The shortest distance between the lines x-32=y-23=z-1-1 and x+32=y-61=z-53 is:

Question

The shortest distance between the lines $\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-1}{-1}$ and $\dfrac{x+3}{2}=\dfrac{y-6}{1}=\dfrac{z-5}{3}$ is:

Accepted Answer

$${L_1}:{{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$

$${L_2}:{{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$

Now, $$\overrightarrow p  \times \overrightarrow q  = \left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   2 & 3 & { - 1}  \
   2 & 1 & 3  \

\end{pmatrix} } \right| = 10\widehat i - 8\widehat j - 4\widehat k$$

and $${\overrightarrow a _2} - {\overrightarrow a _1} = 6\widehat i - 4\widehat j - 4\widehat k$$

$$\therefore$$ $$S.D. = \left| {{{60 + 32 + 16} \over {\sqrt {100 + 64 + 16} }}} \right| = {{108} \over {\sqrt {180} }} = {{18} \over {\sqrt 5 }}$$

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