Understanding Engineering Mathematics

A′

B′(l′,m′,n′)

B(l,m,n)

q A

1

O

1

Figure 11.9Angle between two lines.

This follows by applying the cosine rule to the triangleOBB′in Figure 11.10 whereOB,
OB′are of unit length. ThenB,B′have the coordinates(l,m,n)(l′,m′,n′)respectively.
We have, by the cosine rule (179
➤
)

BB′^2 =OB^2 +OB′^2 − 2 OBOB′cosθ

or
(l′−l)^2 +(m′−m)^2 +(n′−n)^2 = 12 + 12 −2cosθ= 2 −2cosθ

Expanding the left-hand side and usingl^2 +m^2 +n^2 =1 for the direction cosines yields

l′^2 +m′^2 +n′^2 +l^2 +m^2 +n^2 − 2 ll′− 2 mm′− 2 nn′

= 1 + 1 − 2 (ll′+mm′+nn′)= 2 −2cosθ

This then gives the required result cosθ=ll′+mm′+nn′.
From this result it follows that if two lines are perpendicular then

ll′+mm′+nn′= 0

because cos(± 90 °)= 0

Exercise on 11.7

Using direction cosines find the acute angles between each pair of lines through the origin
defined by the points in Exercises 11.5.

Answer

Between (i) and (ii), 17.02°; between (i) and (iii), 18.43°; between (ii) and (iii), 35.02°–all
to two decimal places.

11.8 Basis vectors

We can use the coordinate system defined in Section 11.4 to represent vectors in terms of
numerical components by choosing ‘basis vectors’ along the coordinate axes. The standard
notation for this is to leti,j,kdenote the unit vectors along thex-,y-,z-axes as shown
in Figure 11.10.
In terms of these ‘basis vectors’ we can write any vector,a,inthecomponent form

a=a 1 i+a 2 j+a 3 k