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

Answer 1

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


Equation of plane is $2x+3y-4z+22=0$
Equation of line $L$ is $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$
Direction ratios of line are $<1,4,5>$

Equation of line $PQ$ passing through point $P(1,-2,3)$ along line $L$ is
$\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda$
$A\equiv (\lambda +1,4\lambda -2,5\lambda +3)$
Let us say $A$ is the mid point of the line $PQ.$
Therefore, $A$ lies on the plane.
Hence, it should satisfy the equation of the plane.
$\Rightarrow 2(\lambda +1)+3(4\lambda -2)-4(5\lambda +3)+22=0$
$\Rightarrow -6\lambda +6=0$
$\Rightarrow \lambda =1$
Hence, coordinates of $A$ are $(2,2,8)$

Now, $PA=\sqrt{1^2+4^2+5^2}=\sqrt{42}$
$\therefore PQ=2PA=2\sqrt{42}$