3D Geometry: Let the plane 2x+3y+z+20=0 be rotated through a right angle about… | JEE Main 2022

JEE Main 2022 (Online) 26th June Morning ShiftQuestion 1

Let the plane $2x+3y+z+20=0$ be rotated through a right angle about its line of intersection with the plane $x-3y+5z=8$. If the mirror image of the point $\left(2,-\frac{1}{2},2\right)$ in the rotated plane is $B(a,b,c)$, then :

3D Geometry

$\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}$✓ Correct answer$\frac{a}{4}=\frac{b}{5}=\frac{c}{-2}$$\frac{a}{8}=\frac{b}{-5}=\frac{c}{4}$$\frac{a}{4}=\frac{b}{5}=\frac{c}{2}$

Solution

Rotated plane: $(2x+3y+z+20)+k(x-3y+5z-8)=0$, normal $(2+k,3-3k,1+5k)$, perpendicular to $(2,3,1)$: $2(2+k)+3(3-3k)+1+5k=0\Rightarrow 14-2k=0$, $k=7$? $4+2k+9-9k+1+5k=14-2k=0\Rightarrow k=7$. Plane: $9x-18y+36z-36=0\Rightarrow x-2y+4z=4$. $(9,-18,36)$, constant $20-56=-36$, so $x-2y+4z-4=0$. Point $(2,-\tfrac12,2)$: $2+1+8-4=7$, $|n|^2=21$; image $=(2,-\tfrac12,2)-\frac{2\cdot7}{21}(1,-2,4)=(\tfrac43,\tfrac56,-\tfrac23)$ $\propto(8,5,-4)$. Option (A).

3D Geometry - JEE Mathematics question (JEE Main 2022 (Online) 26th June Morning Shift)

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