vectors cross and dot?

Question

how do you prove ||axb||^2=||a||^2||b||^2-||a.b||^2

Answer 1

||a×b||² = (||a||*||b||*sin θ) = ||a||²*||b||²*sin² θ
= ||a||²*||b||²*(1-cos² θ) = ||a||²*||b||² - ||a||²*||b||²*cos² θ
=||a||²||b||² - (||a||*||b||*cos θ)² = ||a||²||b||² - ||a·b||²

Q.E.D.

Answer 2

sorry, it is impossible to read your full question,
may be you could ammend it