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: 6


$\lt -16,29\gt$ and $\sqrt {1097}$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $-2u +5v=-2\lt 3,-2 \gt +5 \lt -2,5 \gt =\lt -16,29 \gt$ and $|\lt -16,29\gt|=\sqrt{(-16)^2+(29)^2}=\sqrt {1097}$ Hence, our final answers are: $\lt -16,29\gt$ and $\sqrt {1097}$
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