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: 16

Answer

$\approx 1.55$ or $\approx 88.88 ^{\circ}$

Work Step by Step

The angle between two planes can be determined as: $ \theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Now, $a=\lt 0,1,0.2 \gt$ and $b=\lt 1,0,0.1 \gt$ Here, $|a|=\sqrt{(0)^2+(1)^2+(0.2)^2}= \sqrt {1.04}$; $|b|=\sqrt{(1)^2+(0)^2+(0.1)^2}=\sqrt {1.01}$ So, $ \theta =\cos ^{-1} (\dfrac{0.2}{ ( \sqrt {1.04})(\sqrt {1.01}}) \approx 1.55$ or $\approx 88.88 ^{\circ}$
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