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 - Chapter Review - Review Exercises - Page 655: 34

Answer

$\sqrt 5+5\approx 7.24$

Work Step by Step

Let us consider two vectors $v=pi+qj$ and $w=xi+yj$; then we have: $v \pm w=(p \pm x)i+(q \pm y)j$ The magnitude of any vector (let us say $v$) can be determined using the formula $||v||=\sqrt{p^2+q^2} $ Therefore, $||v||=\sqrt{(-2)^2+(1)^2} \\ = \sqrt {4+1} \\ = \sqrt 5$ Also, $||v||=\sqrt{(4)^2+(-3)^2} \\ = \sqrt {16+9} \\ = \sqrt {25}\\=5$ Now, $||v||+||w||=\sqrt 5+5$
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