Show subspace Ua = span {a} is square on hyperplane H

Hi.

This is the second question of my assignment... I have previously found a hyperplane, H(a, x0) = (w + 2x - y + 3z - 13) = 0, given the point, x0 = (5,1,0,2), and the normal vector, a = (1,2,-1,3).

Now I need to show that the subspace Ua = span {a} is square on H, so that

"for all" y "belonging to" Ua "for all" x "belonging to" H : x - x0 is square on y.

Okay, so I know the last part was a little confusing, but I'm just really bad at using the math-tag function. And w, x, y, z - I have named u1, u2, u3, u4 on in my own calculations, but it would just be confussing with all those numbers in the hyperplane equation. :o)

Does anyone know how I should proceed?

Thanks in advance, Simon DK.