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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.
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Solution
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 \[\left( \frac{22}{9}, - \frac{11}{9}, \frac{5}{9} \right)\]
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