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

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.

• Just a quick remark: you can't define the multiplication in the reverse order. Once you defined multiplication, you can of course apply that definition in any order you want. What you can do is prove that some formula is true when you reverse the order. Mar 12, 2020 at 8:56

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 $$\begin{split} q_2 q_1 - q_1 q_2 &= V_2 \times V_1 - V_1 \times V_2\\ &= -V_1 \times V_2 - V_1 \times V_2\\ &= -2V_1 \times V_2\\ \end{split}$$
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 $$q_2q_1=q_1q_2-2(V_1 \times V_2)$$.