# How do you normalize <2,0,-1>?

Sep 9, 2017

$\hat{u} = < \frac{2 \sqrt{5}}{5} , 0 , \frac{- \sqrt{5}}{5} >$

#### Explanation:

To normalize a vector is to find unit vector (vector with magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

$\hat{u} = \frac{\vec{v}}{| \vec{v} |}$

Given $\vec{v} = < 2 , 0 , - 1 >$, we can calculate the magnitude of the vector:

$\left\mid \vec{v} \right\mid = \sqrt{{\left({v}_{x}\right)}^{2} + {\left({v}_{y}\right)}^{2} + {\left({v}_{z}\right)}^{2}}$

$\implies = \sqrt{{\left(2\right)}^{2} + {\left(0\right)}^{2} + {\left(- 1\right)}^{2}}$

$\implies = \sqrt{4 + 0 + 1}$

$\implies = \sqrt{5}$

We now have:

$\hat{u} = \frac{< 2 , 0 , - 1 >}{\sqrt{5}}$

$\implies \hat{u} = < \frac{2}{\sqrt{5}} , 0 , - \frac{1}{\sqrt{5}} >$

We can also rationalize the denominator on the $\hat{x}$ $\left(\hat{i}\right)$ and $\hat{z}$ $\left(\hat{k}\right)$ component:

$\implies \hat{u} = < \frac{2 \sqrt{5}}{5} , 0 , \frac{- \sqrt{5}}{5} >$

Hope that helps!