The shortest distance between the lines x-33=y-8-1=z-31 and x+3-3=y+72=z-64 is:

Question

The shortest distance between the lines $\dfrac{x-3}{3}=\dfrac{y-8}{-1}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-6}{4}$ is:

Accepted Answer

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

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