Question

Sol. [Key Concept: Shortest distance $= \frac{|(a_{1} - a_{2}) \cdot (b_{1} \times \ b_{2})|}{{|b}_{1} \times \ b_{2}|}$, using the lines' points $a_{i}$ and directions $b_{i}$.]

$${\ {\overrightarrow{a}}_{1} = \frac{1}{3}\widehat{i} + 2\widehat{j} + \frac{8}{3}\widehat{k} }{{\overrightarrow{a}}_{2} = \frac{- 2}{3}\widehat{i} - \frac{1}{3}\widehat{k} }{{\overrightarrow{b}}_{1} = 2\widehat{i} - 5\widehat{j} + 6\widehat{k},\ {\overrightarrow{b}}_{2} = \widehat{j} - \widehat{k} }{{\overrightarrow{b}}_{1} \times {\overrightarrow{b}}_{2} = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 2 & - 5 & 6 \\ 0 & 1 & - 1 \end{matrix} \right| = - \widehat{i} + 2\widehat{j} + 2\widehat{k} }{{\overrightarrow{a}}_{1} - {\overrightarrow{a}}_{2} = \widehat{i} + 2\widehat{j} + 3\widehat{k} }{S.D = \left| \frac{\left( {\overrightarrow{a}}_{1} - {\overrightarrow{a}}_{2} \right) \cdot \left( {\overrightarrow{b}}_{1} \times {\overrightarrow{b}}_{2} \right)}{\left| {\overrightarrow{b}}_{1} \times {\overrightarrow{b}}_{2} \right|} \right| }{S.D = \left| \frac{(\widehat{i} + 2\widehat{j} + 3\widehat{k}) \cdot ( - \widehat{i} + 2\widehat{j} + 2\widehat{k})}{3} \right| }{= \frac{9}{3} = 3 }$$