University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.2 - Vectors - Exercises - Page 608: 14


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

Work Step by Step

Use formula $v=\lt |v| \cos \theta, |v| \sin \theta \gt$ Here, $|v|=1$ and $\theta=\dfrac{-3\pi}{4}$ Thus, $v=\lt (1) \cos (\dfrac{-3\pi}{4}), (1) \cos (\dfrac{-3\pi}{4}) \gt=\lt \dfrac{-1}{\sqrt 2},\dfrac{-1}{\sqrt 2} \gt$
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.