Question

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

• A. $(1,0,7)$
• B. $(7,0,1)$
• C. $(2,7,0)$
• D. $(-1,-6,-3)$

Answer 1

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


$\text{Let}\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=k$
Then any point on the line $AB$ is
$x=k,y=2k+1,z=3k+2$
Direction ratio of $PQ=(k-1,2k-5,3k-1)$
$\because PQ$ is perpendicular to $AB$.
$\Rightarrow 1(k-1)+2(2k-5)+3(3k-1)=0$
$\Rightarrow k=1$
$\therefore$ Coordinates of $Q$ is $(1,3,5)$.
$\text{Also},Q$ is the mid-point of $PR$.
$\Rightarrow \frac{a+1}{2}=1,\frac{b+6}{2}=3,\frac{c+3}{2}=5$
$\Rightarrow a=1,b=0,c=7$
$\therefore$ The image of $(1,6,3)$ is $(1,0,7)$.