Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.2 - Vectors - 12.2 Exercises - Page 805: 23


$\frac{1}{\sqrt{10}} <3,-1>$

Work Step by Step

Given $\overline a =<6,-2>$ Find the magnitude $\|\overline a\|$: $\|\overline a\| =\sqrt {(6^2+(-2)^2)} =\sqrt{40} = 2\sqrt{10}$ Plug vector and magnitude into unit vector formula, $\frac{\overline a}{\|\overline a\|}$: $\frac{\overline a}{\|\overline a\|} = \frac{1}{\|\overline a\|}\times\overline a$ $=\frac{1}{2\sqrt{10}} <6,-2>$ $=\frac{1}{\sqrt{10}} <3,-1>$
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