Question
Find a unit vector perpendicular to the vectors $\hat{ j }+2 \widehat{ k }$ and $\hat{ i }+\hat{ j }$.

Answer

$\text { Let } \overline{ a }=\hat{ j }+2 \widehat{ k }$ and $ \overline{ b }=\hat{ i }+\hat{ j }$
Then $\overline{ a } \times \overline{ b }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \widehat{ k } \\ 0 & 1 & 2 \\ 1 & 1 & 0\end{array}\right|$
$=(0-2) \hat{ i }-(0-2) \hat{ j }+(0-1) \widehat{ k }$
$=-2 \hat{ i }+2 \hat{ j }-\widehat{ k }$
$\therefore|\overline{ a } \times \overline{ b }|=\sqrt{(-2)^2+2^2+(-1)^2}$
$=\sqrt{4+4+1}$
$=\sqrt{9}$
$=3$
$\therefore$ Unit vector perpendicular to both $\overline{ a }$ and $\overline{ b }$
$= \pm \frac{\overline{ a } \times \overline{ b }}{|\overline{ a } \times \overline{ b }|}= \pm\left(\frac{-2 \hat{ i }+2 \hat{ j }-\widehat{ k }}{3}\right)$
$= \pm\left(\frac{2}{3} \hat{ i }-\frac{2}{3} \hat{ j }+\frac{1}{3} \widehat{ k }\right)$

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