Answer
See the explanation below.
Work Step by Step
Let vector $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$.
The projection of $\mathbf{v}$ onto $\mathbf{i}$ is expressed as follows:
$\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\frac{\mathbf{v}\centerdot \mathbf{i}}{{{\left\| \mathbf{i} \right\|}^{2}}}\mathbf{i}$ (1)
Substituting the value of $\mathbf{v}$ in equation (1) for $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}$, we get
$\begin{align}
& \text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\frac{\left( a\mathbf{i}+b\mathbf{j} \right)\centerdot \mathbf{i}}{{{\left\| \mathbf{i} \right\|}^{2}}}\mathbf{i} \\
& =\frac{\left( a\mathbf{i}+b\mathbf{j} \right)\centerdot \left( \mathbf{i}+0\mathbf{j} \right)}{{{\left( \sqrt{{{1}^{2}}+{{0}^{2}}} \right)}^{2}}}\left( \mathbf{i}+0\mathbf{j} \right) \\
& =\frac{a\left( 1 \right)+b\left( 0 \right)}{1}\mathbf{i}
\end{align}$
That is,
$\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=a\mathbf{i}$ (2)
Now, calculate $\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}$ to get
$\begin{align}
& \left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}=\left( \left( a\mathbf{i}+b\mathbf{j} \right)\centerdot \mathbf{i} \right)\mathbf{i} \\
& =\left( a\left( 1 \right)+b\left( 0 \right) \right)\mathbf{i}
\end{align}$
That is,
$\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}=a\mathbf{i}$ (3)
Comparing equations (2) and (3), we get
$\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}$
Hence, it is proved that $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}$.
