# Chapter 8 - 8.3 - Vectors in the Plane - 8.3 Exercises - Page 586: 66

$\lt 0, 4 \sqrt 3 \gt$

#### Work Step by Step

We are given that $||\overrightarrow{v}|| = 4 \sqrt 3$ and that: $\theta = 90^{\circ}$ We know that $v=||v|| \lt \cos \theta , \sin \theta \gt$ Thus, $v=4 \sqrt 3 \lt \cos 90^{\circ} , \sin 90^{\circ} \gt =4 \sqrt 3 \lt 0, 1 \gt = \lt 0, 4 \sqrt 3 \gt$

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