# Question #a42a8

Mar 18, 2017

Let

$\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$
and
$\vec{b} = 3 \hat{i} + \hat{j} - 2 \hat{k}$

The cross product of these two vectors ($\vec{a} \mathmr{and} \vec{b}$) is a vector $\vec{c}$ perpendicular to both as shown in the figure below

So
$\vec{c} = \vec{a} \times \vec{b} = \left[\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ \hat{i} & - 2 \hat{j} & \hat{k} \\ 3 \hat{i} & \hat{j} & - 2 \hat{k}\end{matrix}\right]$

$= \left(\left(- 2\right) \cdot \left(- 2\right) - 1 \cdot 1\right) \hat{i} + \left(1 \cdot 3 - 1 \cdot \left(- 2\right)\right) \hat{j} + \left(1 \cdot 1 - 3 \cdot \left(- 2\right)\right) \hat{k}$

$= 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$

So unit vector of $\vec{c} = \frac{\vec{c}}{\left\mid \vec{c} \right\mid}$

$= \frac{3 \hat{i} + 5 \hat{j} + 7 \hat{k}}{\sqrt{{3}^{2} + {5}^{2} + {7}^{2}}}$

$= \frac{3 \hat{i} + 5 \hat{j} + 7 \hat{k}}{\sqrt{83}} \textcolor{red}{\to \text{option (2)}}$