# How do you find a unit vector that is perpendicular to both -2i + j + 3k and i - j - 2k?

##### 1 Answer
Jun 27, 2016

$\hat{e} = \left\{\frac{1}{\sqrt{3}} \hat{i} , - \frac{1}{\sqrt{3}} \hat{j} , \frac{1}{\sqrt{3}} \hat{k}\right\}$

#### Explanation:

Given two vectors $\vec{u}$ and $\vec{v}$, if they are linearly indepedend we can a unit vector perpendicular to both as follows.

$\hat{e} = \frac{\vec{u} \times \vec{v}}{\left\lVert \vec{u} \times \vec{v} \right\rVert} = - \frac{\vec{v} \times \vec{u}}{\left\lVert \vec{u} \times \vec{v} \right\rVert}$

In the present case

$\left\{- 2 \hat{i} , 1 \hat{j} , 3 \hat{k}\right\} \times \left\{1 \hat{i} , - 1 \hat{j} , - 2 \hat{k}\right\} = \left\{1 \hat{i} , - 1 \hat{j} , 1 \hat{k}\right\}$

then

$\hat{e} = \left\{\frac{1}{\sqrt{3}} \hat{i} , - \frac{1}{\sqrt{3}} \hat{j} , \frac{1}{\sqrt{3}} \hat{k}\right\}$