Question

Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).

Answer 1

Let D be the foot of the perpendicular drawn from the point A (1, 0, 3) to the line BC.

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

$$\begin{gathered} \frac{x-4}{4-3}=\frac{y-7}{7-5}=\frac{z-1}{1-3}=\lambda \\ \Rightarrow x=\lambda +4 \\ y=2\lambda +7 \\ z=-2\lambda +1 \end{gathered}$$

Let the coordinates of D be $\left(\lambda +4,2\lambda +7,-2\lambda +1\right)$.




The direction ratios of AD are proportional to $\lambda +4-1,2\lambda +7-0,-2\lambda +1-3,\mathrm{i}.\mathrm{e}.\lambda +3,2\lambda +7,-2\lambda -2$.

The direction ratios of the line BC are proportional to 1, 2, $-$2, but AD is perpendicular to the line BC.

$$\begin{gathered} \therefore 1\left(\lambda +3\right)+2\left(2\lambda +7\right)-2\left(-2\lambda -2\right)=0 \\ \Rightarrow \lambda =-\frac{7}{3} \end{gathered}$$
Substituting $\lambda =-\frac{7}{3}$ in $\left(\lambda +4,2\lambda +7,-2\lambda +1\right)$, we get the coordinates of D as $\left(\frac{5}{3},\frac{7}{3},\frac{17}{3}\right)$.