Find the length of the perpendicular from $A (1, - 2, - 3)$ on the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ .

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Solution

Let $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1} = r$
$\Rightarrow x = - 3 r - 1, y = 2 r + 3, z = - r + 2$
So, general general point on the given line is $\Rightarrow (- 3 r - 1, 2 r + 3, - r + 2)$

Let the foot of perpendicular on the given line be $\Rightarrow Q (- 3 r_{1} - 1, 2 r_{1} + 3, - r_{1} + 2)$

Directional ratio of $\to A Q$ are

$- 3 r_{1} - 1 - 1, 2 r_{1} + 3 + 2, - r_{1} + 2 + 3$
i.e. $- 3 r_{1} - 2, 2 r_{1} + 5, - r_{1} + 5$

Since line $\to A Q$ is perpendicular to given line

So,
$\begin{matrix} - 3 (- 3 r_{1} - 2) + 2 (2 r_{1} + 5) - 1 (- r_{1} + 5) = 0 \Rightarrow 14 r_{1} + 11 = 0 \end{matrix}$

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Find the length of the perpendicular from $A (1, - 2, - 3)$ on the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ .