Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.6 Vectors in Space - 9.6 Assess Your Understanding - Page 625: 38

Answer

$2\sqrt{11}.$

Work Step by Step

The magnitude $||v||$ of a vector $v=ai+bj+ck$, can be computed using the formula $||v||=\sqrt{a^2+b^2+c^2}$. Hence, using the formula above gives $||v||=\sqrt{6^2+2^2+(-2)^2}\\||v||=\sqrt{36+4+4}\\||v||=\sqrt{44}\\||v||=2\sqrt{11}.$
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