Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 712: 14


$\dfrac{\pi}{2}$ or $90 ^{\circ}$

Work Step by Step

we have, $\overrightarrow{AC}=\lt 2,4 \gt$ and $\overrightarrow{DB}=\lt -4,2 \gt$ The angle between two planes can be determined as: $ \theta = \cos ^{-1} (\dfrac{\overrightarrow{AC} \cdot \overrightarrow{DB}}{|\overrightarrow{AC}||\overrightarrow{DB}|})$ Now, $|\overrightarrow{AC}|=\sqrt{(2)^2+(4)^2}= \sqrt {20}$ and $|\overrightarrow{DB}|=\sqrt{(-4)^2+(2)^2}=\sqrt {20}$ Thus, $ \theta =\cos ^{-1} (\dfrac{(2)(-4)+(4)(2)}{ ( \sqrt {20})(\sqrt {20})})$ or, $\cos ^{-1} (0)=\dfrac{\pi}{2}$ or $90 ^{\circ}$
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