Answer
$\mathbf{v}\cdot \mathbf{w}=1\text{ }and\text{ }\theta \approx 71.6{}^\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, ${{a}_{1}}=2,{{a}_{2}}=1,{{b}_{1}}=1,{{b}_{2}}=-1$
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{2}^{2}}+{{1}^{2}}} \\
& =\sqrt{4+1} \\
& =\sqrt{5}
\end{align}$
$\begin{align}
& \left\| \mathbf{w} \right\|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \\
& =\sqrt{1+1} \\
& =\sqrt{2}
\end{align}$
So,
$\begin{align}
& \mathbf{v}\cdot \mathbf{w}=2\left( 1 \right)+1\left( -1 \right) \\
& =2-1 \\
& =1
\end{align}$
And also,
$\begin{align}
& \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|} \\
& =\frac{1}{\sqrt{5}\cdot \sqrt{2}} \\
& =\frac{1}{\sqrt{10}}
\end{align}$
Use a calculator to get
$\begin{align}
& \cos \theta =\frac{1}{\sqrt{10}} \\
& \theta ={{\cos }^{-1}}\left( \frac{1}{\sqrt{10}} \right) \\
& \theta \approx 71.6{}^\circ
\end{align}$
Hence, $\mathbf{v}\cdot \mathbf{w}=1$ and the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 71.6{}^\circ $.
