## Calculus: Early Transcendentals 8th Edition

$\frac{1}{\sqrt{10}} <3,-1>$
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, $\frac{\overline a}{\|\overline a\|}$: $\frac{\overline a}{\|\overline a\|} = \frac{1}{\|\overline a\|}\times\overline a$ $=\frac{1}{2\sqrt{10}} <6,-2>$ $=\frac{1}{\sqrt{10}} <3,-1>$