dot product can be defined like this.
The dot product a · b is equal to the signed length of the projection of b onto any line parallel to a, multiplied by the length of a.
a.b = (length of projection of b on a) * |a|
How is projection related with dot product ?. And any proof for this formula.
On
The projection $P_a(b)$ of $b$ on $a$ is in general defined by
$$ \|P_a(b)-b\| = \min_\lambda \| \lambda a - b \|. $$
Now if you have a scalar product on your vector space, you can square the norms and get for the right side
$$ \|\lambda-b\|^2 = \lambda^2\|a\|^2 - 2 \lambda <a,b> + \|b\|^2. $$
which is optimal for
$$ 2 \lambda \|a\|^2 = 2 <a,b> $$
or $$\lambda = \frac{<a,b>}{\|a\|^2}. $$
Hence, we can calculate the projection then by
$$ P_a(b) = \lambda a = <\frac{a}{\|a\|}, b> \frac{a}{\|a\|}. $$
So, if you have a scalar product you can easily calculate the projection, which should be your desired relation.
Now to your formula: The length of the projection is
$$ \|P_a(b)\| = |\lambda|\|a\|= \left|\frac{<a, b>}{\|a\|^2}\right| \|a\| $$
which is
$$ \|P_a(b)\|\|a\| = |<a, b>|. $$
The orthogonal projection of $b$ onto the line generated by $a$ is a vector $ra$. such that $b-ra$ is orthogonal to $a$. Then $0=(b-ra)\cdot a=b\cdot a-r(a\cdot a)$.
Therefore $a\cdot b=r\|a\|^2$. Since the projection is $ra$, its length is $\|ra\|=|r|\|a\|$.
So, $b\cdot a=r\|a\|\|a\|=sign(r)\|ra\|\|a\|$.
In your formula I guess you need to interpret "legnth of the projection" as "signed length of the projection", which is the $r\|a\|$.
For example, if $a=(1,0)$, then the projections of $b_1=(1,1)$ and $b_2=(-1,1)$ onto the line generated by $a$, which is the $x$-axis, have the same length. But $a\cdot b_1=1$, while $a\cdot b_2=-1$.