Answer
The dot product of the two vectors $v\cdot w$ is $-11$ and the angle between the two vectors is ${{170}^{\circ }}$.
Work Step by Step
Considered the vectors,
$v=-2i+j$ And $w=4i-3j$
Now, use dot product formula $v\cdot w={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}$:
$\begin{align}
& v\cdot w=\left( -2 \right)\left( 4 \right)+\left( 1 \right)\left( -3 \right) \\
& =-8-3 \\
& =-11
\end{align}$
Thus, the dot product of the two vectors $v\cdot w$ is $-11$.
Now, use the angle between two vectors formula $\theta ={{\cos }^{-1}}\left( \frac{v\cdot w}{\left\| v \right\|\left\| w \right\|} \right)$:
$\theta ={{\cos }^{-1}}\left( \frac{-11}{\left\| v \right\|\left\| w \right\|} \right)$
Now, use magnitude of vectors formula, $\left\| v \right\|=\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}$:
$\begin{align}
& \left\| v \right\|=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
& =\sqrt{4+1} \\
& =\sqrt{5}
\end{align}$ And $\begin{align}
& \left\| w \right\|=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
& =\sqrt{16+9} \\
& =\sqrt{25} \\
& =5
\end{align}$
Now, put in the formula:
$\begin{align}
& \theta ={{\cos }^{-1}}\left( \frac{-11}{\left( \sqrt{5} \right)5} \right) \\
& ={{\cos }^{-1}}\left( \frac{-11}{\left( 2.236 \right)5} \right) \\
& ={{\cos }^{-1}}\left( 0.9839 \right) \\
& \approx {{170}^{\circ }}
\end{align}$
