University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 11 - Practice Exercises - Page 638: 5

Answer

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

Work Step by Step

Let the components of a vector $v$ be given as $v=\lt v_x,v_y\gt$ Now, $v_x=-1 \cos (\dfrac{\pi}{6})=\dfrac{-\sqrt 3}{2}$ and $v_y=-1 \sin (\dfrac{\pi}{6})=\dfrac{-1}{2}$ Hence, $v=\lt \dfrac{-\sqrt 3}{2}, \dfrac{-1}{2}\gt$
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.