## Precalculus (6th Edition) Blitzer

The vectors $\mathbf{v}$ and $\mathbf{w}$ parallel.
Let the angle between $\mathbf{v}$ and $\mathbf{w}$ be $\theta$ such that the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$ can be obtained using the formula $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right)$ as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -2\mathbf{i}+3\mathbf{j} \right)\cdot \left( -6\mathbf{i}+9\mathbf{j} \right)}{\left( \sqrt{{{\left( -2 \right)}^{2}}+{{3}^{2}}} \right)\left( \sqrt{{{\left( -6 \right)}^{2}}\mathbf{+}{{\left( 9 \right)}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -2 \right)\cdot \left( -6 \right)+3\cdot 9}{\left( \sqrt{13} \right)\left( \sqrt{117} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{12+27}{\sqrt{1521}} \right) \end{align} Solve ahead to get the result as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{12+27}{\sqrt{1521}} \right) \\ & ={{\cos }^{-1}}\left( \frac{39}{39} \right) \\ & ={{\cos }^{-1}}\left( 1 \right) \\ & ={{0}^{{}^\circ }} \end{align} Since, the angle between $\mathbf{v}$ and $\mathbf{w}$ is ${{0}^{{}^\circ }}$, $\mathbf{v}$ and $\mathbf{w}$ are parallel vectors. Hence, $\mathbf{v}$ and $\mathbf{w}$ are parallel vectors.