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प्रश्न
Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.
उत्तर
Let L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line.
The coordinates of a general point on the line
\[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}\] are given by
\[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} = \lambda\]
\[\text{ They can be re - written as } \]
\[\frac{x - 4}{- 2} = \frac{y}{6} = \frac{z - 1}{- 3} = \lambda\]
\[ \Rightarrow x = - 2\lambda + 4\]
\[ y = 6\lambda\]
\[ z = - 3\lambda + 1\]
Let the coordinates of L be
\[\left( - 2\lambda + 4, 6\lambda, - 3\lambda + 1 \right)\]
The direction ratios of PL are proportional to
\[- 2\lambda + 4 - 2, 6\lambda - 3, - 3\lambda + 1 - 4, i . e . - 2\lambda + 2, 6\lambda - 3, - 3\lambda - 3\]
The direction ratios of the given line are proportional to -2,6,-3, but PL is perpendicular to the given line.
\[\therefore - 2\left( - 2\lambda + 2 \right) + 6\left( 6\lambda - 3 \right) - 3\left( - 3\lambda - 3 \right) = 0\]
\[ \Rightarrow \lambda = \frac{13}{49}\]
Substituting
\[ \Rightarrow \lambda = \frac{13}{49}\] in
\[\left( - 2\lambda + 4, 6\lambda, - 3\lambda + 1 \right)\] we get the coordinates of L as
\[\left( \frac{170}{49}, \frac{78}{49}, \frac{10}{49} \right)\]
\[\therefore PL = \sqrt{\left( \frac{170}{49} - 2 \right)^2 + \left( \frac{78}{49} - 3 \right)^2 + \left( \frac{10}{49} - 4 \right)^2}\]
\[ = \sqrt{\frac{44541}{2401}}\]
\[ = \sqrt{\frac{909}{49}}\]
\[ = \frac{3}{7}\sqrt{101}\]
Hence, the length of the perpendicular from P on PL is
\[\frac{3}{7}\sqrt{101} \text{ units } \] .
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