Having difficulty understanding what happens when reversing order of quaternion multiplication.

Question

The textbook I am reading claims that quaternion multiplication works like so:

$ q_1q_2 = V_1 \times V_2 + s_1V_2+s_2V_1+s_1s_2-V_1 \cdot V_2 $

Which is a simplified view of

$ q_1q_2 = (x_1w_2+y_1z_2-z_1y_2+w_1x_2)i\\ +(y_1w_2+z_1x_2+w_1y_2-x_1z_2)j\\ +(z_1w_2+w_1z_2+x_1y_2-y_1x_2)k\\ +(w_1w_2-x_1x_2-y_1y_2-z_1z_2) $

where $q$ is defined as:

$q=xi+yj+zk+w$

Which makes sense to me if you take into consideration the multiplication rule for quaternions:

$i^2=j^2=k^2=ijk=-1$

The thing that I don't understand is how reversing the order of quaternion multiplication works. The textbook defines it as:

$q_2q_1=q_1q_2-2(V_1 \times V_2)$

However I have no idea how to go about obtaining this result.

Answer 1

So we start with the "definition" of multiplication: $$ q_1q_2 = V_1 \times V_2 + s_1V_2+s_2V_1+s_1s_2-V_1 \cdot V_2 $$ and then ask ourselves, what would be $q_2 q_1$, instead? Well, we simply need to swap the indices:

$$\ \begin{split} q_2q_1 &= V_2 \times V_1 + s_2V_1+s_1V_2+s_2s_1-V_2 \cdot V_1\\ &= V_2 \times V_1 + s_2V_1+s_1V_2+s_1s_2-V_1 \cdot V_2\\ \end{split} $$ The last two terms were switched because both scalar multiplication and dot product are commutative. Now you can easily subtract $q_2q_1 - q_1q_2$ to get the result $$ \begin{split} q_2 q_1 - q_1 q_2 &= V_2 \times V_1 - V_1 \times V_2\\ &= -V_1 \times V_2 - V_1 \times V_2\\ &= -2V_1 \times V_2\\ \end{split} $$

Answer 2

In $q_1q_2 = V_1 \times V_2 + s_1V_2+s_2V_1+s_1s_2-V_1 \cdot V_2$ the $s_1V_2+s_2V_1+s_1s_2-V_1 \cdot V_2$ part is symmetrical in $1,2$. Hence won't change when you permute the $q_1q_2$ product.

The only part that changes is $V_1 \times V_2$ which is anticommutative, i.e. $V_2 \times V_1 = -V_1 \times V_2$.

This explains the result $q_2q_1=q_1q_2-2(V_1 \times V_2)$.