respectively and are therefore parallel to theplane. It follows that the vector product

(1, 1,1) (3,7, 1) = (6,4,10)

is perpendicular to the plane. The equation of the plane is then given by 6(x1)4(y1)10(z1) = 0,or 3x + 2y + 5z 10 = 0.

Consider next a line in R3. Suppose that (x1, y1, z1) R3 is a given point on this line. Suppose furtherthat n = (a,b,c) is a vector parallel to this line. Then for any arbitrary point (x,y,z)

R

3 on this line,

the vector

(x,y,z) (x1, y1, z1) = (x x1, y y1, z z1)

joins one point on the line to another point on the line, and so must be parallel to n = (a,b,c). It followsthat there is some number R such that

(x x1, y y1, z z1) = (a,b,c),

so that

x = x1 + a,

y = y1 + b, (12)

z = z1 + c,

where is called a parameter. Suppose further that a,b,c are all non-zero. Then, eliminating theparameter , we obtain

x x1a

=y y1

b=

z z1c

. (13)

Equations (12) are usually called the parametric form of the equations of a line, while equations (13)

are usually known as the symmetric form of the equations of a line.

Example 4.6.3. Consider the line through the point (2,5, 7) and parallel to the vector (3, 5,4). Here(a,b,c) = (3, 5,4) and (x1, y1, z1) = (2,5, 7). The equations of the line are given in parametric formby

x = 2 + 3,

y = 5 + 5,z = 7 4,

and in symmetric form by

x 23

=y + 5

5= z 7

4.



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Example 4.6.4. Consider the line through the points (3, 0, 5) and (7, 0, 8). Then a vector in the directionof the line is given by

(7, 0, 8) (3, 0, 5) = (4, 0, 3).The equation of the line is then given in parametric form by

x = 3 + 4,

y = 0,

z = 5 + 3,

and in symmetric form by

x 34

=z 5

3and y = 0.

Consider the plane through three fixed points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), not lying on thesame line. Let (x,y,z) be a point on the plane. Then the vectors

(x,y,z) (x1, y1, z1) = (x x1, y y1, z z1),(x,y,z) (x2, y2, z2) = (x x2, y y2, z z2),(x,y,z) (x3, y3, z3) = (x x3, y y3, z z3),

each joining one point on the plane to another point on the plane, are all parallel to the plane. Usingthe vector product, we see that the vector

(x x2, y y2, z z2) (x x3, y y3, z z3)is perpendicular to the plane, and so perpendicular to the vector (x x1, y y1, z z1). It follows thatthe scalar triple product

(x x1, y y1, z z1) ((x x2, y y2, z z2) (x x3, y y3, z z3)) = 0;in other words,

det

x x1 y y1 z z1x x2 y y2 z z2

x x3 y y3 z z3

= 0.

This is another technique to find the equation of a plane through three fixed points.

Example 4.6.5. We return to the plane in Example 4.6.2, through the three points (1, 1, 1), (2, 2, 0)and (4,

6, 2). The equation is given by

det

x 1 y 1 z 1x 2 y 2 z 0

x 4 y + 6 z 2

= 0.

The determinant on the left hand side is equal to 6x 4y 10z + 20. Hence the equation of the planeis given by 6x 4y 10z + 20 = 0, or 3x + 2y + 5z 10 = 0.

We observe that the calculation for the determinant above is not very pleasant. However, the techniquecan be improved in the following way by making less reference to the unknown point ( x,y,z). Note thatthe vectors

(x,y,z) (x1, y1, z1) = (x x1, y y1, z z1),(x2, y2, z2) (x1, y1, z1) = (x2 x1, y2 y1, z2 z1),(x3, y3, z3) (x1, y1, z1) = (x3 x1, y3 y1, z3 z1),



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each joining one point on the plane to another point on the plane, are all parallel to the plane. Usingthe vector product, we see that the vector

(x2 x1, y2 y1, z2 z1) (x3 x1, y3 y1, z3 z1)is perpendicular to the plane, and so perpendicular to the vector (x x1, y y1, z z1). It follows thatthe scalar triple product

(x x1, y y1, z z1) ((x2 x1, y2 y1, z2 z1) (x3 x1, y3 y1, z3 z1)) = 0;

in other words,

det

x x1 y y1 z z1x2 x1 y2 y1 z2 z1

x3 x1 y3 y1 z3 z1

= 0.

Example 4.6.6. We return to the plane in Examples 4.6.2 and 4.6.5, through the three points (1, 1, 1),(2, 2, 0) and (4,6, 2). The equation is given by

det

x 1 y 1 z 12 1 2 1 0 1

4 1 6 1 2 1

= 0.

