Answer
$${\text{8;}}\,{120^ \circ }$$
Work Step by Step
$$\eqalign{
& \left\langle { - 4,4\sqrt 3 } \right\rangle \cr
& {\text{The magnitude of the vector is }} \cr
& \left| {\left\langle { - 4,4\sqrt 3 } \right\rangle } \right| = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( {4\sqrt 3 } \right)}^2}} \cr
& \left| {\left\langle { - 4,4\sqrt 3 } \right\rangle } \right| = \sqrt {16 + 48} \cr
& \left| {\left\langle { - 4,4\sqrt 3 } \right\rangle } \right| = 8 \cr
& {\text{The direction of the vector is }} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{4\sqrt 3 }}{{ - 4}}} \right) + {180^ \circ } \cr
& \theta = {120^ \circ } \cr} $$
