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: 2


$\lt 4,-10\gt$ and $2 \sqrt {29}$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $-2v=-2\lt -2,5 \gt =\lt 4,-10\gt$ and $|\lt 4,-10\gt|=\sqrt{4^2+(-10)^2}=\sqrt {116}=2 \sqrt {29}$ Hence, our final answers are: $\lt 4,-10\gt$ and $2 \sqrt {29}$
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