JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
The plane passing through the line $L:lx-y+3(1-l)z=1$, $x+2y-z=2$ and perpendicular to the plane $3x+2y+z=6$ is $3x-8y+7z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis, then $415\cos^2\theta$ is equal to ________.
3D Geometry
Answer$125$
Solution
Plane through $L$: $(lx-y+3(1-l)z-1)+k(x+2y-z-2)=0$, normal $(l+k,\,2k-1,\,3-3l-k)$. It must be parallel to $(3,-8,7)$, giving $l=\tfrac23,\ k=-\tfrac16$ (this normal $(\tfrac12,-\tfrac43,\tfrac76)\parallel(3,-8,7)$ and is perpendicular to $(3,2,1)$). Then $L$: $\tfrac23x-y+z=1$, $x+2y-z=2$. Direction $=(\tfrac23,-1,1)\times(1,2,-1)=(-1,\tfrac53,\tfrac73)\parallel(-3,5,7)$. $\cos\theta=\dfrac{5}{\sqrt{9+25+49}}$, so $\cos^2\theta=\tfrac{25}{83}$. $415\cos^2\theta=415\cdot\tfrac{25}{83}=125$.