7.

5 The scalar product 241

O
E

Figure 7.24
A charge at O gives rise to an electric field E.

If a second charge is placed in this field it experiences a force. An important quan-

tity is the electric field strength, E. This is a vector field which describes the force
experienced by a unit charge.
A related quantity is the electric displacement, D, also called the electric flux
density, defined as D = εE, where ε is called the permittivity of the medium in
which the field is located. Note that D is a scalar multiple of E.

Engineering application 7.6

Electrostatic potential V
An important electrostatic field is the electrostatic potential V . This is an example
of a scalar field. The difference between the potential measured at any two points in
the field is equal to the work which needs to be done to move a unit charge from one
point to the other. Later, in Chapter 26, we will see that the scalar field V is closely
related to the vector field E.

7.5 THE SCALAR PRODUCT

Given any two vectors a and b, there are two ways in which we can define their product.
These are known as the scalar product and the vector product. As the names suggest, the
result of finding a scalar product is a scalar whereas the result of finding a vector product
b is a vector. The scalar product of a and b is written as
u a·b
a
This notation gives rise to the alternative name dot product. It is defined by the formula
Figure 7.25
Two vectors a and b a · b = |a!b| cos θ
separated by
angle θ . where θ is the angle between the two vectors as shown in Figure 7.25.
242 Chapter 7 Vectors

From the definition of the scalar product, it is possible to show that the following
rules hold:

a·b = b·a the scalar product is commutative

k(a · b) = (ka · b) where k is a scalar
(a + b) · c = (a · c) + (b · c) the distributive rule

It is important at this stage to realize that notation is very important in vector work. You
should not use a × to denote the scalar product because this is the symbol we shall use
for the vector product.

Example 7.12 If a and b are parallel vectors, show that a · b = |a!b|. If a and b are orthogonal show
that their scalar product is zero.

Solution If a and b are parallel then the angle between them is zero. Therefore a · b =
|a!b| cos 0◦ = |a||b|. If a and b are orthogonal, then the angle between them is 90◦
and a · b = |a||b| cos 90◦ = 0.
Similarly we can show that if a and b are two non-zero vectors for which a · b = 0,
then a and b must be orthogonal.

If a and b are parallel vectors, a · b = |a!b|.

If a and b are orthogonal vectors, a · b = 0.

An immediate consequence of the previous result is the following useful set of formulae:

i·i = j·j = k·k = 1

i·j = j·k = k·i = 0

Example 7.13 If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k show that a · b = a1 b1 + a2 b2 + a3 b3 .

Solution We have
a · b = (a1 i + a2 j + a3 k) · (b1 i + b2 j + b3 k)
= a1 i · (b1 i + b2 j + b3 k) + a2 j · (b1 i + b2 j + b3 k)
+ a3 k · (b1 i + b2 j + b3 k)
= a1 b1 i · i + a1 b2 i · j + a1 b3 i · k + a2 b1 j · i + a2 b2 j · j + a2 b3 j · k
+ a3 b1 k · i + a3 b2 k · j + a3 b3 k · k
= a1 b1 + a2 b2 + a3 b3
as required. Thus, given two vectors in component form their scalar product is the sum
of the products of corresponding components.

The result developed in Example 7.13 is important and should be memorized:

If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k,
then a · b = a1 b1 + a2 b2 + a3 b3 .
 7.5 The scalar product 243

Example 7.14 If a = 5i − 3j + 2k and b = −2i + 4j + k find the scalar product a · b.

Solution Using the previous result we find
a · b = (5i − 3j + 2k) · (−2i + 4j + k)
= (5)(−2) + (−3)(4) + (2)(1)
= −10 − 12 + 2
= −20

Example 7.15 If a = a1 i + a2 j + a3 k, find

(a) a · a (b) |a|2

Solution (a) Using the previous result we find

a · a = (a1 i + a2 j + a3 k) · (a1 i + a2 j + a3 k)
= a21 + a22 + a23
!
(b) From Example 7.8 we know that the modulus of r = xi + yj + zk is x2 + y2 + z2
and therefore
"
|a| = a21 + a22 + a23

so that

|a|2 = a21 + a22 + a23

We note the general result that

a · a = |a|2

Example 7.16 If a = 3i + j − k and b = 2i + j + 2k find a · b and the angle between a and b.

