Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.1 Vectors in the Plane - 11.1 Exercises - Page 768: 35

Answer

$2\sqrt 2$

Work Step by Step

We are given: $u=(3,-4)$ $v=(1,1)$ $w=(-1,0)$ Determine $|-2v|$: $|-2v|=|-2(1,1)|=|(-2,-2)|=\sqrt{(-2)^2+(-2)^2}=\sqrt{8}=2\sqrt 2$
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