#### Question

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

(i) *l *+ *m* + *n* = 0 and *l*^{2} + *m*^{2} − *n*^{2} = 0

#### Solution

\[\left( i \right) \text{ Given } : \]

\[l + m + n = 0 . . . (1) \]

\[ l^2 + m^2 - n^2 = 0 . . . (2)\]

\[\text{ From } \left( 1 \right), \text{ we get } \]

\[m = - l - n\]

\[\text { Substituting } m = - l - n \text{ in} \left( 2 \right), \text { we get } \]

\[ l^2 + \left( - l - n \right)^2 - n^2 \]

\[ \Rightarrow l^2 + l^2 + n^2 + 2\ln - n^2 = 0\]

\[ \Rightarrow 2 l^2 + 2\ln = 0\]

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

\[ \Rightarrow l = 0 , l = - n\]

\[\text{ If } l = 0, \text{ then by substituting } l = 0 \text { in } \left( 1 \right), \text { we get } m = - n . \]

\[\text{ If } l = - n, \text { then by substituting } l = - n \text { in }\left( 1 \right), \text { we get } m = 0 . \]

\[\text{ Thus, the direction ratios of the two lines are proportional to } 0, - n, \text { and } - n, 0, n \text{ or } 0, - 1, 1 \text { and } - 1, 0, 1 . \]

\[\text{ Vectors parallel to these lines are } \]

\[ \vec{a} = 0 \hat{i} - \hat{j} + \hat{k} \]

\[ \vec{b} = - \hat{i} + 0 \hat{j} + \hat{k} \]

\[\text{ If } \theta \text{ is the angle between the lines, then } \theta \text{ is also the angle between } \vec{a} \text{ and } \vec{b} . \]

\[\text { Now }, \]

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

\[ = \frac{1}{\sqrt{0 + 1 + 1} \sqrt{1 + 0 + 1}} \]

\[ = \frac{1}{2} \]

\[ \Rightarrow \theta = \frac{\pi}{3}\]