anchor.diff:  [stem]
   ++++
   \begin{align*}
  -\vec{a}×\vec{b}&=\frac{1}{{\left|\vec{v_1}×\vec{v_2}\right|}^{2}}(cos\frac{θ_1}{2}sin\frac{θ_2}{2}((\vec{v_1}×\vec{v_2})×(\vec{v_2}×(\vec{v_1}×\vec{v_2}))) - sin\frac{θ_1}{2}cos\frac{θ_2}{2}((\vec{v_1}×(\vec{v_1}×\vec{v_2}))×(\vec{v_1}×\vec{v_2}) - sin\frac{θ_1}{2}sin\frac{θ_2}{2}((\vec{v_1}×(\vec{v_1}×\vec{v_2}))×(\vec{v_2}×(\vec{v_1}×\vec{v_2}))))\\
  +\vec{a}×\vec{b}&=\frac{1}{{\left|\vec{v_1}×\vec{v_2}\right|}^{2}}(cos\frac{θ_1}{2}sin\frac{θ_2}{2}((\vec{v_1}×\vec{v_2})×(\vec{v_2}×(\vec{v_1}×\vec{v_2})))\\
  +- sin\frac{θ_1}{2}cos\frac{θ_2}{2}((\vec{v_1}×(\vec{v_1}×\vec{v_2}))×(\vec{v_1}×\vec{v_2})\\
  +- sin\frac{θ_1}{2}sin\frac{θ_2}{2}((\vec{v_1}×(\vec{v_1}×\vec{v_2}))×(\vec{v_2}×(\vec{v_1}×\vec{v_2}))))\\
   &=\frac{1}{{\left|\vec{v_1}×\vec{v_2}\right|}^{2}}(cos\frac{θ_1}{2}sin\frac{θ_2}{2}{\left|\vec{v_1}×\vec{v_2}\right|}^{2}\vec{v_2} + sin\frac{θ_1}{2}cos\frac{θ_2}{2}{\left|\vec{v_1}×\vec{v_2}\right|}^{2}\vec{v_1} - sin\frac{θ_1}{2}sin\frac{θ_2}{2}{\left|\vec{v_1}×\vec{v_2}\right|}^{2}(\vec{v_1}×\vec{v_2}))\\
   &=cos\frac{θ_1}{2}sin\frac{θ_2}{2}\vec{v_2} + sin\frac{θ_1}{2}cos\frac{θ_2}{2}\vec{v_1} - sin\frac{θ_1}{2}sin\frac{θ_2}{2}(\vec{v_1}×\vec{v_2})
   \end{align*}
  
positive.code: 
  Now we can rewrite our flip composition rule in a more specific form:
  
  *_Any 3D rotation equivalent to rotating angle stem:[θ] along axis stem:[\vec{v}] can be represented as a sequence of 2 flips along axis stem:[\vec{a}] and stem:[\vec{b}], such that stem:[\vec{a}·\vec{v}=0], stem:[\vec{b}·\vec{v}=0] and the angle from stem:[\vec{a}] to stem:[\vec{b}]: stem:[<\vec{a},\vec{b}>=\frac{θ}{2}]._*
  
  This representation means if we fully understand flip, which is easier to visualize, we can fully understand rotation and quaternions, since any quaternion can be broken down to flips.
  
  
negative.code: 
  We slightly touched quaternion multiplication, and we proved an important equation stem:[{q_1^{-1}}{q_2^{-1}}=({q_2}{q_1})^{-1}].
  
  === Appendix: Derive Quaternion Multiplication
  
  This part is not very important for understanding quaternion. It is a bit calculation heavy and is more for fun. Feel free to skip.
  
  We can actually derive the general quaternion multiplication from this special flip break down. That is if we define flip multiplication stem:[{q_a}{q_b}=(\vec{a},0)(\vec{b},0)=(\vec{a}×\vec{b},-\vec{a}·\vec{b})] directly, we can proof what general quaternion multiplication stem:[{q_1}{q_2}=(sin⁡\frac{θ_1}{2}\vec{v_1},cos⁡\frac{θ_1}{2})(sin⁡\frac{θ_2}{2}\vec{v_2},cos⁡\frac{θ_2}{2})] would look like.
  