JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
Let $ax+by+cz+3=0$ be the plane which bisects the acute angle between the planes $2x-y-2z+3=0$ and $3x-2y+6z+8=0$. Then $a+b+c$ is equal to
3D Geometry
Solution
Both constants positive ($3,8$). Bisectors: $\frac{2x-y-2z+3}{3}=\pm\frac{3x-2y+6z+8}{7}$. Taking $+$: $7(2x-y-2z+3)-3(3x-2y+6z+8)=5x-y-32z-3=0$. Taking $-$: $23x-13y+4z+45=0$. Acute bisector makes the smaller angle with the plane normals' direction: $\cos$ of angle between $(5,-1,-32)$ and $(2,-1,-2)$ is larger (about $39.5^\circ$ vs $50.5^\circ$), so $5x-y-32z-3=0$ is the acute-angle bisector. Rewrite as $-5x+y+32z+3=0$: $a+b+c=-5+1+32=28$. Option (B).