Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - Chapter Review Exercises - Page 702: 38

Answer

$ \sqrt {3}$

Work Step by Step

We calculate the magnitude of the cross product as follows: $||\textbf{v}\times\textbf{w}||=||\textbf{v}||\,||\textbf{w}||\sin \theta$ $||\textbf{v}||=2$ and $\theta=\frac{\pi}{6}$ $||\textbf{w}||=\frac{\textbf{v}\cdot\textbf{w}}{||\textbf{v}||\cos\theta}=\frac{3}{2\times\frac{\sqrt 3}{2}}=\sqrt 3$ Then, we have $||\textbf{v}\times\textbf{w}||=2\times\sqrt {3}\times\sin\frac{\pi}{6}=2\times\sqrt {3}\times\frac{1}{2}= \sqrt {3}$
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