JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
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 :
3D Geometry
Solution
From $n=l+m$: $l^2+m^2=(l+m)^2$ so $lm=0$. Direction cosines $(0,\frac1{\sqrt2},\frac1{\sqrt2})$ and $(\frac1{\sqrt2},0,\frac1{\sqrt2})$: $\cos\alpha=\frac12$, $\sin^2\alpha=\frac34$. $\sin^4\alpha+\cos^4\alpha=\frac9{16}+\frac1{16}=\frac58$.