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: 8


$\lt -3,\dfrac{70}{13}\gt$ and $\dfrac{\sqrt{6421}}{13}$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $-\dfrac{5}{13}u +\dfrac{12}{13}v=-\dfrac{5}{13}\lt 3,-2 \gt +\dfrac{12}{13} \lt -2,5 \gt =\lt -3,\dfrac{70}{13} \gt$ and $|\lt -3,\dfrac{70}{13}\gt|=\sqrt{(-3)^2+(\dfrac{70}{13})^2}=\dfrac{\sqrt{6421}}{13}$ Hence, our final answers are: $\lt -3,\dfrac{70}{13}\gt$ and $\dfrac{\sqrt{6421}}{13}$
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