JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
For $a,b\in\mathbb{Z}$ and $|a-b|\le10$, let the angle between the plane $P:ax+y-z=b$ and the line $l:x-1=a-y=z+1$ be $\cos^{-1}\left(\frac{1}{3}\right)$. If the distance of the point $(6,-6,4)$ from the plane $P$ is $3\sqrt{6}$, then $a^{4}+b^{2}$ is equal to
3D Geometry
Solution
Normal $(a,1,-1)$, direction $(1,-1,1)$: $\sin\theta=\frac{2\sqrt2}{3}=\frac{|a-2|}{\sqrt{3(a^2+2)}}$ gives $5a^2+12a+4=0$, $a=-2$ (integer). Distance $\frac{|-12-10-b|}{\sqrt6}=3\sqrt6$ gives $b=-4$ (since $|a-b|\le10$). So $a^4+b^2=16+16=32$.