Answer
${{160.30}^{{}^\circ }}$
Work Step by Step
Let the angle between $\mathbf{v}$ and $\mathbf{w}$ be $\theta $ such that the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$ can be obtained using the formula $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right)$ as,
$\begin{align}
& \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\
& ={{\cos }^{-1}}\left( \frac{\left( -3\mathbf{i}+2\mathbf{j} \right)\cdot \left( 4\mathbf{i}-\mathbf{j} \right)}{\left( \sqrt{{{\left( -3 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \right)\left( \sqrt{{{4}^{2}}\mathbf{+}{{\left( -1 \right)}^{2}}} \right)} \right) \\
& ={{\cos }^{-1}}\left( \frac{\left( -3 \right)\cdot 4+2\cdot \left( -1 \right)}{\left( \sqrt{13} \right)\left( \sqrt{17} \right)} \right) \\
& ={{\cos }^{-1}}\left( \frac{-14}{\sqrt{221}} \right)
\end{align}$
Solve ahead to get the result as,
$\begin{align}
& \theta ={{\cos }^{-1}}\left( \frac{-14}{\sqrt{221}} \right) \\
& ={{160.30}^{{}^\circ }}
\end{align}$
Hence, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta ={{160.30}^{{}^\circ }}$.
