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 

\[\frac{x - 0}{2 - 0} = \frac{y + 11}{- 3 + 11} = \frac{z - 3}{1 - 3} = \lambda\]

\[ \Rightarrow x = 2\lambda\]

\[ y = 8\lambda - 11 \]

\[ z = - 2\lambda + 3\]

Let the coordinates of D be

\[\left( 2\lambda, 8\lambda - 11, - 2\lambda + 3 \right)\]

The direction ratios of AD are proportional to 

\[2\lambda - 1, 8\lambda - 11 - 0, - 2\lambda + 3 - 4, i . e . 2\lambda - 1, 8\lambda - 11, - 2\lambda - 1\]

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

\[\therefore 2\left( 2\lambda - 1 \right) + 8\left( 8\lambda - 11 \right) - 2\left( - 2\lambda - 1 \right) = 0\]

\[ \Rightarrow \lambda = \frac{11}{9}\]

Substituting 

\[ \Rightarrow \lambda = \frac{11}{9}\] in 

\[\left( 2\lambda, 8\lambda - 11, - 2\lambda + 3  \right)\] we get the coordinates of D as  \[\left( \frac{22}{9}, - \frac{11}{9}, \frac{5}{9} \right)\]