Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 799: 90



Work Step by Step

Here, $\mathbf{v}=\mathbf{i}-5\mathbf{j}$ So, $\begin{align} & -2\mathbf{v}=-2\left( \mathbf{i}-5\mathbf{j} \right) \\ & =-2\mathbf{i}+10\mathbf{j} \end{align}$ The magnitude of $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$ is given by $\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. So, the magnitude of $-2\mathbf{v}=-2\mathbf{i}+10\mathbf{j}$ is calculated as below: $\begin{align} & \left\| -2\mathbf{v} \right\|=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 10 \right)}^{2}}} \\ & =\sqrt{4+100} \\ & =\sqrt{104} \\ & =2\sqrt{26} \end{align}$ Hence, the scalar $\left\| -2\mathbf{v} \right\|=2\sqrt{26}$.
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