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: 28

Answer

The possible values are $\theta = \frac{\pi }{6}$ or $\theta = \frac{{5\pi }}{6}$.

Work Step by Step

From Theorem 1 the length of ${\bf{e}} \times {\bf{f}}$ is $||{\bf{e}} \times {\bf{f}}|| = ||{\bf{e}}||||{\bf{f}}||\sin \theta $, where $\theta$ is the angle between ${\bf{e}}$ and ${\bf{f}}$. Since ${\bf{e}}$ and ${\bf{f}}$ are unit vectors, and $||{\bf{e}} \times {\bf{f}}|| = \frac{1}{2}$, it follows that $||{\bf{e}} \times {\bf{f}}|| = ||{\bf{e}}||||{\bf{f}}||\sin \theta = \frac{1}{2}$ $\sin \theta = \frac{1}{2}$ By Theorem 1 the angle between ${\bf{e}}$ and ${\bf{f}}$ is $0 \le \theta \le \pi $. So, the possible values are $\theta = \frac{\pi }{6}$ or $\theta = \frac{{5\pi }}{6}$.
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