Prove that the lines $x + 3 = 2 y = - 12 z$ and $x = y + 6 = 6 z - 18$ are parallel. Find the distance between them.

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Solution

For 1st line $L_{1} :$
$x + 3 = 2 y = - 12 z$
$\frac{x + 3}{1} = \frac{y - 0}{\frac{1}{2}} = \frac{z - 0}{\frac{- 1}{2}} = 5.......... (1)$
$\Rightarrow \to r = (- 3 ˆ i) + 5 (ˆ i + \frac{1}{2} ˆ j - \frac{1}{12} ˆ k)$
$\Rightarrow \to r = \to a_{1} + 5 \to a_{1}$
Where,
$\to a_{1} = - 3 ˆ i$
$\to b_{1} = ˆ i + \frac{1}{2} ˆ j - \frac{1}{12} ˆ k$
For $2 n d$ line $L_{2} :$
$x = y + 6 = 6 z - 18$
$\begin{matrix} \frac{x - 0}{1} = \frac{y + 6}{1} = \frac{z - 3}{1 / 6} . . . . .1 \Rightarrow \frac{x - 1}{4} = \frac{y - 2}{2} = \frac{z - 3}{- 4} \end{matrix}$
$\Rightarrow \to r = (- 6 ˆ j + 3 ˆ k) + t (ˆ i + ˆ j + \frac{1}{6} ˆ k)$
when,
$\begin{matrix} \to a_{2} = - 6 ˆ j + 3 ˆ k \to b_{2} = ˆ i + ˆ j + \frac{1}{6} ˆ k \end{matrix}$
It $2$ lines are neither parallel nor intersecting then they must be skew.

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