Question

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

2l − m + 2n = 0 and mn + nl + lm = 0

Answer 1

 

`\text{ Given } : `

\[2l - m + 2n = 0 . . . (1)\]

\[mn + nl + lm = 0 . . . (2)\]

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

\[m = 2l + 2n\]

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

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

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

\[ \Rightarrow 2 l^2 + 5ln + 2 n^2 = 0 \]

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

\[ \Rightarrow l = - 2n , - \frac{n}{2}\]

\[\text { If } l = - 2n, \text { then by substituting } l = - 2n \text { in } \left( 1 \right), \text { we get } m = - 2n . \]

\[\text { If } l = - \frac{n}{2}, \text { then by substituting } l = - \frac{n}{2} in \left( 1 \right), \text { we get } m = n . \]

\[\text{ Thus, the direction ratios of the two lines are proportional to } - 2n, - 2n, n \text { and }  - \frac{n}{2}, n, n or - 2, - 2, 1 \text{ and }- \frac{1}{2}, 1, 1 . \]

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

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

\[ \vec{b} = - \frac{1}{2} \hat{i}     + \hat{j} + \hat{k} \]

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

\[\text{ Now }, \]

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

\[ = \frac{1 - 2 + 1}{\sqrt{4 + 4 + 1} \sqrt{ 1/4 + 1 + 1}} \]

\[ = 0 \]

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