Cross product associative triples

[nikcs123]

cross product "associative triples"

Homework Statement

We know that the cross product is not associative, i.e., the identity
(1) (\vec{a}\times\vec{b})\times\vec{c} = \vec{a}\times(\vec{b}\times\vec{c}) is not true in general. However, certain special triples \vec{a};\vec{b};\vec{c}
of vectors do satisfy (1). For example, if one of the vectors is the zero vector, then (1)
holds trivially, but there are also less obvious examples. Call a triple \vec{a};\vec{b};\vec{c} for which (1)
holds "associative". Characterize all nonzero associative triples by some simple geometric
condition. (An example of a possible condition (though not the correct one) would be that
the three vectors are pairwise perpendicular.)

The Attempt at a Solution

I attempted to use the properties/identities of cross products to deduce a relationship between the components of each vector and try to piece it together that way, ended up with a huge mess and no progress... Just need a nudge in the right direction on this one.

 

[arkajad]

Did you try to use the triple product expansion formulas:

a\times(b\times c)=b(a,c)-c(a,b)

(a\times b)\times c=-c\times (a\times b)=c\times (b\times a)=...

 

[nikcs123]

Thank you arkajad.

(\vec{a}\times\vec{b})\times\vec{c}=\vec{a}\times(\vec{b}\times\vec{c})

Working with the left side,

(\vec{a}\times\vec{b})\times\vec{c}=-\vec{c}\times(\vec{a}\times\vec{b})=\vec{c}\times(\vec{b}\times\vec{a})

So then

\vec{c}\times(\vec{b}\times\vec{a})=\vec{a}\times(\vec{b}\times\vec{c})

For the above to be true, \vec{a}=\vec{c}. Furthermore, the above also holds true when \vec{a} and \vec{c} are parallel.

That is correct, right?

 

[arkajad]

For the above to be true, \vec{a}=\vec{c}. Furthermore, the above also holds true when \vec{a} and \vec{c} are parallel.

That is correct, right?

Why don't you bring in the equality involving scalar products and justify your conclusion? Perhaps you have missed something?