Question

The foot of the perpendicular drawn from the point $(1,8,4)$ on the line joining the points $(0,-11,4)$ and $(2,-3,1)$ is

• A. (4, 5, 2)
• B. (-4, 5, 2)
• C. (4, -5, 2)
• D. (4, 5, -2)

Answer 1

Answer: (D)

Equation of line joining points $(0,-11,4)$ and $(2,-3,1)$ is
$\frac{x-2}{2}=\frac{y+3}{8}=\frac{z-1}{-3}=\lambda$

Let $P$ is any point of the above line then coordinate of $P$ is
$(2\lambda +2,8\lambda -3,-3\lambda +1)$
$\therefore D{R}^{\prime }s$ of $PQ$ is $(2\lambda +1,8\lambda -11,-3\lambda -3)$
Now, $(2\lambda +1)(2)+(8\lambda -11)(8)+(-3\lambda -3)(-3)=0$
$\Rightarrow 4\lambda +2+64\lambda -88+9\lambda +9=0$
$\Rightarrow 77\lambda -77=0$
$\Rightarrow \lambda =1$
$\therefore$ Required foot of perpendicular,