Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 646: 38


$2\sqrt {11}$

Work Step by Step

Let us consider that vector $v$ is given by: $v=pi+qj+rz$ The magnitude of a vector can be determined using the formula $||v||=\sqrt{p^2+q^2+r^2} ...(1)$ We will use the formula (1) to obtain: $||v||=\sqrt{(6)^2+(2)^2+(-2)^2}\\ =\sqrt{36+4+4}\\ =\sqrt{44} \\=2\sqrt {11}$
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