Consider the line L given by x-32=y-11=z-21.

Question

Consider the line $L$ given by $\dfrac{x-3}{2}=\dfrac{y-1}{1}=\dfrac{z-2}{1}$. Let $Q$ be the mirror image of the point $(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$, and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?

Accepted Answer

Line point $A=(3,1,2)$, direction $\vec d=(2,1,1)$. $\overrightarrow{A,(2,3,-1)}=(-1,2,-3)$.
$$t=\frac{(-1)(2)+2(1)+(-3)(1)}{6}=\frac{-3}{6}=-\frac12 \implies M=A+t\vec d=(2,0.5,1.5).$$
$Q=2M-(2,3,-1)=(2,-2,4).$
Plane $P\perp L$ through $Q$ has normal $(2,1,1)$:
$$2(x-2)+1(y+2)+1(z-4)=0\Rightarrow 2x+y+z-6=0\Rightarrow 2x+y+z=6.$$
Testing $D(1,2,2)$: $2+2+2=6$  (others fail).

Solutions · 1