The determinant on the left hand side is equal to

det

x 1 y 1 z 11 1 1

3 7 1

= 6(x 1) 4(y 1) 10(z 1) = 6x 4y 10z + 20.

Hence the equation of the plane is given by 6x 4y 10z + 20 = 0, or 3x + 2y + 5z 10 = 0.

We next consider the problem of dividing a line segment in a given ratio. Suppose that x1 and x2 aretwo given points in R3.

We wish to divide the line segment joining x1 and x2 internally in the ratio 1 : 2, where 1 and 2are positive real numbers. In other words, we wish to find the point x on the line segment joining x1and x2 such that

x x1x x2 =

12

,

as shown in the diagram below:

| | |

x1 x x2

x x1 x x2

Since x x1 and x2 x are both in the same direction as x2 x1, we must have

2(x x1) = 1(x2 x), or x = 1x2 + 2x11 + 2

.

We wish next to find the point x on the line joining x1 and x2, but not between x1 and x2, such that

x x1x x2 =

12

,



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where 1 and 2 are positive real numbers, as shown in the diagrams below for the cases 1 < 2 and1 > 2 respectively:

| | | | | |x x1 x2 x1 x2 x

| |

Since x x1 and x x2 are in the same direction as each other, we must have

2(x x1) = 1(x x2), or x = 1x2 2x11 2 .

Example 4.6.7. Let x1 = (1, 2, 3) and x2 = (7, 11, 6). The point

x =2x2 + x1

2 + 1=

2(7, 11, 6) + (1, 2, 3)

3= (5, 8, 5)

divides the line segment joining (1, 2, 3) and (7, 11, 6) internally in the ratio 2 : 1, whereas the point

x =4x2 2x1

4 2 =4(7, 11, 6) 2(1, 2, 3)

2= (13, 20, 9)

satisfies

x x1x x2 =

4

2.

Finally we turn our attention to the question of finding the distance of a plane from a given point.We shall prove the following analogue of Proposition 4F.

PROPOSITION 4F. The perpendicular distance D of a plane ax + by + cz + d = 0 from a point(x0, y0, z0) is given by

D =|ax0 + by0 + cz0 + d|

a2 + b2 + c2.

Proof. Consider the following diagram:

| | | | | |

| |

P

n = (a,b,c)

Q

O

D

ax+by+cz=0

u



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Suppose that (x1, x2, x3) is any arbitrary point O on the plane ax + by + cz + d = 0. For any other point(x,y,z) on the plane ax + by + cz + d = 0, the vector (x x1, y y1, z z1) is parallel to the plane. Onthe other hand,

(a,b,c) (x x1, y y1, z z1) = (ax + by + cz) (ax1 + by1 + cz1) = d + d = 0,

so that the vector n = (a,b,c), in the directionOQ, is perpendicular to the plane ax + by + cz + d = 0.

Suppose next that the point (x0, y0, z0) is represented by the point P in the diagram. Then the vector

u = (x0x1, y0y1, z0z1) is represented by OP, and OQ represents the orthogonal projection projnuof u on the vector n. Clearly the perpendicular distance D of the point (x0, y0, z0) from the planeax + by + cz + d = 0 satisfies

D = projn

u =u nn2 n

= |(x0 x1, y0 y1, z0 z1) (a,b,c)|a2 + b2 + c2=|ax0 + by0 + cz0

ax1

by1

cz1

|a2 + b2 + c2=|ax0 + by0 + cz0 + d

|a2 + b2 + c2

as required.

A special case of Proposition 4F is when (x0, y0, z0) = (0, 0, 0) is the origin. This show that theperpendicular distance of the plane ax + by + cz + d = 0 from the origin is

|d|a2 + b2 + c2

.

Example 4.6.8. Consider the plane 3x + 5y 4z + 37 = 0. The distance of the point (1, 2, 3) from theplane is

|3 + 10 12 + 37|9 + 25 + 16

=38

50=

19

2

5.

The distance of the origin from the plane is

|37|9 + 25 + 16

=37

50.

Example 4.6.9. Consider also the plane 3x+5y4z1 = 0. Note that this plane is also perpendicular tothe vector (3, 5,4) and is therefore parallel to the plane 3x+5y4z+37 = 0. It is therefore reasonable tofind the perpendicular distance between these two parallel planes. Note that the perpendicular distancebetween the two planes is equal to the perpendicular distance of any point on 3x + 5y 4z 1 = 0 fromthe plane 3x + 5y 4z + 37 = 0. Note now that (1, 2, 3) lies on the plane 3x + 5y 4z 1 = 0. It followsfrom Example 4.6.8 that the distance between the two planes is 19

2/5.

