Chapter 1

Vectors in Three Dimensions

1 Introduction

1.1 Vectors and Scalars

A scalar quantity, λ, is completely represented by a single number (with units).Here all scalars will be real numbers, so we write λ ∈ R. A non-scalar has a scalarquantity, its magnitude, and additional information, such as a direction. If sucha non-scalar obeys certain rules it is called vector quantity, denoted a, with itsmagnitude denoted by either a, |a|, or ‖a‖ ∈ R. Note length, speed and mass areall scalar quantities, but displacement, velocity and weight are all vector quantities.

1.2 Directed Line Segments

A geometrical representation of a vector is the directed line segment, a =−−→AB, drawn

between two fixed points A and B in 3-D space. This has a magnitude, its lengthAB = ‖a‖, and a direction, with the sense of direction indicated by an arrowhead.

A s¡¡

¡¡¡

µ

sB

a =−−→AB

A s¡¡

¡¡¡

ª

sB

b =−−→BA

The directed line segment b =−−→BA represents a vector of equal magnitude but of

opposite direction. We write b = −a so that ‖−a‖ = ‖a‖.

1.3 Classification of Vectors

We define the zero vector, denoted by 0, to be a vector of zero magnitude. It isthe only vector without an associated direction. Two vectors are equal if they haveboth the same magnitude and direction.

1.3.1 Position Vectors

We often define a point O to be the origin (of vectors). The position vector of apoint A, relative to a given origin O, is the directed line segment a =

−→OA

Os»»»»»»»»»»»»

»»»»»»:a =

−→OA

sA

¡¡

¡

¡¡µb =

−−→OB

sB

.

1.3.2 Free Vectors

A free vector is one whose tail is not fixed Thus, unlike the position vector, thegeometrical representation of a free vector is not unique, since a =

−−→AB or a =

−−→CD.

As©©©©©©

*

sB

a =−−→AB

Cs©©©©©©

*

sD

a =−−→CD

1

1.4 Scalar Multiplication

Given a scalar, λ ∈ R, we define the vector λa to lie in the same direction as a forλ > 0, and in the opposite sense for λ < 0, with a magnitude |λ| times that of a.

sA

sB

sC

sC

a =−−→AB

λa =−→AC, λ > 0λa =

−→AC, λ < 0 »»»»»»»»»»»»»

»»»»»»»»»»

»»:»»»»»:»»»»»9

Thus if a =−−→AB and λa =

−→AC, then AC =

∥∥∥−→AC∥∥∥ = |λ| a, where a = ‖a‖.

1.5 Vector Addition

The triangle law states that if the free vectors a =−−→AB and b =

−−→BC represent two

sides of a triangle ABC, then the vector sum a + b represents−→AC. That is

−−→AB +

−−→BC =

−→AC (1.1)

Alternatively the parallelogram law states that if a =−−→AB and b =

−−→AD represent

two adjacent sides of a parallelogram ABCD then a + b represents−→AC.

sA

sB

sC

b

a

a + b

-

µ:

sA

sB

sD sC

b

a

a + b

a

b

-

µ :

-

µ

Moreover vector addition is commutative, in that it satisfies

a + b = b + a

Note the vector difference b− a = b + (−a) represents−−→BD.

Example 1.1 Prove that if L, M and N are the mid-points of the sides AB, BCand CA of the triangle ABC

¡¡

¡¡

¡¡

HHHHHHHHHHH

sA sB

sC

L

MN

then2−−→AB + 3

−−→BC +

−→CA = 2

−→LC

Solution (a) We have−−→AB = 2

−→AL,

−−→BC =

−→BL +

−→LC = −−→AL +

−→LC,

−→CA =

−→CL +

−→LA = −−→LC −−→AL

Therefore

2−−→AB + 3

−−→BC +

−→CA = 4

−→AL + 3

−→BL + 3

−→LC +

−→CL +

−→LA

= 4−→AL− 3

−→AL + 3

−→LC −−→LC −−→AL = 2

−→LC

2

2 Rectangular Cartesian Coordinates

2.1 Rectangular Cartesian Axes

Three mutually perpendicular axes Ox, Oy, Oz are drawn through the origin Osuch that the rectangular cartesian axes Oxyz form a right-handed set. That is, ifwe put thumb, index finger and middle finger, of the right hand, at right angles inthe most natural way and let the index finger point along Ox, the middle fingerpoint along Oy, then the thumb points along Oz.

sO XXXXXXXXXz

6

©©©©©©¼xy

z

£££££££££

£££££±

sP

r

©©*x©©¼XXXzyXXXy

6

z

?

sXXz6

©©¼ı

k

©©©©©©

©©©*

sQa

©©*a1©©¼XXXza2

XXXy?

a3

6

The position of any point P is then specified by its Cartesian coordinates (x, y, z)determined by reference to the three axes. We refer to P as the point P : (x, y, z).

2.2 The Position Vector

Let ı, , k denote unit vectors in the directions of the positive x-, y-, and z-axes

respectively, ‖ı‖ =∥∥

∥∥ =∥∥∥k

∥∥∥ = 1, then the position vector of P relative to O canbe uniquely written as

r =−−→OP = xı + y + zk or r = (x, y, z) ∈ R3 (1.2)

and by Pythagoras’ theorem, the distance, r, of the point P from the origin O is

r = ‖r‖ = OP =√

x2 + y2 + z2 (1.3)

2.3 Components of a Vector

We can write any vector a uniquely in the form

a = a1 ı + a2 + a3k or a = (a1, a2, a3) ∈ R3 (1.4)

where a1, a2 and a3 are called the components of a in the x-, y- and z-directions.We define the zero vector by 0 = (0, 0, 0). The magnitude of a is then given by

a = ‖a‖ =√

a21 + a2

2 + a23 (1.5)

For the vector a =−−→PQ, with initial point P : (xp, yp, zp) and end-point Q :

(xq, yq, zq), we have

a1 = xq − xp, a2 = yq − yp, a3 = zq − zp

so by Pythagoras’s theorem the length PQ = ‖a‖

3

2.4 Vector Algebra

Consider vectors a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3) and scalars λ, µ ∈ R.

