QJEEAdv08Ma213e: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ma213e Solution: The unit vector perpendicular to both and is :

Vector in the direction of ${L_1} = \overrightarrow {{n_1}} = 3\widehat i + \widehat j + 2\widehat k$

Vector in the direction of ${L_2} = \overrightarrow {{n_2}} = \widehat i + 2\widehat j + 3\widehat k$

$\therefore$ Vector perpendicular to both L $_1$ and L $_2$

$\overrightarrow {{n_1}} \times \overrightarrow {{n_2}}$

$$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 1& 2& 3\end{array}\right|$$

$= - \widehat i - 7\widehat j + 5\widehat k$

$\therefore$ Required unit vector

$= \widehat n = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {1 + 49 + 25} }} = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$