4.7. Application to Mechanics

Let u = (ux, uy) denote a vector in R2, where the components ux and uy are functions of an independent

variable t. Then the derivative of u with respect to t is given by

du

dt=

duxdt

,duydt

.



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Example 4.7.1. When discussing planar particle motion, we often let r = (x, y) denote the position ofa particle at time t. Then the components x and y are functions of t. The derivative

v = drdt =

dxdt , dydt

represents the velocity of the particle, and its derivative

a =dv

dt=

d2x

dt2,

d2y

dt2

represents the acceleration of the particle. We often write r = r, v = v and a = a.

Suppose that w = (wx, wy) is another vector in R2. Then it is not difficult to see that

d

dt

(u

w) = u

dw

dt

+du

dt w. (14)

Example 4.7.2. Consider a particle moving at constant speed along a circular path centred at theorigin. Then r = r is constant. More precisely, the position vector r = (x, y) satisfies x2 + y2 = c1,where c1 is a positive constant, so that

r r = (x, y) (x, y) = c1. (15)

On the other hand, v = v is constant. More precisely, the velocity vector

v =

dx

dt,

dy

dt

satisfies

dx

dt

2+

dy

dt

2= c2,

where c2 is a positive constant, so that

v v =

dx

dt,

dy

dt

dx

dt,

dy

dt

= c2. (16)

Differentiating (15) and (16) with respect to t, and using the identity (14), we obtain respectively

r v = 0 and v a = 0. (17)

Using the properties of the scalar product, we see that the equations in (17) show that the vector vis perpendicular to both vectors r and a, and so a must be in the same direction as or the oppositedirection to r. Next, differentiating the first equation in (17), we obtain

r a + v v = 0, or r a = v2 < 0.

Let denote the angle between a and r. Then = 0 or = 180. Since

r a = ra cos ,

it follows that cos < 0, and so = 180. We also obtain ra = v2, so that a = v2/r. This is a vectorproof that for circular motion at constant speed, the acceleration is towards the centre of the circle andof magnitude v2/r.

Let u = (ux.uy, uz) denote a vector in R3, where the components ux, uy and uz are functions of anindependent variable t. Then the derivative of u with respect to t is given by

du

dt=

duxdt

,duydt

,duzdt

.



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Suppose that w = (wx, wy, wz) is another vector in R3. Then it is not difficult to see that

d

dt

(u

w) = u

dw

dt

+du

dt w. (18)

Example 4.7.3. When discussing particle motion in 3-dimensional space, we often let r = (x,y,z)denote the position of a particle at time t. Then the components x, y and z are functions of t. Thederivative

v =dr

dt=

dx

dt,

dy

dt,

dz

dt

= (x, y, z)

represents the velocity of the particle, and its derivative

a =dv

dt=

d2x

dt2,

d2y

dt2,

d2z

dt2

= (x, y, z)

represents the acceleration of the particle.

Example 4.7.4. For a particle of mass m, the kinetic energy is given by

T = 12 m(x2 + y2 + z2) = 12 m(x, y, z) (x, y, z) = 12 mv v.

Using the identity (18), we have

dT

dt= ma v = F v,

where F = ma denotes the force. On the other hand, suppose that the potential energy is given by V.

Using knowledge on functions of several real variables, we can show that

dV

dt=

V

x

dx

dt+

V

y

dy

dt+

V

z

dz

dt=

V

x,

V

y,

V

z

v = V v,

where

V =

V

x,

V

y,

V

z

is called the gradient of V. The law of conservation of energy says that T + V is constant, so that

dT

dt

+dV

dt

= (F +

V)

v = 0

holds for all vectors v, so that F(r) = V(r) for all vectors r.

Example 4.7.5. If a force acts on a moving particle, then the work done is defined as the product ofthe distance moved and the magnitude of the force in the direction of motion. Suppose that a force Facts on a particle with displacement r. Then the component of the force in the direction of the motionis given by F u, where

u =r

ris a unit vector in the direction of the vector r. It follows that the work done is given by

r

F rr

= F r.



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For instance, we see that the work done in moving a particle along a vector r = (3,2, 4) with appliedforce F = (2,1, 1) is F r = (2,1, 1) (3,2, 4) = 12.

