anchor.diff:  ++++
   \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}}
  +q_v'&={q_a}{q_v}{q_a^{-1}}
   \end{align*}
   ++++  
   
  
positive.code: 
  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})]
  
  It is naturally derived from the general form, and we will be only using this multiplication in Part 1.
  
  
negative.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.
  
  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}].
  
  