Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - 13.4 The Cross Product - Exercises - Page 677: 24



Work Step by Step

By using the right-hand rule for the cross product of the vectors, we see that the direction of $\textbf{v}\times\textbf{w}$ is along the negative y-axis as the vectors are in the xz-plane. So, the x-component and z-component are both 0 and the magnitude of $\textbf{v}\times\textbf{w}$ is nothing but the component along the negative y-axis. $||\textbf{v}\times\textbf{w}||=||\textbf{v}||\,||\textbf{w}||\,\sin\theta=3\times3\times\sin\frac{\pi}{6}=\frac{9}{2}$ Component along y-axis =$ -\frac{9}{2}$ Therefore, the required vector is $\textbf{v}\times\textbf{w}=⟨0, -\frac{9}{2},0⟩$
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