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

Dec 24, 2016

$\hat{u} = < - \frac{\sqrt{6}}{6} , - \frac{\sqrt{6}}{6} , \frac{\sqrt{6}}{3} >$. See below.

#### Explanation:

To normalize a vector is to find a unit vector in the same direction as the given vector, where a unit vector, $\hat{u}$, is a vector with a magnitude of $1$. In other words, the normalized vector $\hat{u}$ of a non-zero vector u is a unit vector in the direction of u.

We can find a unit vector in the same direction as the given vector by dividing the given vector by its magnitude.

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

Where the magnitude of a vector is given by:

$| u | = \sqrt{{\left({u}_{x}\right)}^{2} + {\left({u}_{y}\right)}^{2} + {\left({u}_{z}\right)}^{2}}$

Given the vector, u $= < - 1 , - 1 , 2 >$,

$| u | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 1\right)}^{2} + {\left(2\right)}^{2}}$

$| u | = \sqrt{1 + 1 + 4}$

$| u | = \sqrt{6}$

We then have

$\hat{u} = \frac{< - 1 , - 1 , 2 >}{\sqrt{6}}$

$\implies \hat{u} = < - \frac{1}{\sqrt{6}} , - \frac{1}{\sqrt{6}} , \frac{2}{\sqrt{6}} >$

Or equivalently, by rationalizing the denominator:

$\implies \hat{u} = < - \frac{\sqrt{6}}{6} , - \frac{\sqrt{6}}{6} , \frac{\sqrt{6}}{3} >$