Answer
The dot product of two vectors is $\mathbf{a}\cdot \mathbf{b}=\left\| a \right\|\text{ }\left\| b \right\|\cos \theta $.
Work Step by Step
Example: Information:
Let vector $\mathbf{a}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, such that the angle between their directions $\theta ={{0}^{\circ }}$.
Now, magnitude of vector a:
$\begin{align}
& \left\| \mathbf{a} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
& =\sqrt{4+1+9} \\
& =\sqrt{14}
\end{align}$
Therefore, $\left\| \mathbf{a} \right\|=\sqrt{14}$
Similarly, magnitude of vector b:
$\begin{align}
& \left\| \mathbf{b} \right\|=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
& =\sqrt{1+4+1} \\
& =\sqrt{6}
\end{align}$
Therefore, $\left\| \mathbf{b} \right\|=\sqrt{6}$
Now, the cosine of the angle between their directions is $\cos \left( {{0}^{\circ }} \right)=1$
Hence, the dot product of the two vectors is
$\begin{align}
& \mathbf{a}\cdot \mathbf{b}=\left( \sqrt{14} \right)\left( \sqrt{6} \right)\left( 1 \right) \\
& =9.165
\end{align}$
