Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 8 - 8.3 - Vectors in the Plane - 8.3 Exercises - Page 586: 48


$ \lt \dfrac{-9}{4}, \dfrac{-15}{16} \gt $

Work Step by Step

Since $v$ has the same direction as $u$ and since it is a multiple of a unit vector $n$, we can write it as: $\overrightarrow{v} = < -12 n, -5n>$ To find the unit vector, we must first find the magnitude: $\sqrt {((-12n)^{2} + (-5n)^{2}} = 3$ or, $144 \ n^2+25 \ n^2 =9$ This implies that $n^2=\dfrac{9}{169} \implies n=\dfrac{3}{13}$ Now, $\overrightarrow{v} = < -12 (\dfrac{3}{13}), -5 (\dfrac{3}{13})> = \lt \dfrac{-9}{4}, \dfrac{-15}{16} \gt $
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