2.4.1 Scalar Multiplication

The operation of vector addition, in terms of components, is defined by

λa = (λa1, λa2, λa3) (1.6)

Thus λa is |λ| times as long as a, since

‖λa‖ =√

λ2(a21 + a2

2 + a23) = |λ|

√a21 + a2

2 + a23 = |λ| ‖a‖ (1.7)

and if λ > 0 then λa is parallel to a but if λ < 0 it is anti-parallel to a. Also

0 = 0a, a = 1a, −a = (−1) a = (−a1,−a2,−a3), λ (µa) = µ (λa)

2.4.2 Vector Addition

The operations of vector addition, in terms of components, is defined by

a + b = (a1 + b1, a2 + b2, a3 + b3) (1.8)

Vectors addition is commutative and associative since it satisfies

a + b = b + a, (a + b) + c = a + (b + c)

Vector addition is also distributive over scalar multiplication, since it satisfies

(λ + µ)a = λa + µa, λ (a + b) = λa + λb

Given any vector a there exists vectors 0 = (0, 0, 0) (the additive identity) and−a = (−a1,−a2,−a3) (the additive inverse) such that

a + 0 = a, a + (−a) = 0

We can then define vector subtraction by

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

2.5 Unit Vectors

A unit vector is a vector of unit magnitude, which we will shall denote by a caret(hat). A unit vector in the direction of a vector a = (a1, a2, a3) is

a =a

‖a‖ =(

a1

‖a‖ ,a2

‖a‖ ,a3

‖a‖)

, ‖a‖ = 1

We call the quantities a1/ ‖a‖, a2/ ‖a‖, a3/ ‖a‖ the direction cosines of the vectora. Any vector a can be represented by the scalar multiplication of its magnitudeand a unit vector, a = ‖a‖ e. A unit vector designates a direction but does nothave any physical dimension (no units). Note

ı = (1, 0, 0), = (0, 1, 0), k = (0, 0, 1)

2.6 Linear Combinations

A vector u is a linear combination of three other vectors a, b, c if there are scalarsλ, µ, ν such that

u = λa + µb + νc

4

3 The Equation of a Straight line

Let r =−−→OP be the position vector of the general point P lying on the line l.

sO

sA

sPsBa

r

b¢¢¢¢¢¢¢¢

©©©©©©©©©©©©

»»»»»»»»»»»»»»»»»»

¢¢¢¢

©©©©©©*

»»»»»»»»»:

XXXXXXXXXXXXXXXXXXXXXXXXXl

XXXXzT

3.1 Vector Equation I

Suppose we are given a point A on a line l with position vector a =−→OA and a vector

T which is parallel to the line. From the triangle rule

r =−−→OP =

−→OA +

−→AP

but we know that−→AP is tangential to the vector T . Therefore

−→AP = λT for some

λ ∈ R Thus we can writer = a + λT (1.9)

This is the (parametric) vector equation of the line l which is parallel to the vectorT and passes through the point A, for some parameter λ.

3.2 Vector Equation II

Suppose we are given two points A and B on a line l with position vectors a =−→OA

and b =−−→OB. Since AB lies on the line l, the vector

−−→AB is parallel to l. Further

−−→AB =

−→AO +

−−→OB =

−−→OB −−→OA = b− a

By the same argument as above we have, for some real λ,

r =−−→OP =

−→OA +

−→AP =

−→OA + λ

−−→AB = a + λ (b− a)

which we can write asr = (1− λ)a + λb (1.10)

This is the (parametric) vector equation of the line l which passes through twopoints A and B, for some parameter λ.

3.3 The Cartesian Equation

Let us use the vector equation (1.9) where we assume that

r = (x, y, z), a = (a1, a2, a3), T = (c1, c2, c3)

then the component form of (1.9) is

x = a1 + λc1, x = a2 + λc2, z = a3 + λc3

and upon eliminating the parameter λ we can writex− a1

c1=

y − a2

c2=

z − a3

c3(= λ)

This is the cartesian equation of the line l which is parallel to the vector (c1, c2, c3)and passes through the point (a1, a2, a3).

5

3.4 Intersections of Lines

If two lines l and m intersect at a point P , then the position vector r =−−→OP must

satisfy the vector equation of both lines.

sO

sP1

sPsP2

r1

r

r2

¢¢¢¢

©©©©©©©©©©©©

»»»»»»»»»»»»»»»»»»»

¢¢

©©©©©©*

»»»»»»»»»»:m

l

Example 1.2 Find, if possible, the intersection of the following lines

(i) The line l through (4, 5, 1) parallel to the vector (1, 1, 1) and the line m through(5,−4, 0) parallel to the vector (2,−3, 1).

(ii) The line l passing through the two points (−1, 1, 2) and (0,−2, 4) and the linem passing through the two points (1, 1, 3) and (3, 0, 4).

