Cross product associative triples

Click For Summary

Homework Help Overview

The discussion revolves around the properties of the cross product, specifically focusing on the non-associative nature of the operation and the conditions under which certain triples of vectors can be considered associative. The original poster seeks to characterize nonzero associative triples of vectors that satisfy a specific identity involving the cross product.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use properties of cross products to find a geometric condition for associative triples but expresses difficulty in making progress. Other participants suggest using triple product expansion formulas and explore relationships between the vectors involved.

Discussion Status

Participants are actively engaging with the problem, offering suggestions and exploring different approaches. There is a focus on manipulating the expressions involving the cross product to derive conditions for associativity, but no consensus has been reached on the correct characterization of the associative triples.

Contextual Notes

The original poster mentions that one trivial case for associativity is when one of the vectors is the zero vector. There is also a suggestion to consider scalar products in relation to the conclusions drawn, indicating potential gaps in the reasoning presented.

nikcs123
Messages
4
Reaction score
0
cross product "associative triples"

Homework Statement



We know that the cross product is not associative, i.e., the identity
(1) [tex](\vec{a}\times\vec{b})\times\vec{c}[/tex] = [tex]\vec{a}\times(\vec{b}\times\vec{c})[/tex] is not true in general. However, certain special triples [tex]\vec{a}[/tex];[tex]\vec{b}[/tex];[tex]\vec{c}[/tex]
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 [tex]\vec{a}[/tex];[tex]\vec{b}[/tex];[tex]\vec{c}[/tex] 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.
 
Last edited:
Physics news on Phys.org


Did you try to use the triple product expansion formulas:

[tex]a\times(b\times c)=b(a,c)-c(a,b)[/tex]

[tex](a\times b)\times c=-c\times (a\times b)=c\times (b\times a)=...[/tex]
 


Thank you arkajad.

[tex](\vec{a}\times\vec{b})\times\vec{c}=\vec{a}\times(\vec{b}\times\vec{c})[/tex]

Working with the left side,

[tex](\vec{a}\times\vec{b})\times\vec{c}=-\vec{c}\times(\vec{a}\times\vec{b})=\vec{c}\times(\vec{b}\times\vec{a})[/tex]

So then

[tex]\vec{c}\times(\vec{b}\times\vec{a})=\vec{a}\times(\vec{b}\times\vec{c})[/tex]

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

That is correct, right?
 


nikcs123 said:
For the above to be true, [tex]\vec{a}=\vec{c}[/tex]. Furthermore, the above also holds true when [tex]\vec{a}[/tex] and [tex]\vec{c}[/tex] 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?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
26
Views
2K
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K