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: 53

Answer

$ \frac{8\sqrt 5}{5}i+\frac{4\sqrt 5}{5}j$ or $ -\frac{8\sqrt 5}{5}i-\frac{4\sqrt 5}{5}j$

Work Step by Step

1. Let $\vec v=ai+bj$, we have $a=2b$, thus $\vec v=2bi+bj$ 2. Use its magnitude, we have $\sqrt {(2b)^2+b^2}=4$, thus $5b^2=16$ and $b=\pm\frac{4\sqrt 5}{5}$ 3. Thus we have he vector(s) $\vec v=\frac{8\sqrt 5}{5}i+\frac{4\sqrt 5}{5}j$ or $\vec v=-\frac{8\sqrt 5}{5}i-\frac{4\sqrt 5}{5}j$
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