Question

The image of 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

• A. $(-2,2,8)$
• B. $(-3,6,13)$
• C. $(2,2,8)$
• D. $(3,6,13)$

Answer 1

The correct option is D $(3,6,13)$

Given line is $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$
Line parallel to this and passing through $P(1,-2,3)$ is
$\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda \left(\text{say}\right)$
Let the image be $Q$ which lies on the line, so
$Q\equiv (\lambda +1,4\lambda -2,5\lambda +3)$

Midpoint of $P$ and $Q$ is $R\equiv (\frac{\lambda +2}{2},\frac{4\lambda -4}{2},\frac{5\lambda +6}{2})$
Midpoint of $P$ and $Q$ lies on the plane $2x+3y-4z+22=0$
Putting it in the equation of plane, we get
$\lambda +2+3(2\lambda -2)-2(5\lambda +6)+22=0\Rightarrow -3\lambda +6=0\Rightarrow \lambda =2\therefore Q\equiv (3,6,13)$