# Chapter 8 - 8.3 - Vectors in the Plane - 8.3 Exercises - Page 586: 43

$\lt \dfrac{- \sqrt 2}{2}, \dfrac{\sqrt 2}{2} \gt$

#### 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.

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