Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $4 x - 3 y + 1 = 0 = 5 x - 3 z + 2$ are interesting lines. Also find point of intersection.

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Solution

$L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} L_{2} : 4 x - 3 y + 1 = 0 = 5 x - 3 z + 2$
Let $l m n$ be directions of $L_{2}$
$∴ 4 l - 3 m = 0 - (i) ∴ 5 l - 3 n = 0 - (i i)$
Applying cramer's rule for $(i)$ & $(i i)$ ,
$\frac{l}{9} = \frac{- m}{- 12} = \frac{n}{+ 15} ∴ (l, m, n) = (3, 4, + 5)$
Let $(x, y, z)$ satisfy $l_{2}$
$∴$ By hit and trail we get $(x, y, z) = (0, \frac{1}{3}, \frac{2}{3})$
$∴ L_{2} : \frac{x - 0}{3} = \frac{y - \frac{1}{3}}{4} = \frac{z - \frac{2}{3}}{+ 5} = r$
Say $3 r, 4 r + \frac{1}{3}, + 5 r + \frac{2}{3}$ is point of intersection hence it must satisfy $L_{1}$
$\frac{3 r - 1}{2} = \frac{4 r + \frac{1}{3} - 2}{3} = \frac{+ 5 r + \frac{2}{3} - 3}{4} r = \frac{- 1}{3}$
$∴$ Point of intersection $= (- 1, - 1, - 1)$

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