Answer
The magnitude is 3 and the direction angle is $180{}^\circ $.
Work Step by Step
Simplify the provided vector as:
$\begin{align}
& \mathbf{v=}\left( 7\mathbf{i}-3\mathbf{j} \right)-\left( 10\mathbf{i}-3\mathbf{j} \right) \\
& =\left( 7-10 \right)\mathbf{i}+\left( -3+3 \right)\mathbf{j} \\
& =-3\mathbf{i}+0\mathbf{j}
\end{align}$
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 then,
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 0 \right)}^{2}}} \\
& =\sqrt{9+0} \\
& =\sqrt{9} \\
& =3
\end{align}$
Hence, the magnitude of the vector $\mathbf{v}$ is $3$.
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,
$\begin{align}
& \theta ={{\tan }^{-1}}\left( \frac{0}{-3} \right) \\
& ={{\tan }^{-1}}\left( -0 \right) \\
& =180{}^\circ
\end{align}$.
Hence, the angle is $180{}^\circ $.
