Answer
The magnitude is 6 and the direction angle is $90{}^\circ $.
Work Step by Step
First, simplify the provided vector to obtain,
$\begin{align}
& \mathbf{v}=\left( 4\mathbf{i}-2\mathbf{j} \right)-\left( 4\mathbf{i}-8\mathbf{j} \right) \\
& =\left( 4-4 \right)\mathbf{i}+\left( -2-\left( -8 \right) \right)\mathbf{j} \\
& =\left( -2+8 \right)\mathbf{j} \\
& =6\mathbf{j}
\end{align}$
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,
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 6 \right)}^{2}}} \\
& =\sqrt{0+36} \\
& =\sqrt{36} \\
& =6
\end{align}$
Hence, the magnitude of the vector $\mathbf{v}$ is $6$.
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,
$\begin{align}
& \theta ={{\tan }^{-1}}\left( \frac{6}{0} \right) \\
& ={{\tan }^{-1}}\left( \infty \right) \\
& =90{}^\circ \\
\end{align}$.
Hence, the angle is $90{}^\circ $.
