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Because of the cross product of two vectors being another vector I can calculate $\vec a\times(\vec b\times\vec c)$ as well as $(\vec a\times\vec b)\times\vec c$. I know that the cross product is not associative--- $$\vec a\times(\vec b\times\vec c)\neq(\vec a\times\vec b)\times\vec c~,$$ ---but is there a way to calculate $$\vec a\times\vec b\times\vec c~?$$ If there is no way, is there something like a convention that tells me whether I have to calculate $\vec a\times\vec b$ first or $\vec b\times\vec c$?

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No, $\def\bfa{{\bf a}}\def\bfb{{\bf b}}\def\bfc{{\bf c}}\bfa \times \bfb \times \bfc$ simply doesn't make sense for vectors in $\Bbb R^3$, precisely because $\times$ is nonassociative. The cross product does, however, satisfy the so-called Jacobi identity, $$\bfa \times (\bfb \times \bfc) + \bfb \times (\bfc \times \bfa) + \bfc \times (\bfa \times \bfb) = {\bf 0}.$$ In fact, this (together with the linearity and antisymmetry of $\times$) makes $(\Bbb R^3, \times)$ a Lie algebra, in this case one isomorphic to $\mathfrak{so}(3, \Bbb R)$. This identity follows, for example, from the iterated cross product identity $$\bfa \times (\bfb \times \bfc) = (\bfa \cdot \bfc) \bfb - (\bfa \cdot \bfb) \bfc.$$

Furthermore, there are some related notions of trilinear products, that is, those that have three arguments and are linear in each: In $\Bbb R^3$ probably the best known is the vector triple product, $$(\bfa, \bfb, \bfc) \mapsto \bfa \cdot (\bfb \times \bfc),$$ which can also be written as the determinant of the matrix produced by adjoining $\bfa, \bfb, \bfc$ (regarding those as column vectors): $$\bfa \cdot (\bfb \times \bfc) = \det \pmatrix{\bfa & \bfb & \bfc}.$$

Given any (not-necessarily associative) algebra $\Bbb A$, one can form the associator, namely, $$[a, b, c] := (ab)c - a(bc)$$ (or perhaps its negative), which measures the failure of associativity of the product, in that, by construction, it is the zero map iff $\Bbb A$ is associative. In the case of the usual cross product on $\Bbb R^3$, it follows from the Jacobi identity that the associator is nothing more than the iterated cross product in a certain order, namely $$[\bfa, \bfb, \bfc] = \bfb \times (\bfc \times \bfa),$$ and this rearrangement, together with the anticommutativity of $\times$, suggests that we can view the Jacobi identity as a sort of Leibniz (product) rule for $\times$.

Finally, there is a notion of a cross product that takes as an argument three vectors and produces another vector, but only in dimensions $4$ and $8$. These "$n$-ary" cross products are typically denoted something like $X(\bfa, \bfb, \bfc)$ or $\times(\bfa, \bfb, \bfc)$, rather than $\bfa \times \bfb \times \bfc$, which suggests repeated use of a binary operator. See the recent question Using Gram-Schmidt to compute the cross product of $3$ vectors in $\Bbb R^4$ for a description of how to compute such a product in dimension $4$.

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  • $\begingroup$ This sounds very interesting :) That there is a cross product with three vectors in 4 dimensions sounds logically to me, but why in 8 dimensions? Wouldn't a cross product with 8 dimensions require 7 vectors? $\endgroup$ – Cubi73 Jul 5 '15 at 14:03
  • $\begingroup$ @Cubinator73 There is a cross product in $8$ dimensions that requires $7$ vectors, but there are binary cross products in $7$ dimensions and trinary cross products in $8$ dimensions, all of which are connected in various ways to the octonions, a very special algebra that is connected to all sorts of "exceptional" objects in mathematics, that is objects that, like the special cross products in dimensions $7$ and $8$, don't belong to nice, usually infinite families of example. $\endgroup$ – Travis Willse Jul 5 '15 at 14:08
  • $\begingroup$ @Cubinator73 You might also like to see my answer to this related, older problem here: math.stackexchange.com/a/1350088/155629 $\endgroup$ – Travis Willse Jul 5 '15 at 14:19
  • $\begingroup$ Perhaps the downvoter would care to voice their objection? $\endgroup$ – Travis Willse Jul 5 '15 at 15:15
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The notation $\vec{a}\times \vec{b}\times \vec{c}$ would be meaningful if and only if you could say that $$(\vec{a}\times \vec{b})\times \vec{c}=\vec{a}\times(\vec{b}\times \vec{c})$$ for any three vectors $\vec{a},\vec{b}$ and $\vec{c}$ but it is not true in general as you have noted in your answer. In fact, due to the non-associativity of the products it is not even clear whether $\vec{a}\times\vec{b}\times \vec{c}$ is a vector or scaler !

And for your second question, the answer is that we don't have any preference in the calculation of $(\vec{a}\times \vec{b})$ and $(\vec{b}\times \vec{c})$. In general, it is matter of convenience as to which product you would evaluate first.

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No, there is no such convention and, since the cross product is not associative, "$\vec{a}\times\vec{b}\times\vec{c}$" simply has no meaning. You must write either $(\vec{a}\times\vec{b})\times\vec{c}$ or $\vec{a}\times(\vec{b}\times\vec{c})$ to distinguish the two.

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