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

$n=\ell +m$

Now, ${\ell }^2+m^2=n^2={\left(\ell +m\right)}^2$

$\Rightarrow 2\ell m=0$

$l^2+m^2+n^2=1$

If $\ell =0\Rightarrow 2{\mathrm{n}}^2=1\Rightarrow n=\pm \frac{1}{\sqrt{2}}$

$m=n=\pm \frac{1}{\sqrt{2}}$

And, If $m=0\Rightarrow n=\ell =\pm \frac{1}{\sqrt{2}}$

$\therefore l^2+m^2=\frac{1}{2}\& l+m=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{2}+2 \mathit{lm}=\frac{1}{2}$

$\therefore l=0, m=\frac{1}{\sqrt{2}}$ or $l=\frac{1}{\sqrt{2}}, m=0$

So, direction cosines of two lines are

$\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)$

Thus,

$\therefore \cos \alpha =0+0+\frac{1}{2}=\frac{1}{2}$

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

$\therefore {\sin }^4\alpha +{\cos }^4\alpha =1-\frac{1}{2}{\sin }^2\left(2\alpha \right)=1-\frac{1}{2}·\frac{3}{4}=\frac{5}{8}$