#### Question

Find the angle between the lines whose direction cosines are given by the equations

2*l* + 2*m* − *n* = 0, *mn* + *ln* + *lm* = 0

#### Solution

(iv) The given relations are

2*l* + 2*m* − *n* = 0 .....(1)*mn* + *ln* + *lm* = 0 .....(2)

From (1), we have*n* = 2*l* + 2*m*

Putting this value of *n* in (2), we get

\[m\left( 2l + 2m \right) + l\left( 2l + 2m \right) + lm = 0\]

\[ \Rightarrow 2lm + 2 m^2 + 2 l^2 + 2lm + lm = 0\]

\[ \Rightarrow 2 m^2 + 5lm + 2 l^2 = 0\]

\[ \Rightarrow \left( 2m + l \right)\left( m + 2l \right) = 0\]

\[ \Rightarrow 2m + l = 0 or m + 2l = 0\]

\[ \Rightarrow l = - 2m or l = - \frac{m}{2}\]

\[l = - 2m\] we have

\[n = 2 \times \left( - 2m \right) + 2m = - 4m + 2m = - 2m\]

When

\[l = - \frac{m}{2}\] we have

\[n = 2 \times \left( - \frac{m}{2} \right) + 2m = - m + 2m = m\]

Thus, the direction ratios of two lines are proportional to

\[- 2m, m, - 2m\]

\[- \frac{m}{2}, m, m\]

Or

\[- 2, 1, - 2\] and

-1,2,2

So, vectors parallel to these lines are

\[\vec{a} = - 2 \hat{i} + \hat{j} - 2 \hat{k}\] and

\[\vec{b} = - \hat{i} + 2\hat{j} - 2 \hat{k}\]

Let

\[\therefore \cos\theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right|\left| \vec{b} \right|}\]

\[ = \frac{\left( - 2 \hat{i} + \hat{j} - 2 \hat{k} \right) . \left( - \hat{i} + 2 \hat{j} + 2 \hat{k} \right)}{\sqrt{4 + 1 + 4}\sqrt{1 + 4 + 4}}\]

\[ = \frac{- 2 \times \left( - 1 \right) + 1 \times 2 + \left( - 2 \right) \times 2}{3 \times 3}\]

\[ = \frac{2 + 2 - 4}{9}\]

\[ = 0\]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

Thus, the angle between the two lines whose direction cosines are given by the given relations is

\[\frac{\pi}{2}\]