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