anchor.diff:  
   We slightly touched quaternion multiplication, and we proved an important equation stem:[{q_1^{-1}}{q_2^{-1}}=({q_2}{q_1})^{-1}].
   
  -
  ----
  -[NOTE, indent=2]
  -.Side Note
  -====
  +=== 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.
  
positive.code: 
  With flip multiplication defined, we can rewrite our flip composition rule again:
  
  *_Any 3D rotation stem:[q=(sin⁡\frac{θ}{2}\vec{v},cos⁡\frac{θ}{2})] can be represented as a sequence of 2 flips stem:[q_a=(\vec{a},0)] and stem:[q_b=(\vec{b},0)], such that_*
  [stem]
  ++++
  q=-{q_b}{q_a}
  ++++
  *_where 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}]._*
  
  You might be thinking why it is not stem:[q= {q_a}{q_b}] instead. We will show where the order and the negative sign coming from in the proof.
  
  
negative.code: ++++ 
  which is the definition of quaternion multiplication of 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})].
  
  