Answer
Magnitude: $16$
Direction angle: $315^{\circ}$
Work Step by Step
The magnitude $|u|$ of a vector $u=\left\langle a,b\right\rangle$ is given by $|u|=\sqrt {a^2+b^2}$.
For the given vector
$u=\left\langle 8\sqrt 2,-8\sqrt 2\right\rangle \Rightarrow ~ a=8\sqrt 2, b=-8\sqrt 2.$ Therefore,
$|u|=|\left\langle 8\sqrt 2,-8\sqrt 2\right\rangle|=\sqrt {(8\sqrt 2)^2+(-8\sqrt 2)^2}=\sqrt {8^2(\sqrt {2})^2+8^2(\sqrt {2})^2}=\sqrt{64\times2+64\times2}=\sqrt{256}=16$
Hence, the magnitude of the vector $\left\langle 8\sqrt 2,-8\sqrt 2\right\rangle$ is $16$.
The direction angle $\theta$ of the vector $u=\left\langle a,b\right\rangle$ is given by
$\theta=\tan^{-1}\left(\frac{b}{a}\right)$
Therefore, for $u=\left\langle 8\sqrt 2,-8\sqrt 2\right\rangle $ we have
$\theta=\tan^{-1}\left(\frac{ 8\sqrt 2}{- 8\sqrt 2}\right)=-\tan^{-1}\left(1\right)=-45^{\circ}$ (in quadrant IV)
Adding $360^{\circ}$ will give the direction angle
$\theta=-45^{\circ}+360^{\circ}=315^{\circ}$ with positive $x$-axis.
Hence, the magnitude of $\left\langle 8\sqrt 2,-8\sqrt 2\right\rangle$ is $16$ and the direction angle is $315^{\circ}$.
