Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 46



Work Step by Step

Therefore, $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \\ & =\sqrt{2} \end{align}$ Therefore, $\frac{\mathbf{v}}{\parallel \mathbf{v}\parallel }=\frac{1}{\sqrt{2}}\mathbf{i}-\frac{1}{\sqrt{2}}\mathbf{j}$
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