Solution (i) From (1.9), any point P both on the lines has the position vector

r = r1 = (4, 5, 1) + λ (1, 1, 1) = (4 + λ, 5 + λ, 1 + λ)r = r2 = (5,−4, 0) + µ (2,−3, 1) = (5 + 2µ,−4− 3µ, µ)

These lines intersect only if there are values for λ and µ that make the above twoexpressions for r identical. Hence we need to solve

4 + λ = 5 + 2µ, 5 + λ = −4− 3µ, 1 + λ = µ

Using the third equation to eliminate µ we find upon substitution that

4 + λ = 5 + 2 (1 + λ) ⇒ λ = −3 ⇒ µ = −25 + λ = −4− 3 (1 + λ) ⇒ λ = −3 ⇒ µ = −2

a solution satisfies all equations. Substituting into the equation for r yields

r = (4, 5, 1)− 3 (1, 1, 1) = (1, 2,−2)= (5,−4, 0)− 2 (2,−3, 1) = (1, 2,−2)

(ii) In this case, from (1.10), we have to solve

r = r1 = (1− λ) (−1, 1, 2) + λ(0,−2, 4) = (−1 + λ, 1− 3λ, 2 + 2λ)r = r2 = (1− µ) (1, 1, 3) + µ (3, 0, 4) = (1 + 2µ, 1− µ, 3 + µ)

Hence we need to solve

−1 + λ = 1 + 2µ, 1− 3λ = 1− µ, 2 + 2λ = 3 + µ

For the 2nd equation µ = 3λ. Substitution into the remaining two yields

−1 + λ = 1 + 6λ ⇒ λ = −25

⇒ µ = −65

2 + 2λ = 3 + 3λ ⇒ λ = −1 ⇒ µ = −3

Hence there is no solution to the set of three equations. Hence the two lines do notintersect. That is the lines are either parallel or skew.

6

4 The Scalar Product

4.1 Geometric Definition

The scalar (dot) product of two vectors a =−→OA and b =

−−→OB is defined to be

a · b = ‖a‖ ‖b‖ cos θ = ab cos θ (1.11)

where θ is the angle between OA and OB measured in the plane containing thethree points O, A and B which is perpendicular to the unit vector n.

©©©©©©©

HHHHHHH

©©©©*

HHHHj

6

s

s

s

O

A

Bb

a

n

θI

If θ is acute (|θ| < π/2) then a · b > 0, but if θ is obtuse (|θ| > π/2) then a · b < 0.The term scalar product refers to the fact that this product of two vectors

results in a scalar quantity. Note the scalar product operation is commutative, since

b · a = ‖b‖ ‖a‖ cos(−θ) = ‖a‖ ‖b‖ cos θ = a · bThe scalar product operation is also distributive over addition. That is

a · (b + c) = a · b + a · c, (b + c) · a = b · a + c · a, a · λb = λ(a · b)

4.1.1 length of a vector

By taking the scalar product of a vector a with itself (cos 0 = 1), we have

a · a = ‖a‖ ‖a‖ = ‖a‖2 = a2. (1.12)

4.1.2 Orthogonal Vectors

If a and b are orthogonal (perpendicular) vectors (a ⊥ b), then θ is a right-angle(θ = π/2), and a · b = 0. However if

a · b = 0 ⇒ a = 0 or b = 0 or a ⊥ b

Given that ı, , k are orthogonal, unit vectors we must have

ı · ı = · = k · k = 1, ı · = · k = k · ı = 0

4.2 Algebraic Definition

Using the commutative and distributive properties, we see that

a · b =(a1 ı + a2 + a3k

)·(b1 ı + b2 + b3k

)

= a1b1 ı · ı + a1b2 ı · + a1b3 ı · k + a2b1 · ı + a2b2 · + a2b3 · k + a3b1k · ı + a2b2k · + a3b3k · k

The orthogonality of ı, , k then results in the relation

a · b = a1b1 + a2b2 + a3b3 (1.13)

which confirms our earlier result concerning the magnitude, namely

a2 = ‖a‖2 = a · a = a21 + a2

2 + a23

7

Example 1.3 Given that OACB is a parallelogram, show that it is a rhombus ifand only if the diagonals are at right-angles.

Solution Let a =−→OA and b =

−−→OB. Since OACB is a parallelogram, opposite sides

are equal in length and parallel, so that−−→BC = a and

−→AC = b.

sO

sA

sB sC

b

a¡

¡¡

¡¡

¡¡

¡¡¡

-¡¡¡µ

The diagonals are−−→OC and

−−→BA and from the triangle law

−−→BA =

−−→BO +

−−→OC = −b + a,

−−→OC =

−→OA +

−→AC = a + b

so that−−→BA · −−→OC = (a + b) · (a− b) = a · a + a · b− a · b− b · b

= a · a− b · b = ‖a‖2 − ‖b‖2

This−−→BA · −−→OC = 0 if and only if ‖a‖ = ‖b‖. Thus the diagonals are perpendicular if

and only if all the sides are equal in length, that is the parallelogram is a rhombus.

4.3 Resolving into Components

The orthogonal projection of the line segment OA onto the line segment OB is theline segment ON , where N is the foot of the perpendicular.

sO

sB

sA

sNb

a

¡¡

¡¡

¡¡

¡¡

-¡¡

¡¡µ

θI sO

sA

sBsN

a⊥

a ‖

a

¡¡

¡¡

¡¡

¡¡

-¡¡

¡¡µ 6

θI

If a =−→OA, b =

−−→OB then the component of a vector a in a direction of a vector b is

an = a · b ⇒ an =a · b‖b‖ = ‖a‖ cos θ =

∣∣∣−−→ON∣∣∣ (1.14)

Indeed if a = (a1, a2, a3), or a = a1 ı + a2 + a3k then since

a1 = a · ı, a2 = a · , a3 = a · k (1.15)

which matches our previous definition that a1, a2, a3 are the (Cartesian) compo-nents of a in the directions of ı, , k.

