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

Answer 1

Answer: (A)

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 $Q=(1,3,5)$