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: 48

Answer

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

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: $u=\dfrac{-6i+12j+4k}{\sqrt {(-6)^2+(12)^2+(4)^2}} =\dfrac{-6i+12j+4k}{\sqrt {36+144+16}} \\=-\dfrac{3}{7}i +\dfrac{6}{7}j +\dfrac{2}{7} k$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.