Linear Algebgra and Vectors

Ladies and Gentlemen, some FUN stuff.

Seriously. Just a simple proof. Is is possible to prove that if axb = axc and a.b = a.c, then b = c where a, b and c are non-parallel vectors?

If not, could I get a counter example please? Any help would be much appreciated.

Cheers
~read~

Well, there's certainly a problem with the wording, as a vector is technically parallel to itself, and so b !|| c implies b != c. Plus, taking a=0 gives infinite counter examples....


Aside from that, assuming we're in R3;
Note case where b=0. Here axb=axc => c=0 (again, assuming a!=0). Disregard.

axb=axc => b,c coplanar.
axb=axc => |b|sin(Angle ab) = |c|sin(Angle ac)
(use axb =|a||b|sin(Angle)n)
a.b=a.c => |b|cos(Angle ab) = |c|cos(Angle ac)
(use a.b = |a||b|cos(Angle)).

Note that the, in general, sin(t) != cos(t), and so this system of equations is true arbitrarily iff |b|=|c|.
We then cancel to get sin(Angle ab) = sin(Angle ac) and cos(Angle ab) = cos(Angle ac) which together imply Angle ab = Angle ac
Thus b and c are equal in magnitude and direction [combine coplanar, angle with a], thus are equal.

Ahh...

...good point. ^_^'

I think I meant to say "non-zero" vectors, NOT non-parallel. ^_^' Silly me...

But thank you, thank you! I had it all written out in R3 with components, and that's pretty odd to try and prove it. your method and proof is much cleaner.

Cheers
~read~