Answer
$\mathbf{v}\cdot \mathbf{w}=2\text{ and }\theta \approx 86.1{}^\circ $.
Work Step by Step
Dot product of vectors:
For the 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, ${{a}_{1}}=2,{{a}_{2}}=7,{{b}_{1}}=3,{{b}_{2}}=-4$
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{2}^{2}}+{{3}^{2}}} \\
& =\sqrt{4+9} \\
& =\sqrt{13}
\end{align}$
$\begin{align}
& \left\| \mathbf{w} \right\|=\sqrt{{{7}^{2}}+{{\left( -4 \right)}^{2}}} \\
& =\sqrt{49+16} \\
& =\sqrt{65}
\end{align}$
So,
$\begin{align}
& \mathbf{v}\cdot \mathbf{w}=2\left( 7 \right)+3\left( -4 \right) \\
& =14-12 \\
& =2
\end{align}$
And also,
$\begin{align}
& \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|} \\
& =\frac{2}{\sqrt{13}\cdot \sqrt{65}} \\
& =\frac{2}{\sqrt{845}}
\end{align}$
Using a calculator we get
$\begin{align}
& \cos \theta =\frac{2}{\sqrt{845}} \\
& \theta ={{\cos }^{-1}}\left( \frac{2}{\sqrt{845}} \right) \\
& \theta \approx 86.1{}^\circ
\end{align}$
Hence, $\mathbf{v}\cdot \mathbf{w}=2$ and the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 86.1{}^\circ $.
