Question

The image of the point $(3,-1,11)$ w.r.t the line $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is:

• A. $(2,5,7)$
• B. $(1,11,3)$
• C. $(0,0,0)$
• D. $(0,2,3)$

Answer 1

The correct option is B $(1,11,3)$

Given line is $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}=r\left(\text{assume}\right)$
Let $P$ be the foot of the perpendicular from $A(3,-1,11)$ to given line then
$P=(2r,3r+2,4r+3)$
D.r's of $\overrightarrow{AP}$ are $(2r-3,3r+3,4r-8)$
$\overrightarrow{AP}$ is perpendicular to the given line
$\Rightarrow 2(2r-3)+3(3r+3)+4(4r-8)=0\Rightarrow r=1$
Hence, $P=(2,5,7)$
Reflection and object will be equidistant from line.
If $B$ is the image of $A$, then
$\overrightarrow{P}=\frac{\overrightarrow{A}+\overrightarrow{B}}{2}$
$\overrightarrow{B}=2\overrightarrow{P}-\overrightarrow{A}=2(2\hat{i}+5\hat{j}+7\hat{k})-(3\hat{i}-1\hat{j}+11\hat{k})$
$\overrightarrow{B}=\hat{i}+11\hat{j}+3\hat{k}$
$\therefore$ Image $=B(1,11,3)$