University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 47

Answer

$\dfrac{\pi}{4}$

Work Step by Step

The formula to calculate the angle between two planes is: $ \theta = \cos ^{-1} (\dfrac{p \cdot q}{|p||q|})$ Here, $p=\lt 1,1,0\gt$ and $q=\lt 2,-1,-2 \gt$ $|p|=\sqrt{1^2+1^2+0^2}=\sqrt 2$ and $|q|=\sqrt{2^2+1^2+(-2)^2}=\sqrt 9=3$ Thus, $ \theta = \cos ^{-1} (\dfrac{p \cdot q}{|p||q|})=\cos ^{-1} (\dfrac{3}{\sqrt 2 (3)})=\cos ^{-1} (\dfrac{\sqrt 2}{2})$ Hence, $ \theta =\dfrac{\pi}{4}$
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