# How do you normalize <-3,0,1>?

Jan 24, 2017

See explanation.

#### Explanation:

To normalize a vector you have to divide all vector's coordinates by the vector's norm.

## $| | v | | = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

Here we have:

$| | v | | = \sqrt{{\left(- 3\right)}^{2} + {0}^{2} + {1}^{2}} = \sqrt{9 + 1} = \sqrt{10}$

So the normalized vector is:

${v}_{1} = < - \frac{3}{\sqrt{10}} , 0 , \frac{1}{\sqrt{10}} >$

After rationalizing the denomionators you get:

${v}_{1} = < - \frac{3 \sqrt{10}}{10} , 0 , \frac{\sqrt{10}}{10} >$