Answer
$\qquad|U|=5 \sqrt{2} , 315^{\circ} = \theta$
Work Step by Step
\[
V=\langle 5,-5\rangle
\]
Here the $x$ component of the vector is $5=V_{x}$, and the $y$ component of the vector is
\[
-5=V_{y}
\]
The magnitude of the vector $\mathrm{V}$ is
\[
\sqrt{V_{y}^{2}+V_{x}^{2}}=|V|
\]
Replacing $V_{y}$ with $V_{x}$ values
\[
\begin{aligned}
&\sqrt{5^{2}+(-5)^{2}}=|V| \\
&=\sqrt{25+25} \\
&=\sqrt{50} \\
&=5 \sqrt{2}
\end{aligned}
\]
$\theta$ is the angle made with the positive $x-$ axis and the vector.
\[
\tan \theta=\frac{V_{x}}{V_{y}}
\]
\[
=\frac{-5}{5}
\]
$ \tan \theta=-1 $
$ \theta=\tan ^{-1}(-1) $
$ =315^{\circ} $