The direction cosines of two lines are connected by the relations $l+m-n=0$ and $l m-2 m n+n l=0$. If…
The direction cosines of two lines are connected by the relations $l+m-n=0$ and $l m-2 m n+n l=0$. If $\theta$ is the acute angle between those lines then $\cos \theta=$
$\frac{\pi}{6}$
$\frac{1}{\sqrt{7}}$
$\sqrt{\frac{5}{6}}$
$\frac{\pi}{3}$
Solution
$\ell+m-n=0 \Rightarrow n=\ell+m \Rightarrow \ell m-2 m n+n \ell=0$
$\Rightarrow \ell m-2 m(\ell+m)+\ell(n+m)=0$
$\Rightarrow \ell^2-2 m^2=0 \Rightarrow \ell=-\sqrt{2} m, \ell=\sqrt{2} m$
$\Rightarrow n=(1-\sqrt{2}) m, n=(1+\sqrt{2}) m$
$\ell: m: n=-\sqrt{2}: 1:(1-\sqrt{2})$ or $\sqrt{2}: 1: 1+\sqrt{2}$
Angle between these lines
$\cos \theta=\frac{|(\sqrt{2})(-\sqrt{2})+(1)(1)+(1-\sqrt{2})(1+\sqrt{2})|}{\sqrt{2+1+1+2-2 \sqrt{2}} \sqrt{2+1+1+2+2 \sqrt{2}}}$
$=\frac{|-2|}{\sqrt{6-2 \sqrt{2}} \sqrt{6+2 \sqrt{2}}}=\frac{1}{\sqrt{7}}$