# How do you find the unit vector in the direction of the given vector u=<-2,2>?

##### 1 Answer
Jan 19, 2017

$= - \frac{i}{\sqrt{2}} + \frac{j}{\sqrt{2}} \mathmr{and} < - \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} >$

#### Explanation:

$| u | = \sqrt{{\left(- 2\right)}^{2} + {2}^{2}}$
$| u | = \sqrt{4 + 4}$
$| u | = \sqrt{8}$
$| u | = \sqrt{4 \cdot 2}$
$| u | = 2 \sqrt{2}$

Unit Vector $= \frac{1}{|} u | \left(- 2 i + 2 j\right)$
$= \frac{1}{2 \sqrt{2}} \left(- 2 i + 2 j\right)$

$= \frac{1}{\cancel{2} \sqrt{2}} \cdot \cancel{2} \left(- i + j\right)$

$= \frac{1}{\sqrt{2}} \left(- i + j\right)$

$= - \frac{i}{\sqrt{2}} + \frac{j}{\sqrt{2}}$