Answer
$16; 315^{\circ}$
Work Step by Step
The magnitude of a vector $\textbf{u}=\langle a,b \rangle$ is given as $|\textbf{u}|=\sqrt (a^{2}+b^{2})$. Since $\textbf{u}=\langle 8\sqrt 2,-8\sqrt 2 \rangle$, the magnitude is:
$|\textbf{u}|=\sqrt ((8\sqrt 2)^{2}+(-8\sqrt 2)^{2})=\sqrt (128+128)=\sqrt (256)=16$
The direction angle $\theta$ can be found through the equation $\tan\theta=\frac{b}{a}$. Substituting the values of $a$ and $b$ in the formula and solving using a calculator,
$\theta=\tan^{-1} (\frac{8\sqrt 2}{-8\sqrt 2})=\tan^{-1} (-1)=-45^{\circ}$
The vector has a positive horizontal component and a negative vertical component which places it in the fourth quadrant. Since the direction angle is supposed to be the positive angle between the x-axis and the position vector, we need to add $360^{\circ}$ to $-45^{\circ}$ to yield the direction angle $\theta$. Therefore, the direction angle $\theta=315^{\circ}$.