Answer
$\mathbf{v}\cdot \mathbf{w}=-32\text{ and }\theta \approx 124.8{}^\circ $.
Work Step by Step
Dot product of vectors:
For two given vectors $\mathbf{v}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ and $\mathbf{w}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$
$\mathbf{v}\cdot \mathbf{w}={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}$ and
$\cos \theta =\frac{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|}{\mathbf{v}\cdot \mathbf{w}}$
where $\theta $ is the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$.
Here, ${{a}_{1}}=2,{{a}_{2}}=6,{{b}_{1}}=4,{{b}_{2}}=-11$
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{2}^{2}}+{{4}^{2}}} \\
& =\sqrt{4+16} \\
& =\sqrt{20}
\end{align}$
$\begin{align}
& \left\| \mathbf{w} \right\|=\sqrt{{{6}^{2}}+{{\left( -11 \right)}^{2}}} \\
& =\sqrt{36+121} \\
& =\sqrt{157}
\end{align}$
So,
$\begin{align}
& \mathbf{v}\cdot \mathbf{w}=2\left( 6 \right)+4\left( -11 \right) \\
& =12-44 \\
& =-32
\end{align}$
And also,
$\begin{align}
& \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|} \\
& =\frac{-32}{\sqrt{20}\cdot \sqrt{157}} \\
& =\frac{-32}{\sqrt{3140}}
\end{align}$
Using a calculator we will get
$\begin{align}
& \cos \theta =\frac{-32}{\sqrt{3140}} \\
& \theta ={{\cos }^{-1}}\left( \frac{-32}{\sqrt{3140}} \right) \\
& \theta \approx 124.8{}^\circ
\end{align}$
Hence, $\mathbf{v}\cdot \mathbf{w}=-32$ and the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 124.8{}^\circ $.