More generally we often wish to resolve a vector into two component vectorswhich are parallel and perpendicular to a given unit vector b, namely

a =(a · b

)b +

[a−

(a · b

)b]

= a ‖ + a⊥ (1.16)

8

4.4 Angles

To find the angle θ between two vectors a and b, we have, from (1.11)

cos θ =a · b

‖a‖ ‖b‖ (1.17)

Consequently if a = (a1, a2, a3) and b = (b1, b2, b3) we see from (1.13) that

cos θ =a1b1 + a2b2 + a3b3√

a21 + a2

2 + a33

√b21 + b2

2 + b23

(1.18)

4.4.1 Direction Cosines

Suppose that θ1, θ2, θ3 are the angles between the direction of a and those of ı, ,k respectively. Given that from (1.17)-(1.18) we have

cos θ1 =a · ı

‖a‖ ‖ı‖ =a1(1) + a2(0) + a3(0)√

a21 + a2

2 + a33.(1)

=a1

|a|and, using (1.15), we can obtain similar results for cos θ2, cos θ3. Therefore

a =a

‖a‖ =(

a1

‖a‖ ,a2

‖a‖ ,a3

‖a‖)

= (cos θ1, cos θ2, cos θ3)

Thus the components of a in the ı, , k are the cosines of the angles made by thedirection of a with the three Cartesian directions. Since ‖a‖ = 1 we must also have

cos2 θ1 + cos2 θ2 + cos2 θ3 = 1

O u

6

-¡

¡¡ª

ı

k

¡¡

¡¡µ

a

θ3θ2

θ1

IR

K

4.5 The Cosine Rule

Consider the triangle ABC with, a = BC, b = AC and c = AB

¡¡

¡¡

¡¡

HHHHHHHHHHH

sA sB

sC

c

ab

θ

Tthen by the triangle rule−−→BC =

−−→BA +

−→AC =

−→AC −−−→AB. From (1.11)

−−→BC · −−→BC =

(−→AC −−−→AB

)·(−→AC −−−→AB

)=−→AC · −→AC − 2

−−→AB · −→AC +

−−→AB · −−→AB

=∥∥∥−→AC

∥∥∥2

− 2∥∥∥−−→AB

∥∥∥∥∥∥−→AC

∥∥∥ cos θ +∥∥∥−−→AB

∥∥∥2

which is the standard formula for the cosine rule, namely

a2 = b2 + c2 − 2bc cos θ (1.19)

9

4.6 The Equation of a Plane

Consider a plane π, with a vector n orthogonal to π, and A a fixed point on theplane, with position vector a =

−→OA. Let r =

−−→OP be the position vector of a general

point P on the plane.

sO

sAsP

sQ

sB

π

a r

nnn

b

q

r − a

3 *

:666

i

Y

6p

?

sN

φ

The vector−→AP is parallel to the plane π, with

−→AP =

−−→OP −−→OA = r − a

and n is perpendicular to the plane π, so

(r − a) · n = 0 or r · n = d = a · n (1.20)

This is the vector equation of the plane passing through the point A, perpendicularto the vector n, with n pointing away from the origin O if d > 0.

If we assume that r = (x, y, z), a = (a, b, c) and n = (l, m, n) then we have

lx + my + nz = d (1.21)

where d = lx + my + nz, which is the cartesian equation of the plane.

4.7 Distance of a point from a Plane

The shortest distance of a point B from the plane π is the length of the perpendicularBQ to π. Clearly

q =−−→OQ =

−−→OB +

−−→BQ = b + λn

for some scalar λ. But since Q lies on π it also satisfies (1.20), so

q · n = (b + λn) · n = b · n + λ ‖n‖2 = d

so that the scalar λ is given by

λ =(d− b · n)‖n‖2 =

(a− b) · n‖n‖2

Hence is the perpendicular distance of π from the point B is

p =∥∥∥−−→BQ

∥∥∥ =∥∥q − b

∥∥ = |λ| ‖n‖ =|(a− b) · n|

‖n‖ =|d− b · n|‖n‖ (1.22)

Alternatively suppose N is the foot of the perpendicular from the origin O to theplane π, then clearly

−−→ON = µn, for some scalar µ. However if φ is the angle between

a and n then the perpendicular distance ON is given by

|µ| ‖n‖ =∥∥∥−−→ON

∥∥∥ = ‖a‖ |cosφ| = |a · n|‖n‖ =

|d|‖n‖ (1.23)

by angle definition (1.17). Note that by setting b = 0 in (1.22) we recover (1.23).

10

5 The Vector Product

5.1 Definition

The vector (cross) product of two vectors a and b is defined to be

a× b = ‖a‖ ‖b‖ sin θn (1.24)

Here n is a unit vector perpendicular to both a and b, a · n = b · n = 0. θ is the anglefrom a =

−→OA to b =

−−→OB, chosen so 0 ≤ θ ≤ π. For an observer looking along n, θ

increases in a clockwise sense. Thus a, b, n form a right-handed triad of vectors.

sO

sB

sA

b

an

PPPPPP

PPPq

6

©©©*?

θ©©©©©©©

HHHHHHH

©©©©*

HHHHj

6

s

s

s

O

A

B

HHjHHY

|b| cos θ

¡¡µ|b| sin θ

¡¡ª

b

a

a× b

θ

I

I

The vector product operation is not commutative but is distributive over addition

a× a = 0, a× b = −b× a, a× (b + c) = a× b + a× c, a× λb = λ(a× b)

Moreover since sin 0 = 0 and ı, , k are right-handed orthogonal unit vectors

ı× ı = 0, × = 0, k × k = 0, ı× = k, × k = ı, k × ı =

From the above properties the vector product has components

a× b =(a1 ı + a2 + a3k

)×

(b1 ı + b2 + b3k

)

= a1b1 ı× ı + a1b2 ı× + a1b3 ı× k + a2b1× ı + a2b2×

+ a2b3× k + a3b1k × ı + a2b2k × + a3b3k × k

Thus the vector product has components

a× b = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1) (1.25)

which we can concisely write in terms of the determinant

a× b =

∣∣∣∣∣∣ı ka1 a2 a3

b1 b2 b3

∣∣∣∣∣∣(1.26)

5.2 The Scalar ‖a× b‖If a =

−→OA, b =

−−→OB represent two adjacent sides of a parallelogram, the scalar

‖a× b‖ = ‖a‖ ‖b‖ sin θ (1.27)

then since h = ‖b‖ sin θ is the height of an equivalent rectangle

sO

sA

sB

b

aθI -

¸

6

b sin θ

?

