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

Answer 1

The correct option is C $\frac{5}{8}$

Given that $l+m=n....\left(1\right)$
$l^2+m^2=n^2....\left(2\right)$
Squaring equation $\left(1\right)$
$l^2+m^2+2lm=n^2....\left(3\right)$
From equations $\left(2\right)$ and $\left(3\right)$
$lm=0\Rightarrow l=0orm=0$
Case $\left(1\right)l=0$
$\Rightarrow m=n$
$\Rightarrow l^2+m^2+n^2=0$
$\Rightarrow m=n=\pm \frac{1}{\sqrt{2}}$
$\therefore (l,m,n)\equiv (0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\text{or}(0,\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$
Case $\left(2\right)m=0$
$\Rightarrow l=n$
$\Rightarrow l^2+m^2+n^2=0$
$\Rightarrow l=n=\pm \frac{1}{\sqrt{2}}$
$\therefore (l,m,n)\equiv (\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})\text{or}(\frac{-1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}})$


$\Rightarrow \overrightarrow{a}\equiv (0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),\overrightarrow{b}=(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})\cos \alpha =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{1}{2}\sin \alpha =\pm \frac{\sqrt{3}}{2}\Rightarrow {\sin }^4\alpha +{\cos }^4\alpha =\frac{1}{16}+\frac{9}{16}=\frac{5}{8}$