Question
The shortest distance between the lines $\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{- 3}$ and $\frac{x + 2}{2} = \frac{y - 6}{4} = \frac{z - 5}{- 5}$ is:
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Solution
Sol. [Key Concept: Shortest distance between skew lines $= \frac{|(a_{2} - a_{1}) \cdot (d_{1} \times \ d_{2})|}{{|d}_{1} \times \ d_{2}|}$ with the given points and direction vectors.]
SD $= \frac{\left| \begin{matrix} 4 - ( - 2) & 3 - 6 & 2 - 5 \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{matrix} \right|}{\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{matrix} \right|}$
$$\begin{matrix} & SD = \frac{|6( - 10 + 12) - ( - 3)( - 5 + 6) - 3(4 - 4)|}{|2\widehat{i} - \widehat{j}|} \\ & SD = \left| \frac{12 + 3}{\sqrt{5}} \right| \\ & \ = \frac{15}{\sqrt{5}} \\ & SD = 3\sqrt{5} \end{matrix}$$
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