Answer
$8$ and $\theta =315^{\circ}$
Work Step by Step
$\overrightarrow{v} = -4 \sqrt 2 \ \widehat{i} +4 \sqrt 2 \ \widehat{j}$
Magnitude:
$||\overrightarrow{v}|| = \sqrt {(-4 \sqrt 2)^{2} + (4 \sqrt 2)^{2}} = 8$
The direction can now be found:
$\tan \ \theta = \dfrac{-4 \sqrt 2}{4 \sqrt 2} =-1 \implies \theta =315^{\circ}$
