Question

Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l+m-n=0$ and $l^2+m^2-n^2=0.$ Then the value of ${\sin }^4\alpha +{\cos }^4\alpha$ is

• A. $\frac{3}{4}$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$n=\ell +m$
Now, ${\ell }^2+m^2=n^2=(\ell +m{)}^2$
$\Rightarrow 2\ell m=0$
If $\ell =0\Rightarrow m=n=\pm \frac{1}{\sqrt{2}}$
And, If $m=0\Rightarrow n=\ell =\pm \frac{1}{\sqrt{2}}$
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, $\cos \alpha =\frac{1}{2}\Rightarrow \alpha =\frac{\pi }{3}$