Question

If the image of the point $P(1,-2,3)$ in the plane, $2x+3y-4z+22=0$ measured parallel to the line; $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $Q$, then $PQ$ is equal to

• A. $2\sqrt{42}$
• B. $\sqrt{42}$
• C. $6\sqrt{5}$
• D. $3\sqrt{5}$

Answer 1

The correct option is A $2\sqrt{42}$

Given that image of P (1 , -2, 3) in the plane 2x + 3y - 4z + 22 = 0 measured parallel to line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is Q .

Take a look at diagram for understanding better.

As PQ is parallel to $\frac{x}{1}=\frac{y}{4}=\frac{z}{5},$ D.R's also remain same.

Line PQ passing through (1 , -2, 3) can be written as

$\frac{x-1}{1}=\frac{y-(-2)}{4}=\frac{z-3}{5}=\lambda$ (let) ----eq.1

Let A is point on the plane and midpoint of P & Q.

From eq.1, coordinates of A will be of form

$A=(\lambda +1,4\lambda -2,5\lambda +3)$

As A lies on the plane 2x + 3y - 4z + 22 = 0

$\Rightarrow 2(\lambda +1)+3(4\lambda -2)-4(5\lambda +3)+22=0$

$\Rightarrow 6\lambda =6\Rightarrow \lambda =1$

$\Rightarrow A=(2,2,8),P(1,-2,3)$

Now, PA $=\sqrt{(2-1{)}^2+(2-(-2){)}^2+(8-3{)}^2}=\sqrt{1+16+25}=\sqrt{42}$

As PA = AQ, [$\because$ image]

PQ = 2PA $=2\sqrt{42}$