The Chemistry Maths Book, Second Edition

16.5 The scalar (dot) product 459

The following examples demonstrate two applications of scalar products in the

physical sciences.

EXAMPLE 16.13Force and work

Let a body be displaced from positionr

1

1 = 1 (x

1

, y

1

, z

1

)to positionr

2

1 = 1 (x

2

, y

2

, z

2

)

under the influence of a constantforce F. The displacement of the body is

d 1 = 1 r

2

1 − 1 r

1

1 = 1 (x

2

1 − 1 x

1

, y

2

1 − 1 y

1

, z

2

1 − 1 z

1

)

and the work done by the force is

W 1 = 1 F 1

·

1 d 1 = 1 Fd 1 cos 1 θ (16.36)

The quantityF 1 cos 1 θis the component of the force

in the direction of d, and this is the only component

of the force that contributes to the work. Thus, as in

Figure 16.21, the force can be written as

F 1 = 1 F

d

1 + 1 F

⊥

whereF

d

acts along the line of dandF

⊥

acts at right angles to d. The latter cannot

cause a displacement along d.

In terms of cartesian components, the work (16.36) is

W 1 = 1 F

x

d

x

1 + 1 F

y

d

y

1 + 1 F

z

d

z

1 = 1 W

x

1 + 1 W

y

1 + 1 W

z

(16.37)

where, for example, F

x

is the component of Fin the x-direction andW

x

1 = 1 F

x

d

x

is the

work done in moving the body through distanced

x

1 = 1 x

2

1 − 1 x

1

in this direction (‘work

along the x-direction’). We note that the work done bythe body againstthe force is

−W 1 = 1 −F 1
·

1 d.

For example, if the force isF 1 = 1 (2, 1, 0)and the displacement isd 1 = 1 (2, −3, 1), in

appropriate units, the work done is

W 1 = 1 F 1

·

1 d 1 = 121 × 121 + 111 × 1 ( 1 − 1 3) 1 + 101 × 111 = 11

In addition, the component of Fin the direction of dcan be written in terms of the

scalar product as

where is the unit vector in direction d. In the present case, so that

0 Exercise 28

F

d

=114.

d= 14

ˆ

dd= d

FF

d

d

===cos

ˆ

θ

Fd

11

·

θ

F

d

F

d

F

⊥

..

...

..

...

..

...

...

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

...........

..........

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

..........

...........

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

...........

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

...........

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

...........

..........

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

..........

...........

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

...........

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

...........

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

...........

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

...........

..........

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

...........

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

......

.......

.

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

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

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

...........

.....

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

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

...........

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

...........

.....

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

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

..

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

......

....

.....

........

.....

......

.....

......

..

....

.....

......

.....

......

..

Figure 16.21