Solution We have
a · b = (3)(2) + (1)(1) + (−1)(2)
=6+1−2
=5
Furthermore, from the definition of the scalar product a · b = |a||b| cos θ . Now,
√ √ √
|a| = 9 + 1 + 1 = 11 and |b| = 4 + 1 + 4 = 3

Therefore,
a·b 5
cos θ = = √
|a!b| 3 11
from which we deduce that θ = 59.8◦ or 1.04 radians.
244 Chapter 7 Vectors

Engineering application 7.7

Work done by a force

If a force is applied to an object and the object moves, then work is done by the force.
It is possible to use vectors to calculate the work done. Suppose a constant force F
is applied and as a consequence the object moves from A to B, represented by the
displacement vector s, as shown in Figure 7.26.
We can resolve the force into two perpendicular components, one parallel to s and
one perpendicular to s. The work done by each component is equal to the product of
its magnitude and the distance moved in its direction. The component perpendicular
to s will not do any work because there is no movement in this direction. For the
component along s, that is |F| cos θ , we find
work done = |F| cos θ |s|
From the definition of the scalar product we see that the r.h.s. of this expression is
the scalar product of F and s.
F cos u

F
F sin u
u Figure 7.26
The component of F in the direction of s
A B
AB = s is F cos θ .

The work done by a constant force F which moves an object along the vector s is
equal to the scalar product F · s.

Example 7.17 A force F = 3i + 2j − k is applied to an object which moves through a displacement

s = 2i + 2j + k. Find the work done by the force.

Solution The work done is equal to

F · s = (3i + 2j − k) · (2i + 2j + k)
=6+4−1
= 9 joules

Engineering application 7.8

Movement of a charged particle in an electric field

Figure 7.27 shows two charged plates situated in a vacuum. Plate A has an excess of
positive charge, while plate B has an excess of negative charge. Such an arrangement
gives rise to an electric field. An electric field is an example of a vector field.
 7.5 The scalar product 245

X
A E B
+ –
+ ds –
+ –
+ –
+ E –
+ –
+ –
+ –
+ –
+ – Figure 7.27
E Two charged plates situated in a vacuum.

In Figure 7.27 the electric field vector E in the region of space between the charged
plates has a direction perpendicular to the plates pointing from A to B. The magnitude
of the electric field vector is constant in this region if end effects are ignored. If
a charged particle is required to move against an electric field, then work must be
done to achieve this. For example, to transport a positively charged particle from the
surface of plate B to the surface of plate A would require work to be done. This would
lead to an increase in potential of the charged particle.
If s represents the displacement and V the potential it is conventional to write δs
to represent a very small change in displacement, and δV to represent a very small
change in potential.
If a unit positive charge is moved a small displacement δs in an electric field (Fig-
ure 7.27) then the change in potential δV is given by
δV = −E · δs (7.1)
This is an example of the use of a scalar product. Notice that although E and δs are
vector quantities the change in potential, δV , is a scalar.
Consider again the charged plates of Figure 7.27. If a unit charge is moved a small
displacement along the plane X, then δs is perpendicular to E. So,
δV = −E · δs = −|E!δs| cos θ
With θ = 90◦ , we find δV = 0. The surface X is known as an equipotential surface
because movement of a charged particle along this surface does not give rise to a
change in its potential. Movement of a charge in a direction parallel to the electric
field gives rise to the maximum change in potential, as for this case θ = 0◦ .

EXERCISES 7.5

1 If a = 3i − 7j and b = 2i + 4j find a · b, b · a, a · a 5 Find the angle between the vectors 7i + j and 4j − k.

and b · b.
6 Find the angle between the vectors 4i − 2j and 3i − 3j.
2 If a = 4i + 2j − k, b = 3i − 3j + 3k and
c = 2i − j − k, find 7 If a = 7i + 8j and b = 5i find a · b̂.
(a) a · a (b) a · b    
3 5
(c) a · c (d) b · c 8 If r1 =  1  and r2 =  1  find r1 · r1 , r1 · r2
3 Evaluate (−13i − 5j) · (−3i + 4j). 2 0
and r2 · r2 .
4 Find the angle between the vectors p = 7i + 3j + 2k
9 Given that p = 2q simplify p · q, (p + 5q) · q and
and q = 2i − j + k.
(q − p) · p.
246 Chapter 7 Vectors

10 Find the modulus of a = i − j − k, the modulus of 13 Use the scalar product to find a two-dimensional
b = 2i + j + 2k and the scalar product a · b. Deduce vector a = a1 i + a2 j perpendicular to the vector
the angle between a and b. 4i − 2j.
11 If a = 2i + 2j − k and b = 3i − 6j + 2k, find 14 If a = 3i − 2j, b = 7i + 5j and c = 9i − j, show that
|a|, |b|, a · b and the angle between a and b. a · (b − c) = (a · b) − (a · c).
12 Use a vector method to show that the diagonals of the 15 Find the work done by the force F = 3i − j + k
rhombus shown in Figure 7.28 intersect at 90◦ . in moving an object through a displacement
s = 3i + 5j.
B
16 A force of magnitude 14 N acts in the direction
i + j + k upon an object. It causes the object to move
A C from point A(2, 1, 0) to point B(3, 3, 3). Find the
work done by the force.
17 (a) Use the scalar product to show that the
D component of a in the direction of b is a · b̂,
Figure 7.28 where b̂ is a unit vector in the direction of b.
The rhombus ABCD. (b) Find the component of 2i + 3j in the direction of
i + 5j.

