Question

Find the foot of perpendicular grom the point $(2,3,-8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$.Also, find the perpendicular distance from the given point to the line.

Answer 1

Let the given equation be $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\lambda$

This can be written as $\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda$ .........$\left(1\right)$

$\therefore$ the coordinates of any point on the line is

$x=4-2\lambda$,$y=6\lambda$,$z=1-3\lambda$

Let $Q(4-2\lambda ,6\lambda ,1-3\lambda )$ be the foot of perpendicular from the point $P(2,3,-8)$ on line $\left(1\right)$

We know the direction ratios of any line segement $PQ$ is given by $(x_2-x_1,y_2-y_1,z_2-z_1)$

The direction cosines of $PQ$ is given by

$=(-2\lambda +4-2,6\lambda -3,-3\lambda +1+8)$

$=(-2\lambda +2,6\lambda -3,-3\lambda +9)$

Now $Q$ is the foot of the perpendicular of the line $\left(1\right)$

$\overrightarrow{PQ}$ is the perpendicular to the line $\left(1\right)$

hence the sum of the product of this direction ratios is $0$

$=(-2\lambda +2)(-2)+(6\lambda -3).6+(-3\lambda +9)(-3)=0$

$\Rightarrow 4\lambda -4+36\lambda -18+9\lambda -27=0$

$\Rightarrow 49\lambda -49=0$

$\therefore \lambda =1$

Substituting $\lambda =1$ in $Q$ we get

$Q(0,3,6)$