The Chemistry Maths Book, Second Edition

16.5 The scalar (dot) product 457

Proof

To show the equivalence of the definitions (16.30) and (16.31), we apply the cosine

rule to the triangle in Figure 16.19:

(AB)

2

1 = 1 (OA)

2

1 + 1 (OB)

2

1 − 1 2(OA)(OB) 1 cos 1 θ 1 = 1 a

2

1 + 1 b

2

1 − 12 ab 1 cos 1 θ

The vector isb 1 − 1 a, and its length is given by

(AB)

2

1 = 1 (b

x

1 − 1 a

x

)

2

1 + 1 (b

y

1 − 1 a

y

)

2

1 + 1 (b

z

1 − 1 a

z

)

2

= 1 (a

x

2

1 + 1 a

y

2

1 + 1 a

z

2

) 1 + 1 (b

x

2

1 + 1 b

y

2

1 + 1 b

z

2

) 1 − 1 2(a

x

b

x

1 + 1 a

y

b

y

1 + 1 a

z

b

z

)

= 1 a

2

1 + 1 b

2

1 − 1 2(a

x

b

x

1 + 1 a

y

b

y

1 + 1 a

z

b

z

)

Thereforeab 1 cos 1 θ 1 = 1 a

x

b

x

1 + 1 a

y

b

y

1 + 1 a

z

b

z

.

EXAMPLE 16.11Givena 1 = 1 (3, 1, −1)andb 1 = 1 (1, 2, −3)finda 1
·

1 b,b 1

·

1 aand the

angle between the vectors.

By equation (16.31),

a 1

·

1 b 1 = 131 × 111 + 111 × 121 + 1 (−1) 1 × 1 (−3) 1 = 1 8, b 1

·

1 a 1 = 111 × 131 + 121 × 111 + 1 (−3) 1 × 1 (−1) 1 = 18

This example demonstrates that scalar multiplication is commutative,

a 1

·

1 b 1 = 1 b 1

·

1 a (16.32)

By equation (16.30), and the lengths of the vectors area 1 = 1 |a| 1 =
1

andb 1 = 1 |b| 1 =
1

. Therefore

0 Exercises 19–21

The sign of the scalar product is the sign of the cosine, so that the scalar product can

be positive, zero, or negative:

cosθθ=,=cos











≈. ≈

−

8

154

8

154

0 8702

1

49.86°°

14

cosθ= , 11

ab 11

·

ab

AB

.

...

..

...

...

..

...

...

b

a

θ

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π

2

,a

.

b> 0

.

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θ=

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,a

.

b=0

.

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b

a

θ

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θ>

π

2

,a

.

b< 0

Figure 16.20