QJEEM20Ma9492: JEE Main 2020 Mathematics - Class 12 JEEM20Ma9492 Solution: The shortest distance between the lines and is :

Question

Verified Solution • 2025–26 Session moderate

The shortest distance between the lines

{{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}

and

{{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}

is :

Accepted Answer

\overrightarrow a

= < 3, 8, 3 >

\overrightarrow b

= < – 3, – 7, 6 >

\overrightarrow p

= < 3, – 1, 1 >

\overrightarrow q

= < –3, 2, 4 >

\vec{p} \times \vec{q} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix} |

= < -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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