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  4. The foot of the perpendicular from (2, 4, -1) to the line x+5 = (1/4)(y+3) = -(1/9)(z-6)

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Q. The foot of the perpendicular from (2, 4, -1) to the line $x+5 = \frac{1}{4}\left(y+3\right) = -\frac{1}{9}\left(z-6\right)$

Three Dimensional Geometry

Solution:

Given equation of line is
$x+5 = \frac{1}{4}\left(y+3\right) = -\frac{1}{9}\left(z-6\right)$
or $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9} = \lambda \left(say\right)$
$x = \lambda - 5, y = 4 \lambda - 3, z = - 9 \lambda + 6$
$\left(x, y, z\right) \equiv \left( \lambda - 5, 4 \lambda - 3, -9 \lambda + 6\right)\quad \dots \left(i\right)$
Let it is foot of perpendicualr
So, d.r.’s of $\bot$ line is
$\left( \lambda - 5 - 2, 4 \lambda - 3 - 4, - 9 \lambda + 6 + 1\right)$
$\equiv \left( \lambda - 7, 4 \lambda - 7, - 9 \lambda + 7\right)$
D.r.’s of given line is $\left(1, 4, - 9\right)$ and both lines are $\bot$
$\therefore \left(\lambda-7\right). 1 + \left(4\lambda - 7\right). 4 + \left(- 9 \lambda + 7\right) \left(-9\right) = 0$
$\Rightarrow 98 \lambda = 98 \Rightarrow \lambda = 1 $
$\therefore $ Point is $(- 4, 1, - 3)$. [Substituting $\lambda = 1$ in (i)]