Answer
The $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ is $2\mathbf{i}+\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=2\mathbf{i}+\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=0\mathbf{i}+0\mathbf{j}$.
Work Step by Step
The projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ can be obtained as,
$\begin{align}
& \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v=}\frac{\mathbf{v}\cdot \mathbf{w}}{{{\left| \mathbf{w} \right|}^{2}}}\mathbf{w} \\
& =\frac{\left( 2\mathbf{i}+1\mathbf{j} \right)\cdot \left( 6\mathbf{i}+3\mathbf{j} \right)}{{{\left( \sqrt{{{6}^{2}}+{{3}^{2}}} \right)}^{2}}}\left( 6\mathbf{i}+3\mathbf{j} \right) \\
& =\frac{2\cdot 6+1\cdot 3}{45}\left( 6\mathbf{i}+3\mathbf{j} \right) \\
& =\frac{12+3}{45}\left( 6\mathbf{i}+3\mathbf{j} \right)
\end{align}$
Solve ahead to get the result as,
$\begin{align}
& \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=\frac{12+3}{45}\left( 6\mathbf{i}+3\mathbf{j} \right) \\
& =\frac{15}{45}\left( 6\mathbf{i}+3\mathbf{j} \right) \\
& =\frac{1}{3}\left( 6\mathbf{i}+3\mathbf{j} \right) \\
& =2\mathbf{i}+\mathbf{j}
\end{align}$
Now, obtain ${{\mathbf{v}}_{\mathbf{1}}}$ such that ${{\mathbf{v}}_{\mathbf{1}}}$ is parallel to $\mathbf{w}$ as,
$\begin{align}
& {{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v} \\
& =2\mathbf{i}+\mathbf{j}
\end{align}$
Obtain ${{\mathbf{v}}_{\mathbf{2}}}$ such that ${{\mathbf{v}}_{\mathbf{2}}}$ is orthogonal to $\mathbf{w}$ as,
$\begin{align}
& {{\mathbf{v}}_{\mathbf{2}}}=\mathbf{v}-{{\mathbf{v}}_{\mathbf{1}}} \\
& =\left( 2\mathbf{i}+\mathbf{j} \right)-\left( 2\mathbf{i}+\mathbf{j} \right) \\
& =2\mathbf{i}+\mathbf{j}-2\mathbf{i}-\mathbf{j} \\
& =0\mathbf{i}+0\mathbf{j}
\end{align}$
Hence, the projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ is $2\mathbf{i}+\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=2\mathbf{i}+\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=0\mathbf{i}+0\mathbf{j}$.
