## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 43

#### Answer

$\frac{3}{\sqrt{13}}\mathbf{i}-\frac{2}{\sqrt{13}}\mathbf{j}$

#### Work Step by Step

Therefore, \begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{3}^{2}}+{{\left( -2 \right)}^{2}}} \\ & =\sqrt{13} \end{align} So, $\frac{\mathbf{v}}{\parallel \mathbf{v}\parallel }=\frac{3}{\sqrt{13}}\mathbf{i}-\frac{2}{\sqrt{13}}\mathbf{j}$

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