anchor.diff:  ++++
   \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}))))\\
  +&- 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: 
  With flip multiplication defined, we can rewrite our flip composition rule again:
  
  *_Any 3D rotation stem:[\textbf{q=(sin⁡\frac{θ}{2}\vec{v},cos⁡\frac{θ}{2})}] can be represented as a sequence of 2 flips stem:[\textbf{{q_a=(\vec{a},0)}] and stem:[\textbf{q_b=(\vec{b},0)}], such that_*
  [stem]
  ++++
  \textbf{q=-{q_b}{q_a}}
  ++++
  *_where stem:[\textbf{\vec{a}·\vec{v}=0}], stem:[\textbf{\vec{b}·\vec{v}=0}] and the angle from stem:[\textbf{\vec{a}}] to stem:[\textbf{\vec{b}}]: stem:[\textbf{<\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: 
  In order to represent 3D rotation, we have a constraint on the quaternions we use. But before that I want to introduce Euler’s rotation theorem:
  
  *_Any rotation in 3D space is equivalent to a single rotation of angle stem:[θ] along some axis stem:[\vec{v}]._*
  
  We can use quaternion to describe this angle-axis rotation : stem:[q=(sin⁡\frac{θ}{2}\vec{v}.x,sin⁡\frac{θ}{2}\vec{v}.y,sin⁡\frac{θ}{2}\vec{v}.z,cos⁡\frac{θ}{2})], or in a more compact form stem:[q=(sin⁡\frac{θ}{2}\vec{v},cos⁡\frac{θ}{2})]. We call this form the vector form of a quaternion, and we will use this form throughout this article. You might be thinking why we are using stem:[\frac{θ}{2}] other than using stem:[θ] directly. I will explain that in a later section.
  
  It is easy to see the length of this quaternion stem:[\left|q\right|=\sqrt{sin^{2}\frac{θ}{2}\left|\vec{v}\right|^{2}+cos^{2}\frac{θ}{2}}=1]. (Remember the axis stem:[\vec{v}] is a unit vector that stem:[\left|\vec{v}\right|=1]). We call it a unit quaternion if the length stem:[\left|q\right|=1]. So we can rewrite Euler’s rotation theorem in quaternion term:
  
  *_Any 3D rotation is equivalent a unit quaternion stem:[q] that stem:[\left|q\right|=1]._*
  
  ---
  [NOTE, indent=2]
  