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