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: 4


$\lt 10,-25 \gt$ and $5 \sqrt {29}$

Work Step by Step

The magnitude of a vector can be calculated as: $|v|=\sqrt{v_1^2+v_2^2}$ Now, $5v=5\lt 2,-5 \gt =\lt 10,-25 \gt$ and $|\lt 10,-25\gt|=\sqrt{(10)^2+(-25)^2}=\sqrt {725}=5 \sqrt {29}$
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