# How do you normalize  (-2i + -1j + 2k)?

Jun 18, 2018

The answer is $= < - \frac{2}{3} , - \frac{1}{3} , \frac{2}{3} >$

#### Explanation:

To normalize a vector $\vec{u}$, divide the vector by the modulus of the vector

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

Here,

$\vec{u} = < - 2 , - 1 , 2 >$

The modulus is

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

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

$= \sqrt{9}$

$= 3$

Therefore,

$\hat{u} = \frac{1}{3} \cdot < - 2 , - 1 , 2 > = < - \frac{2}{3} , - \frac{1}{3} , \frac{2}{3} >$