anchor.diff:  
   We are not going to use this general quaternion multiplication in Part 1. Here we only need to know a simpler form, the multiplication of flips. Let stem:[q_a=(\vec{a},0)], stem:[q_b=(\vec{b},0)] then
   
  -stem:[{q_a}{q_b}=(\vec{a},0)(\vec{b},0)=(\vec{a}×\vec{b},-\vec{a}·\vec{b})]
  +[stem]
  +++++
  +{q_a}{q_b}=(\vec{a},0)(\vec{b},0)=(\vec{a}×\vec{b},-\vec{a}·\vec{b})
  +++++
   
   It is naturally derived from the general form, and we will be only using this multiplication in Part 1.
   
  
positive.code: *_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.
  
  #### Proof
  
  stem:[\vec{a}·\vec{b}=cos<\vec{a},\vec{b}>\left|\vec{a}\right|\left|\vec{b}\right|=cos\frac{θ}{2}]. Since stem:[\vec{a}·\vec{v}=0], stem:[\vec{b}·\vec{v}=0] and stem:[\left|\vec{v}\right|=1], we have stem:[\vec{a}×\vec{b}=sin<\vec{a},\vec{b}>\left|\vec{a}\right|\left|\vec{b}\right|\vec{v}=sin\frac{θ}{2}\vec{v}].
  
  
negative.code: ++++
  \begin{align*}
  {q_v'}{q_a}{q_a^{-1}}&={q_a}{q_v}{q_a^{-1}}\\
  q_v'&={q_a}{q_v}{q_a^{-1}}
  \end{align*}
  ++++  
  
  