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 8.4 Vectors - 8.4 Assess Your Understanding - Page 626: 35

Answer

$$5$$

Work Step by Step

Let us consider that a vector $v$ is given by: $v=pi+qj$ The magnitude of a vector can be determined using the formula $||v||=\sqrt{p^2+q^2} (1)$ We will use the formula (1) to obtain: $||v||=\sqrt{(3)^2+(-4)^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\=5$
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