## Shortest distance - Perpendicularity

How can do we show that the shortest distance between two sets is always the vector that is perpendicular to both sets?

I remembered it went through inner products, but it's not working for me.

So, if i had
<g-f,g-f>= <g,g> - 2 <g,f> + <f,f>
where g,f are column vectors of $\mathbb{R}^n$ with n being finite.
I got stuck at that part.
I also rewrote it as
<g-f,g-f>= <g,g>-2*||g||*||f|*cos( $\alpha$|)+<f,f>

But I'm missing something really important.

And out of curiosity, is this true for all vectors? Or is it only true for column vectors?