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

$l+m+n=0$

$m+n=-l$

$\therefore {\left(m+n\right)}^2=l{}^2$

$\therefore l{}^2= m^2+n^2+2mn$

But,$m^2+n^2=l{}^2$

$\therefore 2mn=0$

$\therefore$$m = 0$or$n = 0$

$\therefore$$l=-n$or$l=-m$

$\because l{}^2+m^2+n^2=1$

$\therefore$The two direction cosines are$\left(\frac{1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}}\right)$and$\left(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}\text{,}0\right)$

$\theta$is the angle between them.

$\therefore \text{cos }\theta =l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}+0+0=\frac{1}{2}$

$\therefore \theta =\frac{\pi }{3}$.

Note :Taking$l=\frac{-1}{\sqrt{2}}$, will also given the same solution, but the supplementary angle. Both are correct, we choose the one present in the options.