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 12.2 - Vectors - Exercises 12.2 - Page 704: 13


$\lt \dfrac{-1}{2},\dfrac{-\sqrt 3}{2} \gt$

Work Step by Step

Since, we have $v=\lt |v| \cos \theta, |v| \sin \theta \gt$ to compute the components of vector $v$ Given: $|v|=1$ and $\theta=\dfrac{2\pi}{3}$ Thus, $v=\lt (1) \cos (\dfrac{2\pi}{3}), (1) \cos (\dfrac{2\pi}{3}) \gt$ or, $v=\lt \dfrac{-1}{2},\dfrac{-\sqrt 3}{2} \gt$
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