Question

The foot of the perpendicular of the point $(1,6,3)$ to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is?

• A. $(0,-3,5)$
• B. $(2,4,7)$
• C. $(1,3,7)$
• D. $(2,3,5)$

Answer 1

The correct option is A $(0,-3,5)$

Let $P=(1,6,3)$

and $Q$ be any point on line

$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$

$\Rightarrow x=\lambda ,y=2\lambda +1,z=3\lambda +2$

Direction ratio of $PQ=$

$(\lambda -1,2\lambda +1-6,3\lambda +2-3)$

$=(\lambda -1,2\lambda -5,3\lambda -1)$

As $PQ$ is perpendicular to given

line,

$1(\lambda -1)+2(2\lambda -5)+3(3\lambda -1)=0$

$\Rightarrow \lambda -1+4\lambda -10+9\lambda -3=0$

$\Rightarrow 14\lambda -14=0$

$\Rightarrow \lambda =1$

$\therefore$ foot of perpendicular =

$=(1-1,2-5,3-1)$

$=(0,-3,2)$