The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is :
Solution
$L_{1}: \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$
<br/><br/>
$$
\begin{aligned}
& L_{2}: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \\\\
& \text { S.D }=\left|\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\right| \\\\
& =\left|\frac{-2-6-6}{7}\right|=2 \text { units }
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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