Let Q be the foot of the perpendicular from the point P(7,-2,13) on the plane containing t…

Question

Let $Q$ be the foot of the perpendicular from the point $P(7,-2,13)$ on the plane containing the lines $\dfrac{x+1}{6}=\dfrac{y-1}{7}=\dfrac{z-3}{8}$ and $\dfrac{x+1}{3}=\dfrac{y-2}{5}=\dfrac{z-3}{7}$. Then $(PQ)^2$ is equal to

Answer: 96

Accepted Answer

Normal to the plane containing the two lines:
$$(6,7,8)\times(3,5,7)=(49-40,\ 24-42,\ 30-21)=(9,-18,9)\parallel(1,-2,1).$$
Plane through $(-1,1,3)$: $(x+1)-2(y-1)+(z-3)=0\implies x-2y+z=0$.
$$PQ=\frac{|7-2(-2)+13|}{\sqrt{1+4+1}}=\frac{24}{\sqrt6}=4\sqrt6 \implies PQ^2=96.$$

Solutions · 1