Let the plane 2x+3y+z+20=0 be rotated through a right angle about its line of intersection…

Question

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 $\left(2,-\dfrac12,2\right)$ in the rotated plane is $B(a,b,c)$, then:

Accepted Answer

Family: $(2+\lambda)x+(3-3\lambda)y+(1+5\lambda)z+(20-8\lambda)=0$. Perpendicular to the original normal $(2,3,1)$ (rotation by $\pi/2$):
$$2(2+\lambda)+3(3-3\lambda)+(1+5\lambda)=0 \implies 14-2\lambda=0 \implies \lambda=7.$$
$$\text{New plane: }9x-18y+36z-36=0 \implies x-2y+4z-4=0.$$
Foot of perpendicular from $\left(2,-\tfrac12,2\right)$: $t=-\tfrac13$, giving $M=\left(\tfrac53,\tfrac16,\tfrac23\right)$.
$$B=2M-\left(2,-\frac12,2\right)=\left(\frac43,\frac56,-\frac23\right).$$
Checking: $\dfrac{a}{8}=\dfrac{4/3}{8}=\dfrac16$, $\dfrac{b}{5}=\dfrac{5/6}{5}=\dfrac16$, $\dfrac{c}{-4}=\dfrac{-2/3}{-4}=\dfrac16$ --- all equal.

Solutions · 1