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: 1


$\lt9, -6\gt$ and $3 \sqrt {13}$

Work Step by Step

Formula to find the magnitude of a vector can be found as: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $3u=3\lt 3, -2 \gt =\lt9, -6\gt$ and $|\lt9, -6\gt|=\sqrt{9^2+(-6)^2}=\sqrt {117}=3 \sqrt {13}$ Hence, our final answers are: $\lt9, -6\gt$ and $3 \sqrt {13}$
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