Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l+\mathrm{m}-\mathrm{n}=0$ and $l^{2}+\mathrm{m}^{2}-\mathrm{n}^{2}=0$. Then the
value of $\sin ^{4} \alpha+\cos ^{4} \alpha$ is :
Correct Option: , 3
$\mathrm{n}=\ell+\mathrm{m}$
Now, $\ell^{2}+\mathrm{m}^{2}=\mathrm{n}^{2}=(\ell+\mathrm{m})^{2}$
$\Rightarrow 2 \ell \mathrm{m}=0$
If $\ell=0 \Rightarrow \mathrm{m}=\mathrm{n}=\pm \frac{1}{\sqrt{2}}$
And, If $\mathrm{m}=0 \Rightarrow \mathrm{n}=\ell=\pm \frac{1}{\sqrt{2}}$
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, $\cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$
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