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Using geometric algebra, I can easily find the geometric tripleproduct of three vectors $a,b,c \in \mathbb{R}^3$ to be

$$abc = a \left(b \cdot c \right) - b \left( c \cdot a \right) + c \left( a \cdot b \right) + i \left(a \times b\right) \cdot c $$

The imaginary part of this is obviously the directed volume spanned by the three vectors. Though what about the other part? Two of those terms arise as the triple cross product $\left(a \times b \right) \times c$, which,among other places, has been nicely interpreted here already: https://math.stackexchange.com/a/400732/49989 as a vector in a plane spanned by $a$ and $b$ which is normal to $c$. But the term $\left(a \cdot b\right) c$ arises independently.

I constructed all cyclic permutations of these in GeoGebra but it didn't help me much for finding an interpretation. Perhaps it helps any of you for finding or illustrating an answer. You can find it here:

GeoGebra: Geometric Triple Product (I chose to keep all the vectors at unit length)

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kram1032
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  • I asked this question at http://math.stackexchange.com/questions/913320/what-is-vector-veca-cdot-vecb-vecc-vecb-cdot-vecc-veca. – ahala May 15 '15 at 13:28
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    @ahala ah I thought I looked throughly enough but yeah apparently my question is a duplicate of yours. – kram1032 May 15 '15 at 14:09

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To get a gemoetric interpretation, convert the above relation into geometric algebra. Whence; \begin{equation} a \times (b \times c) = -a \rfloor (b \wedge c) \end{equation} The $b \wedge c (\equiv X)$ part is a bivector; it is to be understood as an area element in the $bc$ plane. Then $a \rfloor X$ is a left contraction of $a$ onto $X$, which is dot product in nature in that it projects $a$ onto $X$ and roates it $\pi/4$ in the direction of $X$ (the factor of $-1$).

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    I'm not asking for a geometric algebraic version of the triple cross product but rather what the vector valued portion of the triple geometric product means if it has a nice geometric interpretation. The triple cross product is merely a part of the result. – kram1032 May 15 '15 at 10:45
  • @kram1032 the geometric algrbriac setting allows for a dicussion of the points you raise. – Autolatry May 15 '15 at 10:46
  • ok, so what I have is $a \rfloor \left( b \wedge c \right) + \left(a \cdot b\right) c$. What does that tell me? – kram1032 May 15 '15 at 22:03
  • wait, actually I have $c\left(a\cdot b\right)+c \times \left( a \times b \right) = c\left(a\cdot b\right) -c \rfloor \left( a \wedge b \right)$ – kram1032 May 15 '15 at 22:07