3D Geometry 3 Question 32
####32. If the image of the point $P(1,-2,3)$ in the plane $2 x+3 y-4 z+22=0$ measured parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $Q$, then $P Q$ is equal to
(a) $3 \sqrt{5}$
(b) $2 \sqrt{42}$
(c) $\sqrt{42}$
(d) $6 \sqrt{5}$
(2017 Main)
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Answer:
Correct Answer: 32. (b)
Solution:
- Any line parallel to $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ and passing through $P(1,-2,3)$ is
$ \frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda $
Any point on above line can be written as
$(\lambda+1,4 \lambda-2,5 \lambda+3)$
$\therefore$ Coordinates of $R$ are $(\lambda+1,4 \lambda-2,5 \lambda+3)$.
Since, point $R$ lies on the above plane.
$\therefore \quad 2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$
$\Rightarrow \quad \lambda=1$
So, point $R$ is $(2,2,8)$.
Now, $P R=\sqrt{(2-1)^{2}+(2+2)^{2}+(8-3)^{2}}=\sqrt{42}$
$\therefore \quad P Q=2 P R=2 \sqrt{42}$
