Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

We have, 

$l+m+n=0$   $...\left(\mathrm{i}\right)$

and $l^2=m^2+n^2$   $...\left(\mathrm{ii}\right)$

Putting the value of $l$ in Eq. $\left(\mathrm{ii}\right)$, we get

$(m+n{)}^2=m^2+n^2$

$\Rightarrow m^2+n^2+2mn=m^2+n^2$

$\Rightarrow mn=0$

$\Rightarrow m=0,n=0$

when $m=0,l=-n$

$\therefore$ Direction cosines are $\left(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$.

When $n=0,l=-m$

$\therefore$ Direction cosines are $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$.

Angle between lines are 

$\cos \theta =\left(\frac{-1}{\sqrt{2}}\times -\frac{1}{\sqrt{2}}\right)+\left(0\times \frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}\times 0\right)$

$\cos \theta =\frac{1}{2},\theta =\pi /3$