If we look at it in 2D, you'll get the idea:


                               _                  _
The green line from the tip of u perpendicular to v cuts off the
           _    _      _
projection p of u onto v.
                                    _                                _
We want the vector perpendicular to v that has its tip at the tip of u. 

That has to be the vector coinciding with the green line! Like this:


 
All I did was simply to put an arrowhead on that green line!
I made that green line that cut off the projection vector p into a vector!
I put an arrowhead on the tip of it at the tip of u!
                                            _   _
Now what is that green vector?  it's simply u - p.   See now?

So here's what we do:
                             _    _      _                              _
We first find the projection p of u onto v.  Then we subtract that from u
        _ _                    
1. Find u•v: <0,5,3>•<0,-2,-5> = 0-10-15 = -25
         _                               
2. Find ∥v∥²  = 0²+(-2)²+(-5)² = 29
          _ _     _
3. Divide u•v by ∥v∥² = 
                       _      
4. Multiply that by by v:   
  <-- that's p, the projection of u onto v.
             _   _
5. Subtract  u - p:

   
                                         _
That's the green vector perpendicular to v.

Better check my fraction math.

Edwin