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प्रश्न
Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\] Also, write down the coordinates of the foot of the perpendicular from P.
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उत्तर
Let L be the foot of the perpendicular drawn from the point P (2, 4,-1) to the given line.
The coordinates of a general point on the line
\[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\] are given by
\[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} = \lambda\]
\[ \Rightarrow x = \lambda - 5\]
\[ y = 4\lambda - 3\]
\[ z = - 9\lambda + 6\]
Let the coordinates of L be
\[\left( \lambda - 5, 4\lambda - 3, - 9\lambda + 6 \right)\]
The direction ratios of PL are proportional to
\[\lambda - 5 - 2, 4\lambda - 3 - 4, - 9\lambda + 6 + 1, i . e . \lambda - 7, 4\lambda - 7, - 9\lambda + 7\]
The direction ratios of the given line are proportional to 1, 4, - 9, but PL is perpendicular to the given line.
\[\therefore 1\left( \lambda - 7 \right) + 4\left( 4\lambda - 7 \right) - 9\left( - 9\lambda + 7 \right) = 0\]
\[ \Rightarrow \lambda = 1\]
Substituting
\[ \Rightarrow \lambda = 1\] in
\[\left( \lambda - 5, 4\lambda - 3, - 9\lambda + 6 \right)\] we get the coordinates of L as (-4,1,-3) Equation of the line PL is
\[\frac{x - 2}{- 4 - 2} = \frac{y - 4}{1 - 4} = \frac{z + 1}{- 3 + 1}\]
\[ = \frac{x - 2}{- 6} = \frac{y - 4}{- 3} = \frac{z + 1}{- 2}\]
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