Thus geometrically ‖a× b‖ represents the area of the parallelogram.

11

5.3 The Vector Equation of a Plane

Suppose a plane π passes through three points A, B and C with position vectorsa =

−→OA, b =

−−→OB and c =

−−→OC.

sO

sAsB

s C sP

π

ar

b− a

c− a r − ann

©©©*©©©©©©

:

*

:

3

66

Since b− a and c− a lie in π, a vector normal to the plane is

n = (b− a)× (c− a) .

Thus from (1.20) the vector equation of the plane is

(r − a) · [(b− a)× (c− a)] = 0 (1.28)

Alternatively the vector r − a is coplanar with the vectors b− a and c− a, so thatr − a is expressible as a linear combination of these two vectors. Thus the generalpoint P has the position vector

r = a + λ (b− a) + µ (c− a) (1.29)

for some scalars λ and µ. This is the (parametric) vector equation of the plane.

Example 1.4 Find the vector and Cartesian equations of the plane passing throughthe points A, B and C with coordinates (1, 2, 3), (2,−1, 2) and (1,−1, 1).

Solution With the usual notation

b− a = (2,−1, 2)− (1, 2, 3) = (1,−3,−1)c− a = (1,−1, 1)− (1, 2, 3) = (0,−3,−2)

so thatn = (b− a)× (c− a) = (3, 2,−3)

and the vector equation of the plane, (1.20) or (1.28), is

[r − (1, 2, 3)] · (3, 2,−3) = 0

Since r = (x, y, z) the Cartesian equation is

3x + 2y − 3z = 3 + 4− 9 = −2

Alternatively we have the vector equation

r = (1, 2, 3) + λ (1,−3,−1) + µ (0,−3,−2)

Since r = (x, y, z) we have

x = 1 + λy = 2− 3λ− 3µz = 3− λ− 2µ

⇒λ = x− 13µ = 5− 3x− yz = 2

3 + x + 23y

Thus the Cartesian equation is

3x + 2y − 3z = −2

12

5.4 The Intersection of two Planes

Two planes π and π′ with vector and Cartesian equations

r · n ≡ ax + by + cz = d, r ·m = a′x + b′y + c′z = d′

with normals n = (a, b, c) and m = (a′, b′, c′) , intersect in a straight line l unless (i)the planes are parallel, or (ii) the planes are coincident.

π π′

C

l@@@I

n

@@@I

n¡

¡¡µm

¡¡¡µ

ms

¡¡

¡¡

¡¡

¡¡¡

¡¡

¡¡¡

@@

@@

@@

@@@

@@

@@@

Given that n and m are perpendicular to π and π′, and the line l lies in both planes,then the direction of l must be perpendicular to both n and m. Consequently thevector equation of the line of intersection l is given by

r = c + λ (n×m) (1.30)

where c =−−→OC is the position vector of any point C lying in both π and π′

Example 1.5 Find the equation of the line of intersection of the planes

x− y − 2z = 3 and 2x + 3y + 4z = −2

Solution On the line of intersection both equations must hold. From the firstequation x = y + 2z + 3, and substituting into the second equation yields

2 (y + 2z + 3) + 3y + 4z = −2 ⇒ 5y + 8z = −8

Consequently we have

y = −85(1 + z) ⇒ x = −8

5(1 + z) + 2z + 3 =

15

(7 + 2z)

Thus the position vector

r = (x, y, z) =15

(7,−8, 0) +15z (2,−8, 5) = c + λT

which is the equation of a straight line passing through the point C, with positionvector c =

−−→OC = 1

5 (7,−8, 0) and having a direction parallel to T = (2,−8, 5).Alternatively we note that

n×m =

∣∣∣∣∣∣ı k1 −1 −22 3 4

∣∣∣∣∣∣= (2,−8, 5)

and to find the position of any point in both planes, we can set z = 0, since neitherplane is parallel to the plane z = 0. Consequently

x− y = 32x + 3y = −2 ⇒ x =

75, y = −8

5

which recovers the earlier result c = 15 (7,−8, 0)

13

5.5 Vector equation of a line III

The parametric equation of a line l, with direction T and passing through A, B, is

r = a + λT or r = a + λ (b− a)

where a =−→OA, b =

−−→OB. Using the vector product to eliminate the parameter λ

then yields the vector equation of the line l

r × T = a× T or r × (b− a) = a× b (1.31)

5.6 Distance of a point from a line

The shortest distance from a point P , where p =−−→OP , to a line l with equation

r = a + λT ,

is the length of the perpendicular PQ to l. Here a =−→OA is the position vector of

any point A on l and T is a vector parallel to l.

³³³³³³³³³³³³³³³³³³³ l

sO

sA sP

@@

@@s

Q

θ

³³³³1T

zp− a

Iq − p

¸a

1p

From the right-angled triangle PQA, the perpendicular distance

d =∥∥q − p

∥∥ =∥∥∥−→AP

∥∥∥ sin θ =∥∥p− a

∥∥ sin θ

where θ is the angle between−→AP and T . Hence by definition of (1.27)

d =

∥∥(p− a

)× T∥∥

‖T‖ (1.32)

Example 1.6 Find the distance of the point (4, 2,−1) from the line

r = (2, 3, 2) + λ (1, 2,−2)

Solution A point A on the line has position vector−→OA = (2, 3, 2) so that

−→AP = p− a = (4, 2,−1)− (2, 3, 2) = (2,−1,−3)

