Question

If the image of the point $P\left(1,-2,3\right)$ 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

$2\sqrt{42}$

Explanation for the correct option:

The equation of the line is $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$.

Thus, the parallel line of the given line can be given by $\frac{x-a}{1}=\frac{y-b}{4}=\frac{z-c}{5}$.

The point $P$ is defined as $\left(1,-2,3\right)$.

Thus, the equation of the line $PQ$ is $\frac{x-1}{1}=\frac{y-\left(-2\right)}{4}=\frac{z-3}{5}$.

$\Rightarrow \frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}$

As $Q$ is the image of the point $P$ with respect to the plane, $2x+3y-4z+22=0$, thus the midpoint of $PQ$ is on the given plane.

Let us assume that $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda$.

Thus, the midpoint of $PQ$ can be given by $M\left(\lambda +1,4\lambda -2,5\lambda +3\right)$.

Now, the point $M$ is on the plane $2x+3y-4z+22=0$.

Thus, $2\left(\lambda +1\right)+3\left(4\lambda -2\right)-4\left(5\lambda +3\right)+22=0$

$$\begin{gathered} \Rightarrow 2\lambda +2+12\lambda -6-20\lambda -12+22=0 \\ \Rightarrow -6\lambda +6=0 \\ \Rightarrow \lambda =1 \end{gathered}$$

Thus, $M\equiv \left(\lambda +1,4\lambda -2,5\lambda +3\right)$.

$$\begin{gathered} \Rightarrow M\equiv \left(1+1,4\left(1\right)-2,5\left(1\right)+3\right) \\ \Rightarrow M\equiv \left(2,2,8\right) \end{gathered}$$

Thus, the distance, $PM=\sqrt{{\left(2-1\right)}^2+{\left(2+2\right)}^2+{\left(8-3\right)}^2}$.

$$\begin{gathered} {\Rightarrow PM=\sqrt{1+16+25}}_{} \\ \Rightarrow PM=\sqrt{42} \end{gathered}$$

As $M$ is the midpoint of $PQ$, thus $PQ=2PM$.

$\Rightarrow PQ=2\sqrt{42}$.

Hence, option (A) is the correct answer.