The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Shortest distance }=\left|\frac{\left|\begin{array}{ccc}3-(-3) & 8-(-7) & 3-6 \ 3 & -1 & 1 \ -3 & 2 & 4\end{array}ight|}{\sqrt{(-4-2)^2+(12+3)^2+(6-3)^2}}ight| \ & =\frac{|6 	imes(-6)-15 	imes(15)-3 	imes(3)|}{\sqrt{36+225+9}} \ & =\frac{270}{\sqrt{270}}=\sqrt{270}=3 \sqrt{30}\end{aligned}$

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and…

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