Answer
The angle between two vectors $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right)$.
Work Step by Step
Take $\mathbf{v}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$, $\mathbf{w}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, and $\mathbf{v}\cdot \mathbf{w}=9.165$.
Now, find the magnitude of vector v as follows:
$\begin{align}
& \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
& =\sqrt{4+1+9} \\
& =\sqrt{14}
\end{align}$
Therefore, $\left\| \mathbf{v} \right\|=\sqrt{14}$.
Similarly, find the magnitude of the vector $w$ as follows:
$\begin{align}
& \left\| \mathbf{w} \right\|=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
& =\sqrt{1+4+1} \\
& =\sqrt{6}
\end{align}$
Therefore, $\left\| \mathbf{w} \right\|=\sqrt{6}$.
Next, find the cosine of the angle between their directions as follows:
$\begin{align}
& \cos \theta =\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right) \\
& =\left( \frac{9.165}{\sqrt{14}\sqrt{6}} \right) \\
& =\left( \frac{9.165}{9.165} \right) \\
& =1
\end{align}$
So, $\cos \theta =1$.
Therefore,
$\begin{align}
& \theta ={{\cos }^{-1}}\left( 1 \right) \\
& ={{0}^{\circ }}
\end{align}$
Hence, the angle between two vectors is ${{0}^{\circ }}$.
