Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.4 Vectors - 9.4 Assess Your Understanding - Page 607: 64



Work Step by Step

If a vector $v=ai+bj$ makes an angle of $\alpha$ with the positive x-axis and has magnitude $||v||=\sqrt{a^2+b^2}$, then $a=||v||cos\alpha$ and $v=||v||sin\alpha$. Hence here $a=||15||cos(315^\circ)=\frac{15\sqrt2}{2}$ and $v=||15||sin(315^\circ)=-\frac{15\sqrt2}{2}$. Therefore $v=\frac{15\sqrt2}{2}i-\frac{15\sqrt2}{2}j.$
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