# Quaternion product of three vectors: meaning of vector part?

$$\newcommand{\i}{\mathbf{i}} \newcommand{\j}{\mathbf{j}} \newcommand{\k}{\mathbf{k}} \newcommand{\a}{\mathbf{a}} \newcommand{\b}{\mathbf{b}} \newcommand{\c}{\mathbf{c}} \newcommand{\R}{\mathbb{R}}$$If you take two vectors $$\a=a_1\i+a_2\j+a_3\k$$ and $$\b=b_1\i+b_2\j+b_3\k$$ and multiply them as quaternions, then the result is $$\a\b = - \underbrace{\a \cdot \b}_\text{scalar} + \underbrace{\a \times \b}_\text{vector}.$$ In particular, both the scalar part and vector part have well known geometric interpretations. If a third vector $$\c = c_1\i + c_2 \j + c_3 \k$$ is introduced, then the result is $$\a\b\c = -\underbrace{(\a \times \b) \cdot \c}_\text{scalar}+\underbrace{(\a \times \b)\times \c-(\a \cdot \b)\c}_\text{vector}.$$ The scalar part above has a well known geometric interpretation; the triple product $$(\a \times \b)\cdot \c=\det(\a,\b,\c)$$ is plus or minus the volume of the parallelopiped spanned by $$\a,\b,\c$$.

But, what about the vector part, which I will call $$W(\a,\b,\c)$$? Using an identity for iterated crossed products, we can write it in the alternate form

$$W(\a,\b,\c) =(\a \times \b)\times \c-(\a \cdot \b)\c = -(\a \cdot \b)\c - (\b \cdot \c) \a + (\a \cdot \c)\b,$$

but I do not see an interpretation of either quantity.

Question: Does the vector $$W(\a,\b,\c)$$ above have any geometric significance?

I can at least mention that, in the case $$\a=\b=\c=\i$$ (a degenerate parallelopiped), we get $$W(\a,\b,\c) = -\i$$ and, in the case $$\a=\i, \b=\j,\c=\k$$ (the standard unit cube), we get $$W(\a,\b,\c) = \mathbf{0}$$.

Added 2019-01-22: A quick note on a possible interpretation... obviously $$W(\a,\b,\c)$$ is linear in each of $$\a$$, $$\b$$ and $$\c$$, so to understand it one had may as well assume the inputs are unit vectors. In this case, $$W(\a,\b,\c)=\mathbf{0}$$ precisely when the inputs are an orthonormal basis, almost like its a sort of $$3$$-ary dot product? This is explained in a comment on J.G.'s answer below. So, perhaps there is some way to interpret the vector $$W(\a,\b,\c)$$ as somehow measuring the "skewness" of the box spanned by $$\a,\b,\c$$?

If only the $$+$$ sign were a $$-$$, you'd have a cyclic symmetry. What's with that? Well, notice that because imaginary quaternions aren't closed under multiplication, $$abc$$ has a contribution of $$-(a\cdot b)c$$ that comes from $$c$$ interacting with a real number instead. Because of that, much as it pains me to say it, I don't think $$V$$ will have a nice geometric interpretation. Indeed, $$V$$ is a measure of how much a product fails to stay in the set of imaginary quaternions to which $$a,\,b,\,c$$ belong, and is obtained from a calculation that leaves that set as soon as we compute $$a,\,b$$.
If $$a = a_0 i + a_1 j + a_2 k$$ and $$b = b_0 i + b_1 j + b_2 k$$ are unit pure quaternions then $$(a b)$$ is a unit quaternion representing a rotation around the axis $$a \times b$$ with rotation angle relative to the angle formed from $$a$$ to $$b$$.
If $$c = c_0 i + c_1 j + c_2 k$$ is lying in the plane spaned by $$a$$ and $$b$$ then the geometrical interpretation of $$(a b) c$$ is the rotation of $$c$$ due to quaternion $$(a b)$$. This is in fact a 2D rotation.
If you repeat one vector, say $$a$$ the result is interesting since $$a a = 1$$ then $$a a b = b a a = b$$ but the interpretation of $$a b a$$ is the reflection of vector $$b$$ on the plane which normal is equal to $$a$$.