Let P be the image of the point Q(7,-2,5) in the line L:(x-1)/(2)=(y+1)/(3)=(z)/(4) and R(…

Question

Let P be the image of the point $Q(7,-2,5)$ in the line $L:\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5,p,q)$ be a point on $L$. Then the square of the area of $	rianglePQR$ is \_\_\_\_\_\_\_\_\_.

Answer: 957

Accepted Answer

Let $R(2\lambda+1,\ 3\lambda-1,\ 4\lambda)$ be the foot of the perpendicular from $Q$ to $L$: with $\vec{QR}\cdot(2,3,4)=0$,
$$4\lambda-12+9\lambda+3+16\lambda-20=0 \implies 29\lambda-29=0\implies\lambda=1... $$
(using the source's working, $\lambda=2$ was found for a related foot computation giving $R'(5,5,8)$ satisfying $x=5$; using $2\lambda+1=5\Rightarrow\lambda=2$, and checking $4(2)-12+9(2)+3+16(2)-20=8-12+18+3+32-20=29\neq0$ --- so instead the foot of perpendicular from $Q$ used for the image $P$ is found at $T$ below).

Let $T(2\lambda+1,\ 3\lambda-1,\ 4\lambda)$ be the foot of perpendicular from $Q(7,-2,5)$: $\vec{QT}=(2\lambda-6,\ 3\lambda+1,\ 4\lambda-5)$, and $\vec{QT}\cdot(2,3,4)=0\Rightarrow \lambda=1$, giving $T=(3,2,4)$.
$P=2T-Q=(6-7,4+2,8-5)=(-1,6,3).$
Given $R(5,3,8)$ (taking $\lambda=2$ on $L$, since $x=5\Rightarrow\lambda=2$, so $R=(5,5,8)$):
$$QT=\sqrt{4+9+16-\ldots}=\sqrt{33},\qquad RT=\sqrt{29}.$$
$$(\text{Area of }\triangle PQR)^2=\left(\frac12\sqrt{29}\cdot2\sqrt{33}\right)^2=29\times33=957.$$

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