Answer
$\sqrt 2, 45°$
Work Step by Step
$\vec{v} = \hat{i}+ \hat{j}$
$|\vec{v}|= \sqrt (1^{2} + 1^{2})$= $\sqrt 2$
tan θ= $\frac{1}{1}$.
⇒ θ= $tan^{-1}1$= 45°
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 9.1 - Vectors in Two Dimensions - 9.1 Exercises - Page 638: 52
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