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