QJEEAdv08Ma14a5: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ma14a5 Solution: The shortest distance between and is :

The shortest distance between L $_1$ and L $_2$ is

${{(\overrightarrow {{a_2}} - \overrightarrow {{a_1}} )(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} )} \over {|\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} |}} = (\overrightarrow {{a_2}} - \overrightarrow {{a_1}} )\,.\,\widehat n$

Where, ${a_1} = - \widehat i - 2\widehat j - \widehat k$

${a_2} = 2\widehat i - 2\widehat j + 3\widehat k$

$\therefore$ $\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = 3\widehat i + 4\widehat k$

$\therefore$ $({\widehat a_2} - {\widehat a_1})\,.\,\widehat n$

$= (3\widehat i + 4\widehat k)\,.\,\left( {{{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}} \right)$

$= {{ - 3 + 20} \over {5\sqrt 3 }} = {{17} \over {5\sqrt 3 }}$