How to Prove the Vector Triple Product Identity?

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Homework Help Overview

The discussion revolves around proving the vector triple product identity, specifically the equation u x (v x w) = (u*w)v - (u*v)w. The subject area is vector algebra, particularly focusing on the properties of the cross product in three-dimensional space.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to prove the identity using algebraic properties of the cross product and considers representing vectors in a general form. Some participants suggest focusing on three-dimensional space due to the limitations of the cross product. Another participant introduces the use of the epsilon notation and summation convention for a more formal approach.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the proof. Some guidance has been offered regarding the use of epsilon notation, while others are questioning the necessity of proving the identity in higher dimensions.

Contextual Notes

There is a mention of constraints related to the definition of the cross product, which is not applicable in dimensions higher than three. This has led to a focus on proving the identity specifically within three-dimensional space.

robierob12
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Homework Statement




Prove that

u x (v x w) = (u*w)v - (u*v)w


Homework Equations



I've been trying to get this one but keep ending up no where.

I've tried the normal algebraic properties of the cross product but they lead me to a dead end.

What I am trying right now is just proving it in three space. Assigning each vector to a general form like (u1, u2, u3) and busting it out to see if I can get the right side identity.

Is there an easyier way to start this out.
Ideas are much appreciated.

Rob


The Attempt at a Solution

 
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cross product isn't defined for dimensions higher than three, so you would just prove it for three space.
 
I usually do these problems like these using the [itex]\varepsilon_{ijk}[/itex] thingy and the summation convention. If you're not familiar with that notation, this may just confuse you. The epsilon thingy is defined by [itex]\varepsilon_{ijk}=1[/itex] and the requirement that it's totally antisymmetric, i.e. that if you swap two indices, it will change sign. This implies for example that [itex]\varepsilon_{132}=-1[/itex] and that [itex]\varepsilon_{122}=0[/itex]. The "summation convention" is that I don't bother writing out the sigmas for summation, since all the indices that we need to sum over always occur exactly twice.

For example the scalar product [itex]u*v[/itex] is [itex]u_i v_i=v_1u_1+v_2u_2+v_3u_3[/itex] (I don't remember the LaTeX code for the scalar product) and the cross product [itex]u\times v[/itex] is [itex]\varepsilon_{ijk}u_jv_ke_i[/itex], where the [itex]e_i[/itex] are the basis vectors of [itex]\mathbb{R}^3[/itex].

The ith component of the left hand side of the equation you're trying to prove is by definition of the cross product

[tex](u\times (v\times w))_i=\varepsilon_{ijk} u_j (v\times w)_k=\varepsilon_{ijk}\varepsilon_{klm}u_j v_l w_m[/tex]
[tex]=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})u_j v_l w_m=u_j v_i w_j-u_j v_j w_i=(u*w)v_i-(u*v)w_i[/tex]

The tricky step is the one where I replaced two epsilons with some Kronecker deltas. The easiest way to see that this identity must hold is to explicitly calculate e.g. [itex]\varepsilon_{k12}\varepsilon_{k12}[/itex] and [itex]\varepsilon_{k12}\varepsilon_{k13}[/itex]. When you've done that, you'll probably understand.
 
Last edited:
Thanks... I don't know why I was thinking of tring to prove it for vectors outside of three space.
 

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