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: 49

Answer

$\mathbf{v}=-6\sqrt{2}\mathbf{i}-6\sqrt{2}\mathbf{j}$

Work Step by Step

Therefore, $\begin{align} & \mathbf{v}=\parallel v\parallel \cos {{225}^{\circ }}\mathbf{i}+\parallel v\parallel \sin {{225}^{\circ }}\mathbf{j} \\ & =\left[ 12\times \left( -\frac{1}{\sqrt{2}} \right) \right]\mathbf{i}+\left[ 12\times \left( -\frac{1}{\sqrt{2}} \right) \right]\mathbf{j} \\ & =-6\sqrt{2}\mathbf{i}-6\sqrt{2}\mathbf{j} \end{align}$
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