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: 7


$\lt \dfrac{1}{5},\dfrac{14}{5}\gt$ and $\dfrac{\sqrt{197}}{5}$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $\dfrac{3}{5}u +\dfrac{4}{5}v=\dfrac{3}{5}\lt 3,-2 \gt +\dfrac{4}{5} \lt -2,5 \gt =\lt \dfrac{1}{5},\dfrac{14}{5} \gt$ and $|\lt \dfrac{1}{5},\dfrac{14}{5}\gt|=\sqrt{(\dfrac{1}{5})^2+(\dfrac{14}{5})^2}=\dfrac{\sqrt{197}}{5}$ Hence, our final answers are: $\lt \dfrac{1}{5},\dfrac{14}{5}\gt$ and $\dfrac{\sqrt{197}}{5}$
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