Solutions

1 −22, −22, 58, 20 9 2|q|2 , 7|q|2 , −2|q|2

√
2 (a) 21 (b) 3 (c) 7 (d) 6 10 3, 3, −1, 101.1◦

3 19 11 3, 7, −8, 112.4◦

4 47.62◦ 13 c(i + 2j), c constant

5 82.11◦ 15 4 J
6 18.4◦ 16 48.5 J
√
7 7 17 17/ 26
8 14, 16, 26

7.6 THE VECTOR PRODUCT

The result of finding the vector product of a and b is a vector of length |a||b| sin θ ,
where θ is the angle between a and b. The direction of this vector is such that it is
perpendicular to a and to b, and so it is perpendicular to the plane containing a and
b (Figure 7.29). There are, however, two possible directions for this vector, but it is
conventional to choose the one associated with the application of the right-handed screw
rule. Imagine turning a right-handed screw in the sense from a towards b as shown. A
right-handed screw is one which, when turned clockwise, enters the material into which
it is being screwed. The direction in which the screw advances is the direction of the
required vector product. The symbol we shall use to denote the vector product is ×.
Formally, we write

a × b = |a!b| sin θ ê
 7.6 The vector product 247

a3b Plane containing

u a and b

a b

a
Figure 7.29
a × b is perpendicular to the plane containing Figure 7.30
a and b. The right-handed screw rule allows Right-handed screw rule allows the
the direction of a × b to be found. direction of b × a to be found.

where ê is the unit vector required to define the appropriate direction, that is ê is a unit
vector perpendicular to a and to b in a sense defined by the right-handed screw rule. To
evaluate b × a we must imagine turning the screw from the direction of b towards that
of a. The screw will advance as shown in Figure 7.30.
We notice immediately that a × b = & b × a since their directions are different. From
the definition of the vector product, it is possible to show that the following rules hold:

a × b = −(b × a) the vector product is not commutative

a × (b + c) = (a × b) + (a × c) the distributive rule
k(a × b) = (ka) × b = a × (kb) where k is a scalar

Example 7.18 If a and b are parallel show that a × b = 0.

Solution If a and b are parallel then the angle between a and b is zero. Therefore, a × b =
|a!b| sin 0 ê = 0. Note that the answer is still a vector, and that we denote the zero vector
0i + 0j + 0k by 0, to distinguish it from the scalar 0. In particular, we note that
i×i=j×j=k×k=0

If a and b are parallel, then a × b = 0.

In particular:

i×i=j×j=k×k=0

Example 7.19 Show that i × j = k and find expressions for j × k and k × i.

Solution We note that i and j are perpendicular so that |i × j| = |i!j| sin 90◦ = 1. Furthermore,
the unit vector perpendicular to i and to j in the sense defined by the right-handed screw
rule is k. Therefore, i × j = k as required. Similarly you should be able to show that
j × k = i and k × i = j.

i×j=k j×k=i k×i=j

j × i = −k k × j = −i i × k = −j
248 Chapter 7 Vectors

Example 7.20 Simplify (a × b) − (b × a).

Solution Use the result a × b = −(b × a) to obtain
(a × b) − (b × a) = (a × b) − (−(a × b))
= (a × b) + (a × b)
= 2(a × b)

Example 7.21 (a) If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, show that

a × b = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k
(b) If a = 2i + j + 3k and b = 3i + 2j + k find a × b.

Solution (a) a × b = (a1 i + a2 j + a3 k) × (b1 i + b2 j + b3 k)

= a1 i × (b1 i + b2 j + b3 k) + a2 j × (b1 i + b2 j + b3 k)
+ a3 k × (b1 i + b2 j + b3 k)
= a1 b1 (i × i) + a1 b2 (i × j) + a1 b3 (i × k) + a2 b1 (j × i) + a2 b2 (j × j)
+ a2 b3 (j × k) + a3 b1 (k × i) + a3 b2 (k × j) + a3 b3 (k × k)
Using the results of Examples 7.18 and 7.19, we find that the expression for a × b
simplifies to
a × b = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k
(b) Using the result of part (a) we have
a × b = ((1)(1) − (3)(2))i − ((2)(1) − (3)(3))j + ((2)(2) − (1)(3))k
= −5i + 7j + k
Verify for yourself that b × a = 5i − 7j − k.

