Answer
$\lt \dfrac{- \sqrt 2}{2}, \dfrac{\sqrt 2}{2} \gt $
Work Step by Step
$\overrightarrow{v} = < -2, 2>$
To find the unit vector, we must first find the magnitude:
$\sqrt {((-2)^{2} + (2)^{2}} = 2 \sqrt 2$
Now, $\dfrac{v}{|v|}=\dfrac{< -2, 2>}{2 \sqrt 2}=\lt \dfrac{- \sqrt 2}{2}, \dfrac{\sqrt 2}{2} \gt $
To verify that it is a unit vector, we find the magnitude:
$\sqrt {(\dfrac{- \sqrt 2}{2})^2+(\dfrac{\sqrt 2}{2})^2}= 1$
Since the magnitude equals 1, it is a unit vector.
