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Question
Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.
Solution
Equation of line AB passing through the points A(0, 6, -9) and B (-3, -6,3) is
\[\frac{x - 0}{- 3 - 0} = \frac{y - 6}{- 6 - 6} = \frac{z + 9}{3 + 9}\]
\[ = \frac{x}{1} = \frac{y - 6}{4} = \frac{z + 9}{- 4}\]
Here, D is the foot of the perpendicular drawn from C (7, 4, -1) on AB . The coordinates of a general point on AB are given by
\[\frac{x}{1} = \frac{y - 6}{4} = \frac{z + 9}{- 4} = \lambda\]
\[ \Rightarrow x = \lambda\]
\[ y = 4\lambda + 6\]
\[ z = - 4\lambda - 9\]
Let the coordinates of D be \[\left( \lambda, 4\lambda + 6, - 4\lambda - 9 \right)\]
The direction ratios of CD are proportional to \[\lambda - 7, 4\lambda + 6 - 4, - 4\lambda - 9 + 1, i . e . \lambda - 7, 4\lambda + 2, - 4\lambda - 8\]
The direction ratios of AB are proportional to 1, 4, -4, but CD is perpendicular to AB.
Substituting \[\lambda = - 1\] in \[\left( \lambda, 4\lambda + 6, - 4\lambda - 9 \right)\]
we get the coordinates of D as (-1,2,-5)
Equation of CD is ,
\[\frac{x - 7}{- 1 - 7} = \frac{y - 4}{2 - 4} = \frac{z + 1}{- 5 + 1}\]
\[ = \frac{x - 7}{4} = \frac{y - 4}{1} = \frac{z + 1}{2}\]
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