In addition we have

‖T‖ = ‖(1, 2,−2)‖ =√

1 + 4 + 4 =√

9 = 3

−→AP × T =

∣∣∣∣∣∣ı k2 −1 −31 2 −2

∣∣∣∣∣∣= (2 + 6) ı− (−4 + 3) + (4 + 1) k

∥∥∥−→AP × T∥∥∥ = ‖(8, 1, 5)‖ =

√64 + 1 + 25 =

√90 = 3

√10

Thus the perpendicular distance of P from the line

d =

∥∥(p− a

)× T∥∥

‖T‖ =√

10

14

6 The Triple Scalar Product

For three vectors a = (a1, a2, a3), b = (b1, b2, b3) and c = (c1, c2, c3), the triplescalar product is the scalar

[a, b, c] = a · (b× c) (1.33)

In terms of the cartesian components, the scalar product is defined by

a · (b× c) = a1b2c3 − a1b3c2 + a2b3c1 − a2b1c3 + a3b1c2 − a3b2c1 (1.34)

or more compactly in the determinant form

[a, b, c] = a · (b× c) =

∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣(1.35)

6.1 Geometrical Interpretation

Consider a parallelepiped with adjacent edges a =−→OA, b =

−−→OB and c =

−−→OC, whose

faces are all parallelograms. Taking OBDC as the base parallelogram with area

A = bc sin θ = ‖b× c‖ where b× c = ‖b‖ ‖c‖ sin θn

and n is a unit vector perpendicular to base. The height of the parallelepiped is

h = a cos φ = ‖a‖ cosφ = |a · n| =∣∣∣∣a ·

b× c

‖b× c‖

∣∣∣∣

sO

sA

sB

sC sD

θφ

6

h

?

b× c

ac

b

I6

-

¸*

The volume V of the parallelepiped equals ’area of base × height’.

V = hA = |abc cosφ sin θ| = |a · (b× c)| (1.36)

6.2 Cyclic Symmetry

Now we can show that

a · (b× c) = b · (c× a) = c · (a× b) (1.37)

Thus the value only depends on the cyclic order of the three vectors

[a, b, c] = [b, c, a] = [c, a, b] = − [a, c, b] = − [c, b, a] = − [b, a, c]

In particular interchanging two vectors reverses the sign. Consequently

[a, a, b] = 0 (1.38)

15

6.3 Linear Independence

We say that the vectors a, b, c are linear independent if the only solution to

λa + µb + νc = 0

is the trivial solution λ = µ = ν = 0. If there is a non-trivial solution we say thatthe vectors are linearly dependent. Moreover, if the set of vectors are linearlydependent, and w.l.o.g. λ 6= 0, then we can write

a = −µ

λb− ν

λc

That is a lies in the same plane as the vectors b and c. Conversely three vectors arelinearly independent if, and only if, they are non-coplanar.

Theorem 1.1 Three non-zero vectors a, b, c are coplanar, if and only if,

[a, b, c] ≡ a · (b× c) = 0 (1.39)

Proof The volume V of the parallelepiped with adjacent edges a, b and c is zero ifand only if the vectors are coplanar. Thus (1.36) shows that a, b, c are co-planar ifand only if a · (b× c) = 0.

Theorem 1.2 Given any three non-zero, non-coplanar (linearly independent) vec-tors a, b, c, any other vector x may be uniquely expressed in the form

x = λa + µb + νc

where λ, µ, ν are scalars.

Proof Let x = (x1, x2, x3), a = (a1, a2, a3), r = (b1, b2, b3), c = (c1, c2, c3) then

x = λa + µb + νc

if and only if there is a solution λ = (λ, µ, ν) to the simultaneous equations

x1

x1

x3

=

a1 a2 a3

b1 b2 b3

c1 c2 c3

λµν

or x = Aλ

In turn these have a solution λ = A−1x, where A−1 =adj(A)/ |A|, if and only if

[a, b, c] ≡ a · (b× c) ≡∣∣∣∣∣∣

a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣≡ |A| 6= 0

6.4 Intersection of two Straight Lines

Suppose that the two straight lines

r = a + λb and r = c + µd

intersect at a single point P . Then at the point of intersection

r = a + λb = c + µd

From which we see thatc− a = λb− µd

Hence the vector c− a lies in the plane containing b and d. Thus from (1.39), thecondition that two lines intersect is

[c− a, b, d] ≡ (c− a) · (b× d) = 0 (1.40)

16

6.5 Minimum distance between two skew lines

Conversely, let us find the minimum distance between two skew lines, l and m. Letus denote P and Q as the points of closest approach, with

p =−−→OP = a + λb, q =

−−→OQ = c + µd

where a =−→OA, c =

−−→OC, for some scalars λ and µ. The situation is as shown

sO

¡¡

¡¡

¡¡

¡¡¡

l

sQ

?q − p

³³³³³³³³³³³³³³³³³³³

m

sP³³³1b

¡¡¡µ

d

¸a

sA

1c

sC

Assuming the two lines are non-parallel, b× d 6= 0, we have−−→PQ = q − p = c− a + µd− λb = k (b× d)

for some scalar k, since−−→PQ must be perpendicular to both b and d. Thus

−−→PQ · (b× d) = (c− a) · (b× d) = k ‖b× d‖2

so that the scalar k is given by the triple scalar product

k =(c− a) · (b× d)

‖b× d‖2 =[c− a, b, d]‖b× d‖2

Hence the minimum distance D is given by

D =∥∥∥−−→PQ

∥∥∥ = |k| ‖b× d‖ =|(c− a) · (b× d)|

‖b× d‖ (1.41)

where, as required, D = 0 if (1.40) is satisfied and intersection occurs.

