Answer
The magnitude is $18.03$ and the angle is $123.7{}^\circ $.
Work Step by Step
If $v=a\mathbf{i}+b\mathbf{j}$ is a vector, then its magnitude is $\left\| v \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
Use this definition of norm to get
$\begin{align}
& \left\| v \right\|=\sqrt{{{\left( -10 \right)}^{2}}+{{\left( 15 \right)}^{2}}} \\
& =\sqrt{100+225} \\
& =\sqrt{325} \\
& =18.03
\end{align}$
Hence, the magnitude of the vector $v$ is 18.03.
If $v=a\mathbf{i}+b\mathbf{j}$ is a vector, then its direction angle $\theta $ is $\theta ={{\tan }^{-1}}\left( \frac{b}{a} \right)$. So, we have
$\begin{align}
& \theta ={{\tan }^{-1}}\left( \frac{15}{-10} \right) \\
& =123.7{}^\circ
\end{align}$.
Hence, the angle is $123.7{}^\circ $.
