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: 2


$\lt -1,-1 \gt$ and $\sqrt {2}$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $u+v=\lt -3,4 \gt + \lt 2,-5 \gt =\lt -1,-1 \gt$ and $|\lt -1,-1\gt|=\sqrt{(-1)^2+(-1)^2}=\sqrt {2}$ Hence, our final answers are: $\lt -1,-1 \gt$ and $\sqrt {2}$
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