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: 15


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

Work Step by Step

The vector $\lt 0,1 \gt $ is equivalent to rotating the unit vector $<1,0>$ $90^{\circ}$ counterclockwise about the origin. So, the unit vector makes an angle with the positive x-axis of $120^{\circ} + 90^{\circ}=210^{\circ}$ Therefore, $\lt 1*\cos (210^{\circ}), 1*\sin (210^{\circ}) \gt=\lt \dfrac{-\sqrt 3}{2},\dfrac{-1}{2} \gt$
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