I understand that the formula for quaternion multiplication of $q_1=(s_1,\vec{v_1})$ by $q_2=(s_2,\vec{v_2})$ $q_1q_2=(s_1s_2-\vec{v_1}\cdot\vec{v_2}, \vec{v_1} \times\vec{v_2} + \vec{v_1}s_2 + \vec{v_2}s_1)$ is derived from the simple expansion of $(a_1+b_1i+c_1j+d_1k)(a_2+b_2i+c_2j+d_2k)$. I also understand that the conjugation of a pure quaternion by a unit quaternion accomplishes a rotation in 3-D about the vector part of the quaternion by $2\theta$ where $\theta$ is the angle between the quaternion as a 4-D vector and our 3-D space. I am trying to understand the geometric significance of the formula for quaternion multiplication in its own right, not as half of a rotation. In other words, why does the $qp$'s projection into 3-D space lie half way between $p$ and $qpq^{-1}$? Is it analogous in some way to the way multiplying imaginary numbers adds their angles?
Update
To be more clear, I'm wondering if there is a geometric interpretation of this formula $q_1q_2=(s_1s_2-\vec{v_1}\cdot\vec{v_2}, \vec{v_1} \times\vec{v_2} + \vec{v_1}s_2 + \vec{v_2}s_1)$. For example, why is it that the new real part is the product of the real parts minus the cosine of the angle between the imaginary vectors (the dot product)? Why is the new imaginary vector the orthogonal vector (cross product) minus each original vector scaled by the real part of the other quaternion?
Alternatively, if someone could do the same sort of explanation for the equivalent Cayley-Dickinson form of a quaternion $w+xi+yj+zk=(w+xi)+(y+zi)j$, and multiplying two quaternions each with two complex numbers like $(a,b)(c,d)=(ac-bd^*, ad+bc^*)$.
Best Answer
The geometric interpretation of quaternion multiplication is fundamentally 4-dimensional (unlike quaternion conjugation, which can be considered as an action on $\Bbb{R}^3$).
Let's start with an easy case. Say $q=a+bi$ with $b \neq 0$, $a^2+b^2=1$. That is, $q$ is a non-real unit quaternion in the subalgebra of $\Bbb{H}$ generated by $i$. What affect does multiplying by $q$ have on an arbitrary quaternion $r$?
First of all, if $r$ also lies in the subalgebra generated by $i$, then we can consider multiplication of $q$ and $r$ to be ordinary complex multiplication; that is, multiplication by $q$ rotates $r$ by $\theta$ in the $\{1, i\}$ plane.
Secondly, if $r$ lies in the orthogonal complement of that subalgebra, $r=cj+dk$, we can write $r=(c+di)j=j(c-di)$. The first of these representations can be used to left-multiply by $q$ via ordinary complex multiplication; the second one can be used to right-multiply. In either case, multiplying by $q$ rotates $r$ by $\theta$ in the $\{j, k\}$-plane; however, the sign difference means that the two multiplications rotate in opposite directions from each other.
We can then find the effect of multiplying $q$ by an arbitrary quaternion by projecting that quaternion into these two planes. That is, an arbitrary quaternion will have its $\{1, i\}$-projection and $\{j, k\}$-projection both rotated by $\theta$ when it is multiplied by $q$; the direction of the $\{j, k\}$-rotation, but not of the $\{1, i\}$-rotation, will be affected by whether we're left-multiplying or right-multiplying by $q$.
The general case works similarly. For any unreal unit quaternion $q$ that makes an angle $\theta$ with the real axis, multiplication by $q$ rotates by $\theta$ in the $\{1, q\}$-plane, and also rotates by $\theta$ in its orthogonal complement. The direction of the first rotation is fixed, but the direction of the second rotation depends on whether we're multiplying by $q$ on the left or on the right. You can see this just by noticing that any unreal quaternion generates a 2-dimensional subalgebra isomorphic to $\Bbb{C}$, making the previous few paragraphs work in general after some relabeling.
This also gives you a way to see why quaternion conjugation works the way it does on $\Bbb{R}^3$. If $q$ makes an angle $\theta$ with the real axis, then the map $r \mapsto qrq^{-1}$:
Note also that multiplying by a general quaternion involves scaling by the norm of that quaternion, but conjugation conveniently causes the norms of $q$ and $q^{-1}$ to cancel.