Question

Find the angle between the lines whose direction cosines are connected by the relations $l+m+n=0$ and $2lm+2nl-mn=0$.

Answer 1

Given,

$l+m+n=0$

$\Rightarrow m=-(l+n)$

$2lm+2nl-mn=0$

$\Rightarrow 2nl+m(2l-m)=0$

$\Rightarrow 2nl-(l+n)(2l-m)=0$

$2l^2-nl-n^2=0$

$(l-n)(2l+n)=0$

$l=n,l=-\frac{n}{2}$

using the value of $l$ we get,

$m=-\frac{n}{2}$

from values of $l$ and $m$ we have,

$l=-\frac{m}{2}=n$ and

$l=m=-\frac{n}{2}$

$l:m:n=1:-2:1$ or $l:m:n=1:1:-2$

$\therefore \cos \theta =\frac{1\times 1+(-2)\times 1+1\times (-2)}{\sqrt{1^2+2^2+1^2}\sqrt{1^2+2^2+1^2}}$

$=\frac{1-2-2}{6}=-\frac{1}{2}$

$\therefore \theta ={\cos }^{-1}(-\frac{1}{2})=-{60}^{\circ }$