Question

Find the co-ordinates of the foot of perpendicular drawn from the point $A(-1,8,4)$ to the line joining the points $B(0,–1,3)$ and $C(2,–3,–1)$.

Answer 1

Let $P$ be the foot of the perpendicular drawn from point $A$ on the line joining points $B$ and $C$.

Equation of the line joining the points $B(0,–1,3)$ and $C(2,–3,–1)$ is
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
$\Rightarrow \frac{x-0}{2-0}=\frac{y+1}{-3+1}=\frac{z-3}{-1-3}$
$\Rightarrow \frac{x-0}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}$

Let $\frac{x-0}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}=\lambda$

General coordinates of $P$ is $(2\lambda ,-2\lambda -1,-4\lambda +3)$
Direction ratios of $AP(2\lambda +1,-2\lambda -9,-4\lambda -1)$

$\because$ Both the lines $AP$ and $BC$ are perpendicular to each other.
$\therefore 2(2\lambda +1)-2(-2\lambda -9)-4(-4\lambda -1)=0$
$\Rightarrow 24\lambda +24=0$
$\Rightarrow \lambda =-1$
$\therefore P(-2,1,7)$

So, the coordinates of the foot of perpendicular drawn from the point $A$ to $BC$ is $P(-2,1,7)$.