Understanding Engineering Mathematics

obtained as

(cos 45°, 0 ,cos 45°)=

(

1

√

2

, 0 ,

1

√

2

)

This is of course what we would obtain from the coordinates (1, 0, 1) themselves, taking
the form
(x
r

,

y

r

,

z

r

)

=

(

1

√

2

,

0

√

2

,

1

√

2

)

=

(

1

√

2

, 0 ,

1

√

2

)

Similarly for:

( 0 , 1 , 1 )the direction cosines are

(

0 ,

1

√

2

,

1

√

2

)

( 1 , 1 , 0 )the direction cosines are

(

1

√

2

,

1

√

2

, 0

)

Finally, the far vertex with coordinates (1, 1, 1) has direction cosines
(x

r

,

y

r

,

z

r

)

=

(

1

√

3

,

1

√

3

,

1

√

3

)

The acute angle made with each axis is therefore

cos−^1 (^13 )= 70. 53 °to2dp

Exercise on 11.6

Calculate the direction cosines of the position vectors defined by the points in the Exercise
on 11.4.

Answer

In order, referring to the Exercise on 11.4, the direction cosines are:

not defined, (0, 0, 1), (0, 1, 0), (1, 0, 0), (−1, 0, 0),

(

1

√

2

,0,−

1

√

2

)
,
(
2
√
14

,−

3

√

14

,

1

√

14

)

11.7 Angle between two lines through the origin

Using direction cosines we can obtain a simple and useful expression for the angle between
two lines intersecting at the origin. In general, the angle between any two lines in space,
not necessarily intersecting, is the angle between any pair of lines parallel to them. In
particular we can always take a pair of lines through the origin. This expression for the
angle between two lines will be useful in discussing the angle between two vectors. This
comes into the different types of products that we can make with vectors (the scalar and
the vector product – see Sections 11.10 and 11.11). Consider Figure 11.9.
Suppose two linesOA,OA′have direction cosinesl,m,nandl′,m′,n′respectively
with respect to axesOxyz. Then we will show that

cosθ=ll′+mm′+nn′