Answer
(a) See graph.
(b) $8$, $150^{\circ}$
Work Step by Step
We are given $\vec u=\langle -4\sqrt 3, 4 \rangle$.
(a) See graph.
(b) The length of the vector is $|\vec u|=\sqrt {(-4\sqrt 3)^2+(4)^2}=8$ and $tan\theta=|\frac{4}{-4\sqrt 3}|=\frac{\sqrt 3}{3}$ which gives $\theta=30^{\circ}$, thus the direction angle $t=180-30=150^{\circ}$
