Comments on: I’m stumped– you give it a try http://www.tangentspace.net/cz/archives/2008/04/im-stumped/ somewhere near the beginning. Tue, 14 Oct 2008 10:44:38 +0000 http://wordpress.org/?v=2.5 By: Alex http://www.tangentspace.net/cz/archives/2008/04/im-stumped/#comment-303378 Alex Wed, 02 Apr 2008 22:00:07 +0000 http://www.tangentspace.net/cz/?p=890#comment-303378 Thanks for the idea, George. Thanks for the idea, George.

]]> By: George http://www.tangentspace.net/cz/archives/2008/04/im-stumped/#comment-303188 George Wed, 02 Apr 2008 19:14:31 +0000 http://www.tangentspace.net/cz/?p=890#comment-303188 For the first of these, it may help to notice this: If (x,u)(y,v)+(z,u)(w,v)=(a,u)(b,v), then, for all u,v, ((a,u)b, v) = ((x,u)y + (z,u)w, v) and ((b,v)a, u) = ((y,v)x + (w,v)z, u). Therefore, (a,u)b = (x,u)y + (z,u)w and (b,v)a = (y,v)x + (w,v)z. If a and b are nonzero, then the first of these two shows that if u is perpendicular to both x and z, then it's also perpendicular to a. So a=Ax+Bz. Similarly, b=Cy+Dw. It shouldn't be too bad from here. For the first of these, it may help to notice this:

If (x,u)(y,v)+(z,u)(w,v)=(a,u)(b,v), then, for all u,v,

((a,u)b, v) = ((x,u)y + (z,u)w, v)

and

((b,v)a, u) = ((y,v)x + (w,v)z, u).

Therefore,

(a,u)b = (x,u)y + (z,u)w

and

(b,v)a = (y,v)x + (w,v)z.

If a and b are nonzero, then the first of these two shows that if u is perpendicular to both x and z, then it’s also perpendicular to a. So a=Ax+Bz. Similarly, b=Cy+Dw. It shouldn’t be too bad from here.

]]>