Example 1.7 Find the minimum distance between the following two skew lines

r = a + λb = (1, 2, 3) + λ(1, 2, 2)r = c + µd = (2, 1, 2) + µ(1, 3, 1)

Solution We have

(c− a) = (2, 1, 2)− (1, 2, 3) = (1,−1,−1)

b× d =

∣∣∣∣∣∣ı k1 2 21 3 1

∣∣∣∣∣∣= (−4, 1, 1)

‖b× d‖ =√

18 = 3√

2(c− a) · (b× d) = (1,−1,−1) · (−4, 1, 1) = −4− 1− 1 = −6

so that the minimum distance between these two skew lines is

D =|(c− a) · (b× d)|

‖b× d‖ =|−6|3√

2=√

2

17

6.6 Moment of a Force

Consider the moment about a point O of a force F acting at a point P , with positionvector r =

−−→OP .

sO ³³³³³³

³³³1¡

¡¡¡µ

sP

dy

z

F

r

Yθ

©©©©©©

HHHHHH

©©©*HHHj

6

sO

F

r

M

θI

I

By definition, the magnitude of the moment is magnitude of the force times theperpendicular distance d to its line of action

M = Fd = ‖F‖ ‖r‖ sin θ

where θ is the angle between the vectors r and F . The force will tend to rotate thepoint P about an axis, through O, perpendicular to both r and F . If we adopt aright-hand screw rule, where the fingers curl in direction of rotation and the thumbindicates the direction for the moment, then r, F and the vector moment M forma right-hand triad of vectors, where

M = r × F (1.42)

6.7 Angular Velocity

For a rigid body is rotating with an angular velocity ω = dφ/dt about an axis l,the right-hand screw rule means we have both a magnitude and a direction n

sO ¤

¤¤¤¤¤¤¤¤¤¤¤

l

ωy

sA:

a

sC XXXXsP¤¤¤¤º

n

¡¡

¡µv

*r µr − aθ

sC ¡¡¡d@

@@Iv

s P

φk

Thus we can represent rigid-body rotation by an angular velocity vector

ω = ωn

Let v be the velocity of a general point P , with position vector r =−−→OP . Each point

P describes a circle, of radius d, that lies in a plane perpendicular to the axis l witha centre C on the axis. Let A be some fixed point on the axis l, with a =

−→OA, such

that r − a makes an angle θ with the axis l, then the magnitude the velocity is

v = ωd = ‖r − a‖ sin θ

The direction of the v is perpendicular to both−−→CP and the axis l, so that

v · (r − a) = v · [(r − c) + ‖r − a‖ cos θn] = 0 + 0 = 0

Given that r − a, v, n make a right-handed triad set of vectors, we see that

v = ω × (r − a) where ω = ωn (1.43)

18

7 The Triple Vector Product

The triple vector product of three vectors a, b and c is the vector defined by

a× (b× c) = (a · c) b− (a · b) c (1.44)

Because b× c is a vector d (say) which is perpendicular to both b and c, and a× dis a vector perpendicular to both a and d, and so lies in the plane of b and cperpendicular to a. We also note that

(a× b)× c = (a · c) b− (b · c) a (1.45)

7.1 Component vectors II

It is sometimes useful to express a vector a as the sum of two component vectors,one parallel and one perpendicular to a given vector b, namely

a = a ‖ + a⊥ = λb + a⊥ where b · a⊥ = 0

From (1.14)-(1.16), or by taking scalar product with b, we have

λ =a · b‖b‖2 =

a · b‖b‖ ⇒ λb =

(a · b

)b

whilst the component perpendicular to b

a⊥ = a− (a · b)‖b‖2 b =

1‖b‖2 {(b · b) a− (a · b) b} =

1‖b‖2 b× (a× b)

from (1.44). Hence

a =1‖b‖2 {(a · b)b + b× (a× b)} (1.46)

Example 1.8 For given vectors a, b solve, for x, the vector equation

x× a = b

Solution Clearly b must be perpendicular to both a and x. Thus if a · b 6= 0 thenno solution exists. Clearly a, b and a× b are mutually perpendicular so that fromTheorem 1.2 we can express x as

x = λa + µ (a× b) + νb where a · b = 0

but since x is perpendicular to b we must have

ν = x · b = 0

Therefore

b = x× a = λ (a× a) + µ (a× b)× a = µ (a× b)× a

= µ[‖a‖ 2 b− (a · b) a

]= µ ‖a‖ 2b

from (1.45). Thus µ = 1/ |a|2. Hence the general solution is given by

x = λa +1

‖a‖2 (a× b) (1.47)

for any scalar λ ∈ R, which is the equation of a straight line.

19

7.2 Distance of a point from a line II

We recall that the shortest distance from a point P , where p =−−→OP , to a line l, is

the length of the perpendicular PQ to l.

³³³³³³³³³³³³³³³³³³³ l

sO

sA sP

@@

@@s

Q

θ

³³³³1T

zp− a

Iq − p

¸a

1p

Suppose a =−→OA for a point A on l and T is a vector parallel to l, then

q =−−→OQ =

−→OA +

−→AQ = a + λT

for some scalar λ. Since−−→PQ is perpendicular to the line l

(q − p

) · T =[(

a− p)

+ λT] · T

=(a− p

) · T + λ ‖T‖2 = 0

which yields the scalar

λ = −(a− p

) · T‖T‖2 ⇒ q = a− 1

‖T‖2[(

a− p) · T ]

T

which in turn yields

q − p =(a− p

)− 1‖T‖2

[(a− p

) · T ]T

Now from the vector product (1.44)

T × [(a− p

)× T]

= ‖T‖2 (a− p

)− [(a− p

) · T ]T

so we can writeq − p =

1‖T‖2 T × [(

a− p)× T

]

from which it follows that we recover the result (1.32)

PQ =∥∥q − p

∥∥ =

∥∥(p− a

)× T∥∥

‖T‖ (1.48)

20

8 Quartic Products

8.1 First Product

Given the dot and cross in a triple scalar product are interchangeable it follows that

(a× b) · (c× d) = a · [b× (c× d)] = a · [(b · d) c− (b · c) d]= (a · c)(b · d)− (a · d)(b · c)

after using (1.44). We can write this in the determinant form

(a× b) · (c× d) =∣∣∣∣

a · c a · db · c b · d

∣∣∣∣ (1.49)

