Question

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

• A. $(1,3,5)$
• B. $(-1,-1,-1)$
• C. $(2,5,8)$
• D. $(-2,-3,-4)$
• E. $(\frac{1}{2},2,\frac{1}{2})$

Answer 1

The correct option is A $(1,3,5)$

Any point on the line can be written as $x=\lambda ,y=2\lambda +1,z=3\lambda +2$
Hence, coordinates of $Q$ are $(\lambda ,2\lambda +1,3\lambda +2)$
DR's of $PQ$ are $\lambda -1,2\lambda +1-6,3\lambda +2-3$
i.e. $\lambda -1,2\lambda -5,3\lambda -1$
Since, $PQ$ is perpendicular to the given lines.
So, $1(\lambda -1)+2(2\lambda -5)+3(3\lambda -1)=0$
$\Rightarrow \lambda -1+4\lambda -10+9\lambda -3=0$
$\Rightarrow 14\lambda =14\Rightarrow \lambda =1$
Therefore, coordinates of $Q=(1,3,5)$.