## 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( 3\mathbf{i}-5\mathbf{j} \right)\cdot \left( 6\mathbf{i}-10\mathbf{j} \right)}{\left( \sqrt{{{3}^{2}}+{{\left( -5 \right)}^{2}}} \right)\left( \sqrt{{{6}^{2}}\mathbf{+}{{\left( -10 \right)}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{3\cdot 6+\left( -5 \right)\cdot \left( -10 \right)}{\left( \sqrt{34} \right)\left( \sqrt{136} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{18+50}{\sqrt{4624}} \right) \end{align} Solve ahead to get the result as, \begin{align} & \theta ={{\cos }^{-1}}\left( \frac{18+50}{\sqrt{4624}} \right) \\ & ={{\cos }^{-1}}\left( \frac{68}{68} \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.