Precalculus (6th Edition) Blitzer

The magnitude is 6 and the direction angle is $90{}^\circ$.
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 $90{}^\circ$.