University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Practice Exercises - Page 638: 3


$\lt 6,-8 \gt$ and $10$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $-2u=-2\lt -3,4 \gt =\lt 6,-8 \gt$ and $|\lt 6,-8\gt|=\sqrt{(6)^2+(-8)^2}=\sqrt {100}=10$ Hence, our final answers are: $\lt 6,-8 \gt$ and $10$
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