Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.4 Vectors - 8.4 Assess Your Understanding - Page 627: 62


$\dfrac{15 \sqrt 2}{2} \ i - \dfrac{15 \sqrt 2}{2} \ j$

Work Step by Step

Let us consider that a vector $v$ is given by $v=pi+qj$ and makes an angle of $\alpha$ with the positive $x$-axis. The magnitude of the vector can be determined using the formula: $||v||=\sqrt{p^2+q^2} $ Then we find $p=||v||cos\alpha$ and $v=||v||sin\alpha$. Here, we have: $p=||15|| \cos \ (315^\circ)= \dfrac{15 \sqrt 2}{2}$ and $v=||15|| \sin \ (315^\circ)=\dfrac{-15 \sqrt 2}{2}$ Therefore $v= \dfrac{15 \sqrt 2}{2} \ i - \dfrac{15 \sqrt 2}{2} \ j$
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