Answer
The vectors $\mathbf{v}$ and $\mathbf{w}$ neither parallel nor orthogonal.
Work Step by Step
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,
$\begin{align}
& \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\
& ={{\cos }^{-1}}\left( \frac{\left( 3\mathbf{i}-5\mathbf{j} \right)\cdot \left( 6\mathbf{i}+10\mathbf{j} \right)}{\left( \sqrt{{{3}^{2}}+{{\left( -5 \right)}^{2}}} \right)\left( \sqrt{{{6}^{2}}\mathbf{+}{{10}^{2}}} \right)} \right) \\
& ={{\cos }^{-1}}\left( \frac{3\cdot 6+\left( -5 \right)\cdot 10}{\left( \sqrt{34} \right)\left( \sqrt{136} \right)} \right) \\
& ={{\cos }^{-1}}\left( \frac{18-50}{\sqrt{4624}} \right)
\end{align}$
Solve ahead to get the result as,
$\begin{align}
& \theta ={{\cos }^{-1}}\left( \frac{18-50}{\sqrt{4624}} \right) \\
& ={{\cos }^{-1}}\left( \frac{-32}{68} \right) \\
& ={{\cos }^{-1}}\left( -0.47 \right) \\
& ={{118}^{{}^\circ }}
\end{align}$
Since, the angle between $\mathbf{v}$ and $\mathbf{w}$ is ${{118}^{{}^\circ }}$, $\mathbf{v}$ and $\mathbf{w}$ are neither parallel nor orthogonal vectors.
Hence, $\mathbf{v}$ and $\mathbf{w}$ are neither parallel nor orthogonal.
