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: 23



Work Step by Step

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