Answer
$\|\mathbf{v}\|=\sqrt{14}$
Work Step by Step
$\mathbf{v}=\mathbf{i}-2\mathbf{j}-3\mathbf{k}$ and so
$\|\mathbf{v}\|=\sqrt{1^2+(-2)^2+(-3)^2}=\sqrt{1+4+9}=\sqrt{14}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 11 - Vectors and the Geometry of Space - 11.2 Exercises - Page 764: 75
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