Answer
See the explanation below.
Work Step by Step
The vector is, $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$.
The magnitude of the vector $\mathbf{v}$ is given by:
$\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
The unit vector $\frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}$ is given by:
$\begin{align}
& \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}=\frac{a\mathbf{i}+b\mathbf{j}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \\
& =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{i}+\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{j}
\end{align}$
The dot product of unit vector $\left( \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} \right)$ with $\mathbf{v}$ is,
$\begin{align}
& \mathbf{v}\cdot \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}=\left( a\mathbf{i}+b\mathbf{j} \right)\cdot \left( \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{i}+\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{j} \right) \\
& =\frac{{{a}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{i}\cdot \mathbf{i}+\frac{{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\mathbf{j}\cdot \mathbf{j} \\
& =\frac{{{a}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}+\frac{{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \\
& =\frac{{{a}^{2}}+{{b}^{2}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}
\end{align}$
Thus, the dot product of unit vector $\left( \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} \right)$ with $\mathbf{v}$ is,
$\mathbf{v}\cdot \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}=\sqrt{{{a}^{2}}+{{b}^{2}}}$
The dot product of unit vector $\left( \frac{\mathbf{v}}{\left\| \mathbf{v} \right\|} \right)$ with $\mathbf{v}$ gives the magnitude of the vector $\mathbf{v}$.
That means the $\frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}$ is the unit vector in the direction of $\mathbf{v}$.
