The Chemistry Maths Book, Second Edition

458 Chapter 16Vectors

For nonzero vectors aand b, the scalar product is zero when the vectors are

perpendicular,

a 1

·

1 b 1 = 10 (16.33)

The vectors are then said to be orthogonal, with aorthogonal to b, and borthogonal

to a.*We note that equation (16.33) shows that it is not possible to cancel vectors in a

vector equation in the same way as is possible for scalars. Thus the equation

a 1

·

1 b 1 = 1 a 1

·

1 c

has the three possible solutions: (i)a 1 = 10 , (ii)b 1 = 1 c, (iii) ais orthogonal to (b 1 − 1 c).

EXAMPLE 16.12Find the value of λfor which a 1 = 1 (2, λ, 1)and b 1 = 1 (4, −2, −2)

are orthogonal.

For orthogonality,a 1

·

1 b 1 = 101 = 121 × 141 + 1 λ 1 × 1 (−2) 1 + 111 × 1 (−2) 1 = 161 − 12 λ. Thereforeλ 1 = 13.

0 Exercises 22, 23

When aand bare the same vector, (16.31) gives

a 1

·

1 a 1 = 1 a

x

2

1 + 1 a

y

2

1 + 1 a

z

2

1 = 1 |a|

2

The length of a vector is therefore given in terms of the scalar product by

(16.34)

The use of cartesian base vectors

The base vectors i, j, and kare orthogonal and of unit length so that, by equations

(16.33) and (16.34),

i 1

·

1 j 1 = 10 j 1

·

1 k 1 = 10 k 1

·

1 i 1 = 1 0 (orthogonality)

(16.35)

i 1

·

1 i 1 = 11 j 1

·

1 j 1 = 11 k 1

·

1 k 1 = 1 1 (unit length)

The expression (16.31) for the scalar product follows from these properties of the base

vectors. Thus, expressing aand bin terms of the base vectors,

a 1

·

1 b 1 = 1 (a

x

i 1 + 1 a

y

j 1 + 1 a

z

k) 1

·

1 (b

x

i 1 + 1 b

y

j 1 + 1 b

z

k)

= 1 a

x

b

x

i 1

·

1 i 1 + 1 a

x

b

y

i 1

·

1 j 1 + 1 a

x

b

z

i 1

·

1 k 1 + 1 a

y

b

x

j 1

·

1 i 1 + 1 a

y

b

y

j 1

·

1 j 1 +1-1+ 1 a

z

b

z

k 1

·

1 k

= 1 a

x

b

x

1 + 1 a

y

b

y

1 + 1 a

z

b

z

0 Exercises 24–27

||aaa= 11

·

*Orthogonal means perpendicular for vectors in ordinary space, but the definition of orthogonality applies

generally to vectors in ‘vector spaces’ of arbitrary dimensions.