User:Nightkhaos

For Hibby
$$\arccos \left( \hat{a} \cdot \hat{b} \right)$$

where $$\arccos = \cos ^{-1}$$ NightKhaos (talk) 21:34, 30 March 2009 (UTC)

$$\sqrt{a_{1}^2 + a_{2}^2}\sqrt{b_{1}^2 + b_{2}^2}\sin{ \left[ \arccos{ \left( a_1 b_1 + a_2 b_2 \right)} \right]}$$ NightKhaos (talk) 21:58, 30 March 2009 (UTC)