Example 4.7.6. We can also resolve a force into components. Consider a weight of mass m hangingfrom the ceiling on a rope as shown in the picture below:

m

Here the rope makes an angle of 60 with the vertical. We wish to find the tension T on the rope. Tofind this, note that the tension on the rope is a force, and we have the following picture of forces:

T1 60

T2

magnitude mg

The force T1 has magnitude T1 = T. Let z be a unit vector pointing vertically upwards. Using scalarproducts, we see that the component of the force T1 in the vertical direction is

T1 z = T1z cos60 = 12 T.

Similarly, the force T2 has magnitude T2 = T, and the component of it in the vertical direction is

T2 z = T2z cos60 = 12 T.

Since the weight is stationary, he total force upwards on it is 12 T +12 Tmg = 0. Hence T = mg.



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Problems for Chapter 4

1. For each of the following pairs of vectors in R2, calculate u + 3v, u v, u v and find the anglebetween u and v:

a) u = (1, 1) and v = (5, 0) b) u = (1, 2) and v = (2, 1)

2. For each of the following pairs of vectors in R2, calculate 2u 5v, u 2v, u v and the anglebetween u and v (to the nearest degree):

a) u = (1, 3) and v = (2, 1) b) u = (2, 0) and v = (1, 2)

3. For the two vectors u = (2, 3) and v = (5, 1) in the 2-dimensional euclidean space R2, determineeach of the following:

a) u v b) uc) u (u v) d) the angle between u and u v

4. For each of the following pairs of vectors inR

3

, calculateu

+ 3v

,u

v,

u

v, find the anglebetween u and v, and find a unit vector perpendicular to both u and v:

a) u = (1, 1, 1) and v = (5, 0, 5) b) u = (1, 2, 3) and v = (3, 2, 1)

5. Find vectors v and w such that v is parallel to (1, 2, 3), v + w = (7, 3, 5) and w is orthogonal to(1, 2, 3).

6. Let ABCD be a quadrilateral. Show that the quadrilateral obtained by joining the midpoints ofadjacent sides of ABCD is a parallelogram.[Hint: Let a, b, c and d be vectors representing the four sides of ABCD.]

7. Suppose that u, v and w are vectors in R3 such that the scalar triple porduct u (vw) = 0. Let

u = v wu (v w) , v

= w uu (v w) , w

= u vu (v w) .

a) Show that u u = 1.b) Show that u v = u w = 0.c) Use the properties of the scalar triple product to find v v and w w, as well as v u, v w,

w u and w v.

8. Suppose that u, v, w, u, v and w are vectors in R3 such that u u = v v = w w = 1 andu v = u w = v u = v w = w u = w v = 0. Show that if u (v w) = 0, then

u =v w

u (v w), v =

w u

u (v w), w =

u v

u (v w).

9. Suppose that u, v and w are vectors in R3.a) Show that u (v w) = (u w)v (u v)w.b) Deduce that (u v) w = (u w)v (v w)u.

10. Consider the three points P(2, 3, 1), Q(4, 2, 5) and R(1, 6,3).a) Find the equation of the line through P and Q.b) Find the equation of the plane perpendicular to the line in part (a) and passing through R.c) Find the distance between R and the line in part (a).d) Find the area of the parallelogram with the three points as vertices.e) Find the equation of the plane through the three points.

f) Find the distance of the origin (0, 0, 0) from the plane in part (e).g) Are the planes in parts (b) and (e) perpendicular? Justify your assertion.



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11. Consider the points (1, 2, 3), (0, 2, 4) and (2, 1, 3) in R3.a) Find the area of a parallelogram with these points as three of its vertices.b) Find the perpendicular distance between (1, 2, 3) and the line passing through (0, 2, 4) and (2, 1, 3).

12. Consider the points (1, 2, 3), (0, 2, 4) and (2, 1, 3) in R3.a) Find a vector perpendicular to the plane containing these points.b) Find the equation of this plane and its perpendicular distance from the origin.c) Find the equation of the line perpendicular to this plane and passing through the point (3, 6, 9).

13. Find the equation of the plane through the points (1, 2,3), (2,3, 4) and (3, 1, 2).

14. Find the equation of the plane through the points (2,1, 1), (3, 2,1) and (1, 3, 2).

15. Find the volume of a parallelepiped with the points (1, 2, 3), (0, 2, 4), (2, 1, 3) and (3, 6, 9) as four ofits vertices.

16. Consider a weight of mass m hanging from the ceiling supported by two ropes as shown in thepicture below:

m

Here the rope on the left makes an angle of 45 with the vertical, while the rope on the right makesan angle of 60 with the vertical. Find the tension on the two ropes.

Ch 4 V 4 f 4

linear algebra chapter 4- vectors

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