Question

Direction cosines of two lines satisfy the equation $\Rightarrow 4\mathcal{l} + m - n = 0$

$$\begin{array}{r} 2mn + 10n\mathcal{l} + 3\mathcal{l}\text{ }m = 0\#(2) \end{array}$$

& we know

$$\begin{array}{r} \Rightarrow \mathcal{l}^{2} + m^{2} - n^{2} = 1\#(3) \end{array}$$

$\Rightarrow n = 4\mathcal{l} + m$ putting in eqn.

$${\Rightarrow n(2\text{ }m + 10\mathcal{l}) + 3\mathcal{l}\text{ }m = 0 }{\Rightarrow (4\mathcal{l} + m)(2\text{ }m + 10\mathcal{l}) + 3\mathcal{l}\text{ }m = 0 }{\Rightarrow 8\mathcal{l}\text{ }m + 40\mathcal{l}^{2} + 2{\text{ }m}^{2} + 10\mathcal{l}\text{ }m + 3\mathcal{l}\text{ }m = 0 }{\Rightarrow 40\mathcal{l}^{2} + 21\mathcal{l}\text{ }m + 2{\text{ }m}^{2} = 0 }{\Rightarrow (8\mathcal{l} + m)(5\mathcal{l} + 2\text{ }m) = 0}$$

Case 1: $8\mathcal{l} + m = 0 \Rightarrow \text{ }m = - 8\mathcal{l}$

Case 2 : $5\mathcal{l} + 2\text{ }m = 0 \Rightarrow \text{ }m = \frac{- 5}{2}\mathcal{l}$

So direction ratio of $L_{1}$ is $\mathcal{l}, - 8\mathcal{l}, - 4\mathcal{l}$

& direction ratio of $L_{2}$ is $\mathcal{l},\frac{- 5\mathcal{l}}{2},\frac{3\mathcal{l}}{2}$

$${cos\theta = \left| \frac{\mathcal{l}^{2} + 20\mathcal{l}^{2} - 6\mathcal{l}^{2}}{\sqrt{\mathcal{l}^{2} + 64\mathcal{l}^{2} + 16\mathcal{l}^{2}}\sqrt{\mathcal{l}^{2} + \frac{25\mathcal{l}^{2}}{4} + \frac{9\mathcal{l}^{2}}{4}}} \right| }{= \frac{15\mathcal{l}^{2}}{(9\mathcal{l})\frac{\sqrt{38}\mathcal{l}}{2}} = \frac{10}{3\sqrt{38}}}$$