8.2 2nd Product

If we consider a× b as a single vector and apply (1.44) then

(a× b)× (c× d) = [(a× b) · d] c− [(a× b) · c] d

= [a · (b× d)] c− [a · (b× c)] d (1.50)

If we consider c× d as a single vector and apply (1.45) then

(a× b)× (c× d) = [a · (c× d)] b− [b · (c× d)] a

= [c · (d× a)] b− [c · (d× b)] a (1.51)

Example 1.9 Show that, in confirmation of Theorem 1.2, provided [a, b, c] 6= 0,any other vector x can always be expressed in the form

x =[x, b, c][a, b, c]

a +[a, x, c][a, b, c]

b +[a, b, x][a, b, c]

c (1.52)

Solution From (1.50) and (1.51) we have

(a× b)× (c× x) = [(a× b) · x] c− [(a× b) · c] x

= [a · (c× x)] b− [b · (c× x)] a

Upon equating these two forms and re-arranging

[(a× b) · c] x = [b · (c× x)] a− [a · (c× x)] b + [(a× b) · x] c

= [b, c, x] a− [a, c, x] b + [x, a, b] c

= [x, b, c] a + [a, x, c] b + [a, b, x] c

where from (1.37) we recall that provided the cyclic order is maintained, the valueis unchanged, but interchanging just two values reverses the sign.

21

8.3 Common Point of three Planes

Suppose that the three planes π1, π2 and π3, defined by

r · n1 = d1, r · n2 = d2, r · n3 = d3

intersect at a common point P .

³³³³³³³³

""

""

""

""

""

""

¢¢

¢¢

¢¢

¢¢

Ps¡

¡¡

¡¡

¡¡¡¡

¡¡

¡¡

¡¡¡

¡¡

¡¡

¡

JJ

JJ

J π1

π2 π3

We know that planes π1, π2 intersect at a line with direction n1×n2. Using (1.44)this line has an equation

r × (n1 × n2) = (r · n2) n1 − (r · n1)n2

= d2n1 − d1n2

where we note that the right-hand side is perpendicular to n1×n2. For intersectionwith the plane π3 some point on this line must also satisfy r ·n3 = d3. Using (1.44)we see that

n3 × (d2n1 − d1n2) = n3 × [r × (n1 × n2)]= [n3 · (n1 × n2)] r − (n3 · r) (n1 × n2)= [n1, n2, n3] r − d3 (n1 × n2)

Hence we have

r =d1 (n3 × n1) + d2 (n2 × n3) + d3 (n1 × n2)

[n1, n2, n3]

where we can easily check that

r · n1 = d1, r · n2 = d2, r · n3 = d3

We see that the condition for the planes to intersect is that

[n1, n2, n3] 6= 0

That is the three normals are non-coplanar.

22

9 Vector Area

We can represent any given plane area π of area A, with a unit normal n, by itsvector (directed) area

A = An (1.53)

The orientation of the unit normal n is often fixed by assigning a sense of directionto the boundary curve of the plane area and applying a right-hand (screw) rule.

Area A

?π

9

:B

BBM

A = An

Note a× b is the vector area of the parallelogram formed by the vectors a and b.

9.1 Projections of Vector Area

Consider the vector area A = An of some plane surface

ª>

-

CCCCO

A = An

A

θ

6

A0 = A0n0

A0

6

¡¡

¡¡ª

-

x

y

z

XXXXX¡¡¡XXXXX

¡¡¡

JJ]

¡¡µ

6A1 A 2

A1 + A 2

z

A1 · k9

A 2 · k

Given that A0 is the (signed) area projection of A along n0

A0 = ±A cos θ = ±A · n0 (1.54)

where the sign depends on whether θ is acute or obtuse. Consider a surface S madeup of planar elements S = ∪iSi. The (signed) area projection of such a surface isequal to the sum of individual projections which, in turn, is equal to the projectionof the boundary curve (rim). Thus we can define the total vector area of thesurface S as the sum of the individual vector areas.

A =∑

i

Ai

In the above example of two segments the area projected onto the xy-plane

A1 · k + A 2 · k = (A1 + A 2) · k = A · kwhere A = A1 + A 2 is the total vector area

We can partition any general curved surface S into many small planar ele-ments and in the limit replace the sum by an integral.

9.2 Closed Surfaces

For a closed surface we normally orientate matters so that the unit normals pointout of the finite enclosed volume. In addition the total vector area for any closedsurface is zero. Thus

A =∑

i

Ai = 0 (1.55)

23

Example 1.10 Express the areas of the side of a tetrahedron in terms of the areaA of the inclined surface with unit outward-drawn normal n = (n1, n2, n3)

sO

6

©©©©©©©*

HHHHHHHj x

y

z

Area Aª

?

©©©¼

HHY

A1

A2

A3

¡¡¡µ

A = An

Solution Given A = An the remaining sides have vector areas

A1 = −A1 ı, A2 = −A2, A3 = −A3k

where from (1.54) the projected areas are

A1 = A · ı = An1, A2 = A · = An2, A3 = A · k = An3

We note then from (1.55) that since this is a closed surface

0 = A +3∑

i=1

Ai = A− (A1, A2, A3) = A−A (n1, n2, n3) = A−An

Example 1.11 What is the vector area of a truncated hollow cone bounded by ahorizontal circle of radius 4 units at z = 0 and a horizontal circle of radius 3 unitsat z = 3.

sO

6

z

z = 3

z = 0

-3¾

-4¾

Solution Let A be the vector area. Consider now the closed surface formedby capping the truncated cone with discs of vector area A1 = 9πk (upper) andA2 = −16πk (lower). Then by (1.55)

∑

i

Ai = A + 9πk − 16πk = 0

Hence we have the resultA = 7πk

24