Answer
$${\text{16;}}\,{315^ \circ }$$
Work Step by Step
$$\eqalign{
& \left\langle {8\sqrt 2 , - 8\sqrt 2 } \right\rangle \cr
& {\text{The magnitude of the vector is }} \cr
& \left| {\left\langle {8\sqrt 2 , - 8\sqrt 2 } \right\rangle } \right| = \sqrt {{{\left( {8\sqrt 2 } \right)}^2} + {{\left( { - 8\sqrt 2 } \right)}^2}} \cr
& \left| {\left\langle {8\sqrt 2 , - 8\sqrt 2 } \right\rangle } \right| = \sqrt {128 + 128} \cr
& \left| {\left\langle {8\sqrt 2 , - 8\sqrt 2 } \right\rangle } \right| = \sqrt {256} \cr
& \left| {\left\langle {8\sqrt 2 , - 8\sqrt 2 } \right\rangle } \right| = 16 \cr
& {\text{The direction of the vector is }} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{ - 8\sqrt 2 }}{{8\sqrt 2 }}} \right) + {360^ \circ } \cr
& \theta = {315^ \circ } \cr} $$