Question

The image of the point $(1,6,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$

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

Answer 1

The correct option is B $(1,0,7)$

Let $P$ be the foot of the perpendicular from $A(1,6,3)$ to given line then,
$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=r$
$P=(r,2r+1,3r+2)$
D.R's of $\overrightarrow{AP}$ are $(r-1,2r-5,3r-1)$
$\overrightarrow{AP}$ is perpendicular to the given line
$\Rightarrow 1(r-1)+2(2r-5)+3(3r-1)=0\Rightarrow r=1$
Hence, $P=(1,3,5)$
If $B$ is the image of $A$, then
So, $P$ is midpoint of image and object
$\overrightarrow{P}=\frac{\overrightarrow{A}+\overrightarrow{B}}{2}$
$\overrightarrow{B}=2\overrightarrow{P}-\overrightarrow{A}=2(\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+6\hat{j}+3\hat{k})=\hat{i}+0\hat{j}+7\hat{k}$
Image is $(1,0,7)$