The angle between lines whose direction cosines are given by $\ell+m+n=0$, $\ell^2+m^2-n^2=0$, is :
Answer & explanation
Correct answer: option 4
$\ell+m+n=0, \ell^2+m^2-n^2=0$
We also have
$\ell^2+m^2+n^2=1 $
$\Rightarrow 2 n^2=1 $
$\Rightarrow n=\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$
Also, $\ell^2+m^2=n^2=(-(\ell+m))^2$
$\Rightarrow \ell m=0$ and $\ell+m= \pm \frac{1}{\sqrt{2}}$
Hence, direction cosines are lines are
$\left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right), $
$\left(0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Angle between these lines in both cases is zero.