Question

Find the direction cosines of the lines, connected by the relations: l + m +n = 0 and 2lm + 2ln − mn = 0.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ l+m+n=0...\left(1\right) \\ 2lm\mathit{}+2\mathit{ln}\mathit{}-nm\mathit{}=0\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\left(2\right) \\ \mathrm{From}\left(1\right),\mathrm{we}\mathrm{get} \\ l=\mathit{-}m\mathit{-}n \\ \mathrm{Substituting}l=-m-n\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 2\left(-m-n\right)m+2\left(-m-n\right)n-mn=0 \\ \Rightarrow -2m^2-2mn-2mn-2n^2-mn=0 \\ \Rightarrow 2m^2+2n^2+5mn=0 \\ \Rightarrow \left(m+2n\right)\left(2m+n\right)=0 \\ \Rightarrow \mathrm{m}=-2n,-\frac{n}{2} \\ \mathrm{If}m=-2n,\mathrm{then}\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}\mathit{}l=n. \\ \mathrm{If}m=-\frac{n}{2},\mathrm{then}\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}\mathit{}l=-\frac{n}{2}. \\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{are}\mathrm{proportional}\mathrm{to}n,-2n,n\mathrm{and}-\frac{n}{2},-\frac{n}{2},n,\mathrm{i}.\mathrm{e}.1,-2,1\mathrm{and}-\frac{1}{2},-\frac{1}{2},1. \\ \mathrm{Hence},\mathrm{their}\mathrm{direction}\mathrm{cosines}\mathrm{are} \\ \pm \frac{1}{\sqrt{6}},\pm \frac{-2}{\sqrt{6}},\pm \frac{1}{\sqrt{6}} \\ \pm \frac{-1}{\sqrt{6}},\pm \frac{-1}{\sqrt{6}},\pm \frac{2}{\sqrt{6}} \end{gathered}$$