# How do you normalize  <3,-6,4>?

Mar 4, 2017

$u = < \frac{3}{\sqrt{61}} , - \frac{6}{\sqrt{61}} , \frac{4}{\sqrt{61}} >$

#### Explanation:

In normalizing the vector we are finding a unit vector (magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

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

Given $v = < 3 , - 6 , 4 >$, we can calculate the magnitude of the vector:

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

$| v | = \sqrt{{\left(3\right)}^{2} + {\left(- 6\right)}^{2} + {\left(4\right)}^{2}}$

$| v | = \sqrt{9 + 36 + 16}$

$| v | = \sqrt{61}$

We now have:

$u = \frac{< 3 , - 6 , 4 >}{\sqrt{61}}$

$\implies u = < \frac{3}{\sqrt{61}} , - \frac{6}{\sqrt{61}} , \frac{4}{\sqrt{61}} >$

Hope that helps!