Shortest Distance Between 2 Skew Lines

Method I
(Use of Magic \(\vartriangle \) without finding \(P\)

\(\left| \overrightarrow{AB}\cdot \frac{\overrightarrow{PQ}}{\left| \overrightarrow{PQ} \right|} \right|\) \(\overrightarrow{AB}=\left( \begin{matrix} 1 \\ 3 \\ 2 \\ \end{matrix} \right)-\left( \begin{matrix} -1 \\ 1 \\ 3 \\ \end{matrix} \right)\)

Shortest Distance
\(=\)Projection length of \(AB\) onto \(\overrightarrow{PQ}\)
\(=\left| \overrightarrow{AB}\cdot \frac{\overrightarrow{PQ}}{\left| \overrightarrow{PQ} \right|} \right|\); \(\overrightarrow{PQ}=\left( \begin{matrix} -2 \\ 2 \\ 1 \\ \end{matrix} \right)\times \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right)=\left( \begin{matrix} 11 \\ 15 \\ -8 \\ \end{matrix} \right)\)
\(=\frac{\left| \left( \begin{matrix} 2 \\ 2 \\ -1 \\ \end{matrix} \right)\cdot \left( \begin{matrix} 11 \\ 15 \\ -8 \\ \end{matrix} \right) \right|}{\sqrt{{{11}^{2}}+{{15}^{2}}+{{\left( -8 \right)}^{2}}}}=\frac{\left| 22+30+8 \right|}{\sqrt{410}}\)
\(=\frac{60}{\sqrt{410}}\cdot \frac{\sqrt{410}}{\sqrt{410}}=\frac{60}{410}\cdot \sqrt{410}\)

\(=\frac{6}{41}\sqrt{410}\)units

Method II (Find \(P\) and \(Q\) first)

For some \(\lambda \in \mathbb{R}\): \(\overrightarrow{OP}=\left( \begin{matrix} -1 \\ 1 \\ 3 \\ \end{matrix} \right)+\lambda \left( \begin{matrix} -2 \\ 2 \\ 1 \\ \end{matrix} \right)\)

For some \(\mu \in \mathbb{R}\): \(\overrightarrow{OQ}=\left( \begin{matrix} 1 \\ 3 \\ 2 \\ \end{matrix} \right)+\mu \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right)\)
\(\overrightarrow{PQ}=\left[ \left( \begin{matrix} 1 \\ 3 \\ 2 \\ \end{matrix} \right)+\mu \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right) \right]\)

\(-\left[ \left( \begin{matrix} -1 \\ 1 \\ 3 \\ \end{matrix} \right)+\lambda \left( \begin{matrix} -2 \\ 2 \\ 1 \\ \end{matrix} \right) \right]\)
=\(\left( \begin{matrix} 2 \\ 2 \\ -1 \\ \end{matrix} \right)+\mu \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right)+\lambda \left( \begin{matrix} 2 \\ -2 \\ -1 \\ \end{matrix} \right)\)

\(\overrightarrow{PQ}\cdot {{\mathbf{d}}_{1}}=0\)
\(\left[ \left( \begin{matrix} 2 \\ 2 \\ -1 \\ \end{matrix} \right)+\mu \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right)+\lambda \left( \begin{matrix} 2 \\ -2 \\ -1 \\ \end{matrix} \right)\cdot \left( \begin{matrix} -2 \\ 2 \\ 1 \\ \end{matrix} \right) \right]=0\)
\(\left( -4+4-1 \right)+\mu \left( -6+2+6 \right)+\lambda \left( -4-4-1 \right)=0\)
\(2\mu -9\lambda =1\)------- \(1\).

\(\overrightarrow{PQ}\cdot {{\mathbf{d}}_{2}}=0\)
\(\left[ \left( \begin{matrix} 2 \\ 2 \\ -1 \\ \end{matrix} \right)+\mu \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right)+\lambda \left( \begin{matrix} 2 \\ -2 \\ -1 \\ \end{matrix} \right)\cdot \left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right) \right]=0\)
\(\left( \left( 6+2-6 \right)+\mu \left( 9+1+36 \right)+\lambda \left( 6-2-6 \right) \right)=0\)
\(46\mu -2\lambda =-2\)--------\(2\).

Solve \(1\And 2\):
\(\mu =-\frac{2}{41}\), \(\lambda =-\frac{5}{41}\)

\(\overrightarrow{OP}=\left( \begin{matrix} -1 \\ 1 \\ 3 \\ \end{matrix} \right)-\frac{5}{41}\left( \begin{matrix} -2 \\ 2 \\ 1 \\ \end{matrix} \right)\)
\(=\left( \begin{matrix} -\frac{31}{41} \\ \frac{31}{41} \\ \frac{118}{41} \\ \end{matrix} \right)=\frac{1}{41}\left( \begin{matrix} -31 \\ 31 \\ 118 \\ \end{matrix} \right)\)

\(\overrightarrow{OQ}=\left( \begin{matrix} 1 \\ 3 \\ 2 \\ \end{matrix} \right)-\frac{2}{41}\left( \begin{matrix} 3 \\ 1 \\ 6 \\ \end{matrix} \right)\)
\(\left( \begin{matrix} \frac{35}{41} \\ \frac{121}{41} \\ \frac{70}{41} \\ \end{matrix} \right)=\frac{1}{41}\left( \begin{matrix} 35 \\ 121 \\ 70 \\ \end{matrix} \right)\)

\(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\)
\( =\frac{1}{41}\left( \begin{matrix} 35 \\ 121 \\ 70 \\ \end{matrix} \right)-\frac{1}{41}\left( \begin{matrix} -31 \\ 31 \\ 118 \\ \end{matrix} \right)\)
\(=\frac{1}{41}\left( \begin{matrix} 66 \\ 90 \\ -48 \\ \end{matrix} \right)\)

\(\left| \overrightarrow{PQ} \right|\)
\( =\sqrt{{{\left( \frac{1}{41} \right)}^{2}}\left( {{66}^{2}}+{{90}^{2}}+{{48}^{2}} \right)}\)
\( =\sqrt{\frac{14760}{{{41}^{2}}}}=\sqrt{\frac{360}{41}}\)
\(=\sqrt{\frac{36\times 10}{41}}\)
\(=\frac{6\sqrt{10}}{\sqrt{41}}\cdot \frac{\sqrt{41}}{\sqrt{41}}\)

\( =\frac{6}{41}\sqrt{410}\) units