7.6.1 Using determinants to evaluate vector products

Evaluation of a vector product using the previous formula is very cumbersome. A more
convenient and easily remembered method is now described. The vectors a and b are
written in the following pattern:
' '
' i j k ''
'
'a a a '
' 1 2 3'
'b b b '
1 2 3

This quantity is called a determinant. A more thorough treatment of determinants is

given in Section 8.7. To find the i component of the vector product, imagine crossing
out the row and column containing i and performing the following calculation on what
is left, that is
a2 b3 − a3 b2
The resulting number is the i component of the vector product. The j component is found
by crossing out the row and column containing j, performing a similar calculation, but
now changing the sign of the result. Thus the j component equals
−(a1 b3 − a3 b1 )
 7.6 The vector product 249

The k component is found by crossing out the row and column containing k and per-
forming the calculation
a1 b2 − a2 b1
We have
If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, then
' '
' i j k ''
'
a × b = '' a1 a2 a3 ''
'b b b '
1 2 3

= (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k

Example 7.22 Find the vector product of a = 2i + 3j + 7k and b = i + 2j + k.

Solution The two given vectors are represented in the following determinant:
' '
' i j k'
' '
'2 3 7'
' '
'1 2 1'
Evaluating this determinant we obtain
a × b = (3 − 14)i − (2 − 7)j + (4 − 3)k = −11i + 5j + k
You will find that, with practice, this method of evaluating a vector product is simple to
apply.

7.6.2 Applications of the vector product

The following three examples illustrate applications of the vector product.

Engineering application 7.9

Magnetic flux density B and magnetic field strength H

It is possible to model the effect of magnetism by means of a vector field. A magnetic
field with magnetic flux density B is a vector field which is defined in relation to the
force it exerts on a moving charged particle placed in the field. Consider Figure 7.31.
If a charge q moves with velocity v in a magnetic field B it experiences a force F
given by
F = qv × B
Note that this force is defined using a vector product. The unit of magnetic flux den-
sity is the weber per square metre (Wb m−2 ), or tesla (T). The direction of this force
is at right angles to both v and B, in a sense defined by the right-handed screw rule.
Its magnitude, or modulus, is
F = qvB sin θ
➔
250 Chapter 7 Vectors

B
q Figure 7.31
Force, F, exerted on a particle with charge, q, when moving with
v velocity, v, in a magnetic field, B.

where θ is the angle between v and B, v is the modulus of v and B is the modulus
of B.
F
Note that if θ = 90◦ , sin θ = 1, then B = .
qv
These formulae are useful because they can be used to calculate the forces exerted
on a conductor in an electric motor. They are also used to analyse electricity gener-
ators in which the motion of a conductor in a magnetic field leads to movement of
charges within the conductor, thus generating electricity.
A related quantity is the magnetic field strength, or the magnetic field intensity,
H, defined from
B = µH
µ is called the permeability of a material and has units of webers per ampere per
metre (Wb A−1 m−1 ). The units of H are amperes per metre (A m−1 ). Confirm for
yourself that the units match on both sides of the equation.

Engineering application 7.10

Magnetic field due to a moving charge

A charge q moving with velocity v gives rise to a magnetic field with magnetic flux
density B in its vicinity. As a result of this, another moving charge placed in this field
will experience a magnetic force. The magnetic flux density is given by
qµ0
B= (v × r̂)
4πr2
where r is a vector from the charge to the point at which B is measured, and µ0 is a
constant called the permeability of free space.
This equation can be used to find the magnetic field due to a current in a wire. Sup-
pose a small portion of wire has length δs and contains a current I. By writing δs as a
vector of length δs in the direction of the wire, it can be shown that the corresponding
contribution to the magnetic flux density is given by
µI
δB = 0 2 (δs × r̂)
4πr
This is the Biot--Savart law. Techniques of integration are required in order to com-
plete the calculation, but using this it is possible to show, for example, that the mag-
µI
netic flux density a distance r from a long straight wire has magnitude B = 0 .
2πr
 7.6 The vector product 251

Engineering application 7.11

The Hall effect in a semiconductor

A frequent requirement in the semiconductor industry is to be able to measure the
density of holes in a p-type semiconductor and the density of electrons in an n-type
semiconductor. This can be achieved by using the Hall effect. We will consider the
case of a p-type semiconductor but the derivation for an n-type semiconductor is
similar.

area, A

z
y
x L +
I V
hole
B –
EH
B p-type
semiconductor

+ –

Meter
Figure 7.32
VH Hall effect in a p-type semiconductor.

