Answer
The vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)$ is $0\mathbf{i}+0\mathbf{j}$.
Work Step by Step
The vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)$ can be obtained as,
$\begin{align}
& \text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)\mathbf{=}\frac{\left( \mathbf{v}-\mathbf{w} \right)\cdot \mathbf{u}}{{{\left| \mathbf{u} \right|}^{2}}}\mathbf{u} \\
& =\frac{\left[ \left( 3\mathbf{i}-2\mathbf{j} \right)-\left( -5\mathbf{j} \right) \right]\cdot \left( -\mathbf{i}+\mathbf{j} \right)}{{{\left( \sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}} \right)}^{2}}}\left( -\mathbf{i}+\mathbf{j} \right) \\
& =\frac{\left( 3\mathbf{i}+3\mathbf{j} \right)\cdot \left( -\mathbf{i}+\mathbf{j} \right)}{\sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}}}\left( -\mathbf{i}+\mathbf{j} \right) \\
& =\frac{3\cdot \left( -1 \right)+3\cdot 1}{2}\left( -\mathbf{i}+\mathbf{j} \right)
\end{align}$
Solve ahead to get the result as,
$\begin{align}
& \text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)=\frac{-3+3}{2}\left( -\mathbf{i}+\mathbf{j} \right) \\
& =0\left( -\mathbf{i}+\mathbf{j} \right) \\
& =0\mathbf{i}+0\mathbf{j}
\end{align}$
Hence, the vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)$ is $0\mathbf{i}+0\mathbf{j}$.
