Question

Show that the straight lines whose direction cosines are given by 2l+2m-n=0 and mn+nl+lm=0 are at right angles.

Answer 1

We have, 2l+2m-n=0 ...(i) and
mn+nl+lm=0 ...(ii)

Eliminating m from the both equations, we get

$m=\frac{n-2l}{2}[fromEq.(i\left)\right]\Rightarrow \left(\frac{n-2l}{2}\right)n+nl+l\left(\frac{n-2l}{2}\right)=0\Rightarrow \frac{n^2-2nl+2nl+nl-2l^2}{2}=0\Rightarrow n^2+nl-2l^2=0\Rightarrow n^2+2nl-nl-2l^2=0\Rightarrow (n+2l)(n-l)=0\Rightarrow n=-2landn=l\therefore m=\frac{-2l-2l}{2},m=\frac{l-2l}{2}\Rightarrow m=-2l,m=\frac{-l}{2}$

Thus,the direction ratios of two lines are proportional to l,-2l,-2 and l,$\frac{-l}{2}$,l.

$\Rightarrow 1,-2,-2and1,\frac{-1}{2},1\Rightarrow 1,-2,-2and2,-1,2$

Also,the vectors parallel to these lines are $\overrightarrow{a}=\hat{i}-2\hat{j}-2\hat{k}and\overrightarrow{b}=2\hat{i}-\hat{j}+2\hat{k}$ respectively.

$\therefore cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{(\hat{i}-2\hat{j}-2\hat{k}).(2\hat{i}-\hat{j}+2\hat{k})}{3.3}=\frac{2+2-4}{9}=0\therefore \theta =\frac{\pi }{2}[\therefore cos\frac{\pi }{2}=0]$