anchor.diff:  .Side Note
   ****
   This claim actually has 2 sides. Let me go a little be more in details in math term: 
  -(1). For any 3D rotation equivalent to a rotation angle stem:[θ] along axis stem:[\vec{v}], there exists a unit quaternion stem:[q=(sin⁡\frac{θ}{2}\vec{v},cos⁡\frac{θ}{2})] to describe this rotation. 
  -(2). For any unit quaternion stem:[q=(x,y,z,w)], it describes a rotation of angle stem:[θ=2cos^{-1}w] along axis stem:[\vec{v}=\frac{(x,y,z)}{\sqrt{1-w^{2}}}].
  +(1). For any 3D rotation equivalent to a rotation angle stem:[θ] along axis stem:[\vec{v}], there exists a unit quaternion stem:[q=(\sin⁡\frac{θ}{2}\vec{v},\cos⁡\frac{θ}{2})] to describe this rotation. 
  +(2). For any unit quaternion stem:[q=(x,y,z,w)], it describes a rotation of angle stem:[θ=2\cos^{-1}w] along axis stem:[\vec{v}=\frac{(x,y,z)}{\sqrt{1-w^{2}}}].
   
   ****
   ---
  
positive.code: 
  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 the special flip break down stem:[q_1=-{q_c}{q_a}], stem:[q_2=-{q_b}{q_c}], we used to visualize the result of rotation composition. 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.If you don’t remember this, see “Rotation Composition” section in Part 1.
  
  Here are some equations we will be using:
  [stem]
  
negative.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}
  