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$\newcommand{\i}{\mathbf{i}} \newcommand{\j}{\mathbf{j}} \newcommand{\k}{\mathbf{k}} \newcommand{\a}{\mathbf{a}} \newcommand{\b}{\mathbf{b}} \newcommand{\c}{\mathbf{c}} \newcommand{\R}{\mathbb{R}}$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 $\a,\b,\c$?

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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 $a,\,b$.

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  • $\begingroup$ Maybe, but I'm not so pessimistic myself... for example, assuming a,b,c to be nonzero, it seems V(a,b,c)=(a×b)×c−(a⋅b)c vanishes if and only if the vectors are pairwise orthogonal. Clearly pairwise orthogonality implies the vanishing. Conversely, if (a×b)×c=(a⋅b)c then, since the LHS is perpendicular to the RHS, both sides vanish which implies a is perpendicular to b and also that c is parallel to a×b, whence they are pairwise orthogonal. So, at least V(a,b,c)=0 has some geometric meaning $\endgroup$ – Mike F Jan 21 at 1:47
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My 2 cents:

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 $a$.

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