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: 7

Answer

See below:

Work Step by Step

The magnitude of the linear combination of the vector is calculated from, $\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$ Here, a is the horizontal component of $\mathbf{v}$ and b is the vertical component of $\mathbf{v}$. The next step is: $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{1}^{2}}+{{(-1)}^{2}}} \\ & =\sqrt{1+1} \\ & =\sqrt{2} \end{align}$ Thus, the magnitude of the vector is $\sqrt{2}$.
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