The shortest distance between the lines

${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$ and

${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$ is :

$\overrightarrow a$ = &lt; 3, 8, 3 &gt; <br><br>$\overrightarrow b$ = &lt; – 3, – 7, 6 &gt; <br><br>$\overrightarrow p$ = &lt; 3, – 1, 1 &gt; <br><br>$\overrightarrow q$ = &lt; –3, 2, 4 &gt; <br><br>$$\overrightarrow p \times \overrightarrow q = \left| {\matrix{ {\widehat i} &amp; {\widehat j} &amp; {\widehat k} \cr 3 &amp; { - 1} &amp; 1 \cr { - 3} &amp; 2 &amp; 4 \cr } } \right|$$ = &lt; -6, -15, 3 &gt; <br><br>Shortest distance = $$\left| {{{\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$$ <br><br>= $$\left| {{{\left( { - 6, - 15,3} \right).\left( { - 6, - 15,3} \right)} \over {\sqrt {36 + 225 + 9} }}} \right|$$ <br><br>= $\left| {{{36 + 225 + 9} \over {\sqrt {36 + 225 + 9} }}} \right|$ <br><br>= $\sqrt {270}$ = $3\sqrt {30}$