Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Practice Exercises - Page 734: 3


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

Work Step by Step

The magnitude of a vector can be calculated as: $|v|=\sqrt{v_1^2+v_2^2}$ Now,$-2u=-2\lt -3,4 \gt =\lt 6,-8 \gt$ and $|\lt 6,-8\gt|=\sqrt{(6)^2+(-8)^2}=\sqrt {100}=10$
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