Question

A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.

Answer 1

Point D is the foot of the perpendicular drawn from the point A (1, 0, 4) to the line BC.

The coordinates of a general point on the line BC are given by

$$\begin{gathered} \frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-3}{1-3}=\lambda \\ \Rightarrow x=2\lambda \\ y=8\lambda -11 \\ z=-2\lambda +3 \end{gathered}$$

Let the coordinates of D be $\left(2\lambda ,8\lambda -11,-2\lambda +3\right)$.

The direction ratios of AD are proportional to $2\lambda -1,8\lambda -11-0,-2\lambda +3-4,\mathrm{i}.\mathrm{e}.2\lambda -1,8\lambda -11,-2\lambda -1$.

The direction ratios of the line BC are proportional to 2, 8,$-$2, but AD is perpendicular to the line BC.

$$\begin{gathered} \therefore 2\left(2\lambda -1\right)+8\left(8\lambda -11\right)-2\left(-2\lambda -1\right)=0 \\ \Rightarrow \lambda =\frac{11}{9} \end{gathered}$$

Substituting $\lambda =\frac{11}{9}$ in $\left(2\lambda ,8\lambda -11,-2\lambda +3\right)$, we get the coordinates of D as $\left(\frac{22}{9},-\frac{11}{9},\frac{5}{9}\right)$.