Question

$A(1,8,4)$, $B(0,-11,4)$, $C(2,-3,1)$ are three points and L is the foot of the perpendicular from A to BC. Find the coordinates of L.

Answer 1

The Cartesian equation of a line passing through $B(0,-11,4)$ and $C(2,-3,1)$ is,

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Therefore, equation of line $BC$ is,

$\Rightarrow \frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-4}{1-4}$

$\Rightarrow \frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}$

Consider the diagram shown below. Let $L$ be the foot of the perpendicular from point $A(1,8,4)$ to the given line.

The coordinates of point $L$ on the line $BC$ is given by,

$\frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda$

$\Rightarrow x=2\lambda$

$\Rightarrow y=8\lambda -11$

$\Rightarrow z=-3\lambda +4$

The direction ratios of $AL$ is $(x_2-x_1),(y_2-y_1),(z_2-z_1)$.

$\Rightarrow (2\lambda -1),(8\lambda -11-8),(-3\lambda +4-4)$

$\Rightarrow (2\lambda -1),(8\lambda -19),(-3\lambda )$

Since, both the lines are perpendicular, we know that

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$2(2\lambda -1)+8(8\lambda -19)+(-3)(-3\lambda )=0$

$4\lambda -2+64\lambda -152+9\lambda =0$

$77\lambda =154$

$\lambda =2$

Therefore, coordinates of $L$ are,

$x=2\left(2\right)=4$

$y=8\left(2\right)-11=5$

$z=-3\left(2\right)+4=-2$

Hence, the coordinates of the foot of the perpendicular is $(4,5,-2)$.