Consider the piece of semiconductor shown in Figure 7.32. A d.c. voltage, V , is

applied to the ends of the semiconductor. This gives rise to a flow of current com-
posed mainly of holes as they are the majority carriers for a p-type semiconductor.
This current can be represented by a vector pointing in the x direction and denoted
by I. A magnetic field, B, is applied to the semiconductor in the y direction. The
moving holes experience a force, FB , per unit volume, caused by the magnetic field
given by
1
FB = I×B
A
where A is the cross-sectional area of the semiconductor. This causes the holes to
drift in the z direction and so causes an excess of positive charge to appear on one
side of the semiconductor. This gives rise to a voltage known as the Hall voltage,
VH . As this excess charge builds up it creates an electric field, EH , in the negative z
direction, which in turn exerts an opposing force on the holes. This force is given by
FE = qp0 EH where q = elementary charge = 1.60 × 10−19 C, and p0 = density of
holes (holes per cubic metre). Equilibrium is reached when the two forces are equal
in magnitude, that is |FB | = |FE |. Now,
|I × B| IB
| FE | = qp0 |EH | |FB | = =
A A
➔
252 Chapter 7 Vectors

In equilibrium the magnitude of the electric field, |EH |, is constant and so we can
V
write |EH | = H , where L is the width of the semiconductor. Hence,
L
IB V
= qp0 H
A L
so that
BIL
p0 =
VH qA
So, by measuring the value of the Hall voltage, it is possible to calculate the density
of the holes, p0 , in the semiconductor.

EXERCISES 7.6

1 Evaluate
' ' ' ' 10 If a = 7i − j + k, b = 3i − j − 2k and
' i j k '' ' i j k' c = 9i + j − 3k, show that
' ' '
(a) '' 3 1 2 '' (b) '' −1 2 −3'
'
' 2 1 4 ' ' −4 0 1' a × (b + c) = (a × b) + (a × c)
' ' ' '
' i j k '' ' i j k ''
' ' 11 (a) The area, A, of a parallelogram with base b
(c) '' 0 1 0 '' (d) '' 3 5 2 '' and perpendicular height h is given by A = bh.
' 1 0 4 ' ' −3 −1 4 ' Show that if the two non-parallel sides of the
parallelogram are represented by the vectors a
2 If a = i − 2j + 3k and b = 2i − j − k, find
and b, then the area is also given by
(a) a × b A = |a × b|.
(b) b × a
(b) Find the area of the parallelogram with sides
3 If a = i − 2j and b = 5i + 5k find a × b. represented by 2i + 3j + k and 3i + j − k.
4 If a = i + j − k, b = i − j and c = 2i + k find 12 The volume, V , of a parallelepiped with sides a, b and
(a) (a × b) × c c is given by V = |a · (b × c)|. Find the volume of the
parallelepiped with sides 3i + 2j + k, 2i + j + k and
(b) a × (b × c)
i + 2j + 4k.
5 If p = 6i + 7j − 2k and q = 3i − j + 4k find |p|, |q|
and |p × q|. Deduce the sine of the angle between p 13 Suppose a force F acts through the point P with
and q. position vector r. The moment about the origin, M,
of the force is a measure of the turning effect of the
6 For arbitrary vectors p and q simplify force and is given by M = r × F. A force of 4 N acts
(a) (p + q) × p in the direction i + j + k, and through the point with
coordinates (7, 1, 3). Find the moment of the force
(b) (p + q) × (p − q)
about the origin.
7 If c = i + j and d = 2i + k, find a unit vector
14 In the theory of electromagnetic waves an important
perpendicular to both c and d. Further, find the sine of
quantity associated with the flow of electromagnetic
the angle between c and d.
energy is the Poynting vector S. This is defined as
8 A, B, C are points with coordinates (1, 2, 3), (3, 2, 1) S = E × H where E is the electric field strength and
and (−1, 1, 0), respectively. Find a unit vector H the magnetic field strength. Suppose that in a plane
perpendicular to the plane containing A, B and C. electromagnetic wave
( )
9 If a = 7i − 2j − 5k and b = 5i + j + 3k, find a vector 2πz
E = E0 cos − ωt j
perpendicular to a and b. λ