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 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 47



Work Step by Step

The angle between two plane is given by: $ \theta = \cos ^{-1} (\dfrac{a \cdot b}{|a||b|})$ Here, $a=\lt 1,1,0\gt$ and $b=\lt 2,-1,-2 \gt$ $|a|=\sqrt{1^2+1^2+0^2}=\sqrt{1+1}=\sqrt 2$ and $|b|=\sqrt{2^2+1^2+(-2)^2}=\sqrt 9=3$ Now, $ \theta =\cos ^{-1} (\dfrac{3}{\sqrt 2 (3)})$ or, $=\cos ^{-1} (\dfrac{\sqrt 2}{2})$ or, $=\dfrac{\pi}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.