Answer
θ=48 °11'
Work Step by Step
u·v=1·4+2·0-2·(-3)=10
|u|=$\sqrt (1^2+2^2+(-2)^2$=3
|v|=$\sqrt (4^2+0+(-3)^2$=5
cos θ=$\frac{10}{3·5}$=$\frac{2}{3}$
θ=48 °11'
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: 36
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