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: 47


$\dfrac{3}{7}i-\dfrac{6}{7}j -\dfrac{2}{7} k$

Work Step by Step

The magnitude of any vector (let us say $v$) can be determined using the formula $||v||=\sqrt{p^2+q^2+r^2} $ and the unit vector $u$ in the same direction as $v$ can be found as: $u=\dfrac{v}{||v||}$ Therefore, the unit vector $u$ in the same direction as $v$ is: $u=\dfrac{3i-6j-2k}{\sqrt {(3)^2+(-6)^2+(-2)^2}} =\dfrac{3i-6j-2k}{\sqrt {9+36+4}} = \dfrac{3i-6j-2k}{7}=\dfrac{3}{7}i-\dfrac{6}{7}j -\dfrac{2}{7} k$
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