How to go from $d= \left(\frac{A\cdot B}{B\cdot B} B,A \right)$ to $d=\frac{|A\cdot B|}{|B|}$?

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I'm reading Banchoff/Wermer's: Linear Algebra Through Geometry.

I understand that the distance from a point $A$ to the line through $B$ is:

$$d= \left(\frac{A\cdot B}{B\cdot B} B,A \right)$$

But in the book, the author writes it as:

$$d=\frac{|A\cdot B|}{|B|}$$

But I have no idea on how he went from the first formula to the second. I've tried to write the vectors with 2 components and expand but had no success.

I thought about the following:

$$\frac{A\cdot B}{B\cdot B} B=\frac{A\cdot B}{|B|^2} B=\frac{A\cdot B}{|B|} \frac{B}{|B|}=\frac{|A\cdot B|}{|B|} \left[ \frac{1}{|B|} B \right]$$

As $\displaystyle \left[ \frac{1}{|B|} B \right]$ is the versor of $B$, then It's norm is one. It seems that $\displaystyle \frac{A\cdot B}{|B|}$ is what gives the size of the vector because whenever we a versor, we can multiply it by $|B|$:

$$\displaystyle \left[ \frac{1}{|B|} B \right] |B|=B$$

And then $|B|$ is obviously the norm of $B$, but it seems that whenever we can write:

$$|X| \left[ \frac{1}{|B|} B \right]=B$$

Then $|X|$ is the norm of the vector. In this case, the norm of the vector is precisely:

$$\frac{|A\cdot B|}{|B|}$$