Answer
The magnitude is 8.25 and the direction angle is $284.0{}^\circ $.
Work Step by Step
If $\mathbf{v=}a\mathbf{i}-b\mathbf{j}$ is a vector then its magnitude is $\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Use this definition of norm to get,
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -8 \right)}^{2}}} \\
& =\sqrt{4+64} \\
& =\sqrt{68} \\
& =8.25
\end{align}$
Hence, the magnitude of the vector $v$ is $8.25$.
If $\mathbf{v=}a\mathbf{i}-b\mathbf{j}$ is a vector then its direction angle $\theta $ is $\theta ={{\tan }^{-1}}\left( \frac{b}{a} \right)$. So,
$\begin{align}
& \theta ={{\tan }^{-1}}\left( \frac{-8}{2} \right) \\
& ={{\tan }^{-1}}\left( -4 \right) \\
& =284.0{}^\circ
\end{align}$.
Hence, the angle is $284.0{}^\circ $.
