Question

Find the angle between the two lines whose direction cosines satisfy $l+m+n=0,l^2+m^2-n^2=0$

Answer 1

Put $n=-l-m$ in $l^2+m^2-n^2=0$

$l^2+m^2-{(-l-m)}^2=0$

$\Rightarrow l^2+m^2-{(l+m)}^2=0$

$\Rightarrow l^2+m^2-l^2-m^2-2lm=0$

$\Rightarrow -2lm=0$

$\Rightarrow lm=0$

$\Rightarrow l=0$

Direction ratio's are $(0,m,-m)$,
$m=0,$

direction ratio's are $(l,0,-l)$

The angle between them is ${\cos }^{-1}\frac{l\times 0+0\times m+l\times m}{\sqrt{2m^2.2l^2}}$

$={\cos }^{-1}\frac{0+0+lm}{\sqrt{4m^2{l}^2}}$

$={\cos }^{-1}\frac{lm}{2ml}$

$={\cos }^{-1}\frac{1}{2}$

$=\frac{\pi }{3}$