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: 1


$\lt -17,32 \gt$ and $\sqrt {1313}$

Work Step by Step

The magnitude of a vector can be calculated as: $|v|=\sqrt{v_1^2+v_2^2}$ Now, $3u -4v=3\lt -3,4 \gt -4 \lt 2,-5 \gt =\lt -17,32 \gt$ and $|\lt -17,32\gt|=\sqrt{(-17)^2+(32)^2}=\sqrt {1313}$
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