11 chương 4 véc tơ và các tính chất của véc tơ

If we project the vector u on tothe line OA, then the image of the projection is the vector w, represented by −−→OQ.. On the other hand, if we project the vector u on to a line perpendic

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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 velocity

is north-east and the magnitude of the velocity is 50 kph We can describe our velocity in kph as

50

√

2,

50

√2

,

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 magnitude

of 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 later

to 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 R2are 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 by −−→AB and −−→BC respectively, then the sum

u + v is represented by −→AC as shown in the diagram below:

Linear Algebra !c W W L Chen, 1982, 2006

4.2 Vectors in R2

A vector on the plane R2can 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

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

PROPOSITION 4A (VECTOR ADDITION)

(a) For every u, v ∈ R2, we have u + v ∈ R2

(b) For every u, v, w ∈ R2, we have u + (v + w) = (u + v) + w

(c) For every u ∈ R2, we have u + 0 = u, where 0 = (0, 0) ∈ R2

(d) For every u ∈ R2, there exists v ∈ R2 such that u + v = 0

(e) For every u, 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

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

4.2 Vectors in R2

A vector on the plane R2can 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

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

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Definition For any vector u = (u1, u2) in R2and 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:

Definition For any vector u = (u1, u2) in R2and 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:

AO

B

u

−2u

PROPOSITION 4B (SCALAR MULTIPLICATION)

(a) For every c ∈ R and u ∈ R2, we have cu ∈ R2

(b) For every c ∈ R and u, v ∈ R2, we have c(u + v) = cu + cv

(c) For every a, b ∈ R and u ∈ R2, we have (a + b)u = au + bu

(d) For every a, b ∈ R and u ∈ R2, we have (ab)u = a(bu)

(e) For every u ∈ 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 R2, we define the norm of u to be the non-negative realnumber

d(P, Q) ="(v1− u1)2+ (v2− u2)2

PROPOSITION 4B (SCALAR MULTIPLICATION)

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

Definition For any vector u = (u1, u2) in R2, we define the norm of u to be the non-negative realnumber

d(P, Q) =p(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 kcuk = |c|kuk.Definition Any vector u ∈ R2 satisfying kuk = 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) isp(9 ư 6)2+ (7 ư 3)2= 5

Example 4.2.4 The vectors (1, 0) and (0, ư1) 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 θ, sin θ), where θ ∈ 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 θ ∈ [0, π] representsthe angle between them We define the scalar product u · v of u and v by

u · v =n kukkvkcosθ if u 6= 0 and v 6= 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 in R2, and that

θ ∈ [0, π] represents the angle between them Then

kukkvk cos θ = u1v1+ u2v2

Proof Geometrically, if we represent the two vectors u and v by ư→OA and ưư→OB respectively, then thedifference v ư u is represented by ưư→AB as shown in the diagram below:

(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 ư 6)2+ (7 ư 3)2= 5

Example 4.2.4 The vectors (1, 0) and (0, ư1) 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 θ, sin θ), where θ ∈ 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 θ ∈ [0, π] representsthe angle between them We define the scalar product u · v of u and v by

(1) u · v =" #u##v#cosθ if u &= 0 and v &= 0,

PROPOSITION 4C Suppose that u = (u1, u2) and v = (v1, v2) are non-zero vectors in R2, and that

θ ∈ [0, π] represents the angle between them Then

vưu

θ uv

By the Law of cosines, we have

AB2= OA2+ OB2ư 2OA OB cos θ;

By the Law of cosines, we have

AB2= OA2+ OB2ư 2OA OB cos θ;

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in other words, we have

kv − uk2= kuk2+ kvk2− 2kukkvk cos θ,

Remarks (1) We say that two non-zero vectors in R2are orthogonal if the angle between them is π/2

It follows immediately from the definition of the scalar product that two non-zero vectors u, v ∈ R2areorthogonal 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

2 ,

so that θ = π/6

Example 4.2.8 Suppose that u = (√3, 1) and v = (−√3, 3) Then by the formula (2), we have

u · 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 = u2+ u2 ≥ 0, and that equalityholds precisely when u1= u2

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Consider the diagram below:

(3)

P

QO

a v

w u

Here we represent the two vectors a and u by −→OA and −−→OP respectively If we project the vector u on tothe line OA, then the image of the projection is the vector w, represented by −−→OQ 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 isthe 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 = projau

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

O

D

u ax+by+c=0

(3)Here we represent the two vectors a and u by −→OA and −−→OP respectively If we project the vector u on tothe line OA, then the image of the projection is the vector w, represented by −−→OQ On the other hand,

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 · akak2

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:

(3)

P

QO

a v

w u

Here we represent the two vectors a and u by −→OA and −−→OP respectively If we project the vector u on tothe line OA, then the image of the projection is the vector w, represented by −−→OQ On the other hand,

