# Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 734: 6

$v=\lt \dfrac{\sqrt 3}{2}, \dfrac{1}{2}\gt$

#### Work Step by Step

The components of vector $v$ are: $v=\lt v_x,v_y\gt$ Thus, $v_x=(1) \cos (\dfrac{\pi}{6})=\dfrac{\sqrt 3}{2}$ and $v_y=(1) \sin (\dfrac{\pi}{6})=\dfrac{1}{2}$ Hence, $v=\lt \dfrac{\sqrt 3}{2}, \dfrac{1}{2}\gt$

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