Precalculus (6th Edition) Blitzer

Two vectors are said to be parallel vectors if they have the same direction or are in exactly opposite directions, such that the angle between them is ${{0}^{\circ }}$.
Consider vectors $\mathbf{v}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$, $\mathbf{w}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, and $\mathbf{v}\cdot \mathbf{w}=9.165$. 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} $\left\| \mathbf{v} \right\|=\sqrt{14}$ Similarly, find the magnitude of 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} $\left\| \mathbf{w} \right\|=\sqrt{6}$ Now, 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 that \begin{align} & \theta ={{\cos }^{-1}}\left( 1 \right) \\ & ={{0}^{\circ }} \end{align} Hence, the angle between two vectors is ${{0}^{\circ }}$.Therefore, it can be concluded that the vectors are parallel to each other.