O

D

u ax+by+c=0

Chapter 4 : Vectors page 6 of 24 Chapter 4 : Vectors page 6 of 24

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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 u

on the vector n Clearly the perpendicular distance D of the point (x0, y0) from the line ax + by + c = 0satisfies

D = kprojnuk = knku · n2n |(x0− x1√, y0− y1) · (a, b)|

a2+ b2 = |ax0+ by√0− ax1− by1|

a2+ b2 = |ax√0+ 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

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PROPOSITION 4A’ (VECTOR ADDITION)

(c) For every u ∈ R3, we have u + 0 = u, where 0 = (0, 0, 0) ∈ R3

PROPOSITION 4B’ (SCALAR MULTIPLICATION)

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

kuk =qu2+ u2+ u2

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

kcuk = |c|kuk

Definition Any vector u ∈ R3 satisfying kuk = 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 θ ∈ [0, π]represents the angle between them We define the scalar product u · v of u and v by

u · v =n kukkvkcosθ if u 6= 0 and v 6= 0,

0 if u = 0 or v = 0 (4)Alternatively, we write

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PROPOSITION 4C’ Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are non-zero vectors in R3,and that θ ∈ [0, π] represents the angle between them Then

kukkvk cos θ = u1v1+ u2v2+ u3v3

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 have

u · v = 2 Note now that kuk = 2 and kvk = 2 It follows from the formula (4) that

cos θ = kukkvku · v =24 =12,

so that θ = π/3

Example 4.3.6 Suppose that u = (2, 3, 5) and v = (1, 1, −1) Then by the formula (5), we have

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

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:

PROPOSITION 4C’ Suppose that u = (u1, u2, u3) and v = (v1, v2, v3) are non-zero vectors in R3,and that θ ∈ [0, π] represents the angle between them Then

#u##v# cos θ = u1v1+ u2v2+ u3v3

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 ∈ R3by 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 have

u · v = 2 Note now that #u# = 2 and #v# = 2 It follows from the formula (4) that

cos θ = #u##v#u · v =24 = 12,

so that θ = π/3

Example 4.3.6 Suppose that u = (2, 3, 5) and v = (1, 1, −1) Then by the formula (5), we have

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

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:

(6)

P

QO

a v

w u

Note that this diagram is essentially the same as the diagram (3), the only difference being that whilethe diagram (3) shows the whole of R2, the diagram (6) only shows part of R3 As before, we representthe two vectors a and u by −→OA and −−→OP respectively If we project the vector u on to the line OA, thenthe image of the projection is the vector w, represented by −−→OQ On the other hand, if we project thevector u on to a line perpendicular to the line OA, then the image of the projection is the vector v,represented by −−→OR

(6)Note that this diagram is essentially the same as the diagram (3), the only difference being that whilethe diagram (3) shows the whole of R2, the diagram (6) only shows part of R3 As before, we representthe two vectors a and u by −→OA and −−→OP respectively If we project the vector u on to the line OA, thenthe image of the projection is the vector w, represented by −−→OQ On the other hand, if we project thevector u on to a line perpendicular to the line OA, then the image of the projection is the vector v,represented by −−→OR

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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

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 vectorproduct u × v is defined by the determinant

=

det

u2 u3

v2 v3

, − det

u1 u3

v1 v3

, det

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 in R3 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 u3

v2 v3

, − det

u1 u3

v1 v3

, det

1 2

3 2

, det

PROPOSITION 4H (VECTOR PRODUCT) Suppose that u, v, w ∈ R3 and c ∈ R Then

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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 θ ∈ [0, π] represents the angle between them Then

(a) ku × vk2= kuk2kvk2− (u · v)2; and

(b) ku × vk = kukkvk sin θ

Proof Note that

ku × vk2= (u2v3− u3v2)2+ (u3v1− u1v3)2+ (u1v2− u2v1)2 (7)and

u · v = kukkvk cos θ

Combining with part (a), we obtain

ku × vk2= kuk2kvk2− kuk2kvk2cos2θ = kuk2kvk2sin2θ

Consider now a parallelogram with vertices O, A, B, C Suppose that u and v are represented by −→OAand −−→OC respectively If we imagine the side OA to represent the base of the parallelogram, so thatthe base has length kuk, then the height of the the parallelogram is given by kvk sin θ, as shown in thediagram below:

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 θ ∈ [0, π] represents the angle between them Then

(a) #u × v#2= #u#2#v#2− (u · v)2; and

Combining with part (a), we obtain

#u × v#2= #u#2#v#2− #u#2#v#2cos2θ = #u#2#v#2sin2θ

Part (b) follows !

Consider now a parallelogram with vertices O, A, B, C Suppose that u and v are represented by −→OAand −−→OC respectively If we imagine the side OA to represent the base of the parallelogram, so thatthe base has length #u#, then the height of the the parallelogram is given by #v# sin θ, as shown in thediagram below:

!v! sin θ v

u θ

It follows from Proposition 4J that the area of the parallelogram is given by ku × vk 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 ku × vk

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 handoutwards 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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