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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).
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Solution
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
\[\frac{x - 4}{4 - 3} = \frac{y - 7}{7 - 5} = \frac{z - 1}{1 - 3} = \lambda\]
\[ \Rightarrow x = \lambda + 4\]
\[ y = 2\lambda + 7 \]
\[ z = - 2\lambda + 1\]
Let the coordinates of D be
\[\left( \lambda + 4, 2\lambda + 7, - 2\lambda + 1 \right)\]
The direction ratios of AD are proportional to
\[\lambda + 4 - 1, 2\lambda + 7 - 0, - 2\lambda + 1 - 3, i . e . \lambda + 3, 2\lambda + 7, - 2\lambda - 2\]
The direction ratios of the line BC are proportional to 1, 2,-2, but AD is perpendicular to the line BC.
\[\therefore 1\left( \lambda + 3 \right) + 2\left( 2\lambda + 7 \right) - 2\left( - 2\lambda - 2 \right) = 0\]
\[ \Rightarrow \lambda = - \frac{7}{3}\]
Substituting
\[ \Rightarrow \lambda = - \frac{7}{3}\] in
\[\left( \lambda + 4, 2\lambda + 7, - 2\lambda + 1 \right)\] we get the coordinates of D as \[\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)\] .
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