Answer
θ=116°23'
Work Step by Step
u·v=2·1-2·2-1·2=-4
|u|=$\sqrt (2^2+(-2)^2+(-1)^2)$=3
|v|=$\sqrt (1^2+2^2+2^2)$=3
cosθ=$\frac{-4}{3·3}$=$\frac{-4}{9}$
θ=116°23'
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 9 - Section 9.4 - Vectors in Three Dimensions - 9.4 Exercises - Page 658: 33
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