# How do you normalize (2 i -3j -k)?

Jan 9, 2018

$\hat{u} = < \frac{2}{\sqrt{14}} , - \frac{3}{\sqrt{14}} , - \frac{1}{\sqrt{14}} >$

#### Explanation:

To normalize a vector is to find unit vector (vector with magnitude/length of one) in the same direction as the given vector. This can be accomplished by dividing the given vector by its magnitude.

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

Given $\vec{v} = < 2 , - 3 , - 1 >$, we can calculate the magnitude of the vector:

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

$\implies = \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2} + {\left(- 1\right)}^{2}}$

$\implies = \sqrt{4 + 9 + 1}$

$\implies = \sqrt{14}$

We now have:

$\hat{u} = \frac{< 2 , - 3 , - 1 >}{\sqrt{14}}$

$\implies \hat{u} = < \frac{2}{\sqrt{14}} , - \frac{3}{\sqrt{14}} , - \frac{1}{\sqrt{14}} >$

We can also rationalize the denominator for each of the components:

$\implies \hat{u} = < \frac{\sqrt{14}}{7} , - \frac{3 \sqrt{14}}{14} , - \frac{\sqrt{14}}{14} >$

Which may also be written $\left(\frac{\sqrt{14}}{7} i - \frac{3 \sqrt{14}}{14} j - \frac{\sqrt{14}}{14} k\right)$

Hope that helps!