Answer
$|u| = 4 $, $θ =120°$
Work Step by Step
Length of a vector is defined as:
$ u = ai + bj$
$|u| = \sqrt {a^2 + b^2}$
Direction is $\tan θ = \frac{b}{a}$
Given $u = -2i + 2\sqrt3 j$
$|u| = \sqrt {(-2)^2 + (2\sqrt3)^2} = \sqrt {16} = 4$
$\tan θ = \frac{2\sqrt3}{-2} = -\sqrt3$
Then, $θ = 120°$
