# How do you normalize  (3i – 4j + 4k) ?

Jul 1, 2017

The answer is $= \frac{3}{\sqrt{41}} i - \frac{4}{\sqrt{41}} j + \frac{4}{\sqrt{41}} k$

#### Explanation:

Let $\vec{v} = < 3 , - 4 , 4 >$

The magnitude of $\vec{v}$ is

$| | \vec{v} | | = | | < 3 , - 4 , 4 > | | = \sqrt{{3}^{2} + {\left(- 4\right)}^{2} + {4}^{2}} = \sqrt{9 + 16 + 16} = \sqrt{41}$

Therefore, the normalized vector is

$\hat{v} = \frac{\vec{v}}{| | \vec{v} | |} = \frac{1}{\sqrt{41}} \cdot < 3 , - 4 , 4 >$