Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 666: 24



Work Step by Step

\begin{align*} \cos \theta &=\frac{\langle 5,\sqrt{3}\rangle \cdot \langle \sqrt{3},2\rangle}{\|\langle 5,\sqrt{3}\rangle\|\|\langle \sqrt{3},2\rangle\|}\\ &=\frac{5\sqrt{3}+2\sqrt{3}}{\sqrt{25+3}\sqrt{3+4}}\\ &=\frac{7\sqrt{3}}{\sqrt{28}\sqrt{7}}\\ &=\frac{\sqrt{3}}{2} \end{align*} That is $$\theta=\cos^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{6}.$$
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