Let d be the distance between the feet of perpendiculars of the points P(1,2,-1) and Q(2,-…

Question

Let $d$ be the distance between the feet of perpendiculars of the points $P(1,2,-1)$ and $Q(2,-1,3)$ on the plane $-x+y+z=1$. Then $d^2$ is equal to:

Answer: 26

Accepted Answer

Foot from $P$: $\dfrac{x-1}{-1}=\dfrac{y-2}{1}=\dfrac{z+1}{1}=\dfrac{-(-1+2-1-1)}{3}=\dfrac13$, giving $P'=\left(\dfrac23,\dfrac73,-\dfrac23\right)$.

Foot from $Q$: $\dfrac{x-2}{-1}=\dfrac{y+1}{1}=\dfrac{z-3}{1}=\dfrac{-(-2-1+3-1)}{3}=\dfrac13$, giving $Q'=\left(\dfrac53,-\dfrac23,\dfrac{10}{3}\right)$.
$$P'Q'=\sqrt{1^2+3^2+4^2}=\sqrt{26} \implies d^2=26.$$

Solutions · 1