# Vector Triple Product

## Homework Statement

This isn't a coursework question. Rather, I'm asking for help on a geometric proof of the vector triple product. I find it strange and annoying that I can't find this proof anywhere online, because everyone just uses the messy expansion proof, and I hate that proof because it lacks insight. So, I attempted my own proof. Here goes.

## Homework Equations

Ax(BxC) = (A.C)B - (A.B)C

## The Attempt at a Solution

The following is a diagram that I drew.

[PLAIN]http://img695.imageshack.us/img695/6313/unledwbx.png [Broken]

Both BxC and A are out of the plane. B, C, and Ax(BxC) are in the plane.

B and C span a plane. It's known (and not necessary to prove that) BxC is orthogonal to B and C, so it is not contained in span{B,C}. Let A be any vector. Then Ax(BxC) is in the span of B and C, because B and C span the unique plane that is orthogonal to all planes containing BxC.

Therefore, B and C form a basis for Ax(BxC), so bB+cC = Ax(BxC). Thus, all that's left is to determine the coefficients b and c.

At this step, I wasn't sure how to continue, but I thought it may be worthwhile to find the projection of A onto the plane so that we're only working with quantities in span{B,C}. I'm going to denote the projection of A onto the plane as (proj A), since (proj A) is very messy. Let BxC be one of the two perpendicular bases for A. Then the projection of A onto BxC is as follows:

(A.(BxC)/|BxC|2)(BxC)

It follows that the projection of A onto the plane is:

proj A = A-(A.(BxC)/|BxC|2)(BxC)

Since Ax(BxC) is orthogonal to the projection of A onto the plane (the above), we have the following:

[Ax(BxC)].[proj A]=0

[bB+cC].[proj A] = 0

b[B.(proj A)]+c[C.(proj A)] = 0

c = -b[B.(proj A)]/[C.(proj A)]

Since this is an equation of two variables, in order find a unique solution, I need to find another equation that is linearly independent of the above equation. To obtain a second equation, I know that |Ax(BxC)| = |A||BxC|sin(θ), where θ is the angle between A and BxC. This can be equivalently expressed as the cross product of (proj A) with BxC. This is because |A|sin(θ) is also the length of (proj A). Therefore:

|Ax(BxC)|= |BxC||A|sin(θ) = |BxC||proj A|

|bB+cC|=|BxC||proj A|

Here, I replace c using c = -b[B.(proj A)]/[C.(proj A)]

|bB-b[B.(proj A)/C.(proj A)]C|=|BxC||proj A|

b|B-C[B.(proj A)/C.(proj A)]|=|BxC||proj A|

b=|BxC||proj A|/|B-C[B.(proj A)/C.(proj A)]|

Recall that proj A = A-(A.(BxC)/|BxC|2)(BxC)

b=|BxC||A-(A.(BxC)/|BxC|2)(BxC)|/|B-C[B.(A-(A.(BxC)/|BxC|2)(BxC))/C.(A-(A.(BxC)/|BxC|2)(BxC))]|

Holy crap that's messy.

Now I have a unique b, but I'm lost. How do I reduce the above equation to b=A.C? I am already this far, but I'm just about out of brainpower for the night. I wanted every step to have an obvious geometric implication, but this end step is a tough cookie. I feel like as long as the b=A.C relationship can be shown, I can show that c=-A.B.

Last edited by a moderator:

Okay. So it turns out that I took a wrong step here:

[A×(B×C)].[proj A]=0

Instead of using proj A, I can just use A.

[A×(B×C)].A=0
[bB+cC].A=0
b(B.A)+c(C.A)=0

Therefore, the constants b and c are m(C.A) and -m(B.A), respectively, and it turns out that m=1 (by any vector we choose to test what k is).

-----

Since [A×(B×C)]=(C.A)B-(B.A)C, I obtain the following relation:

∇×∇×C = ∇(∇.C)-(∇.∇)C = ∇(∇.C)-∇.(∇C)
(This isn't really precalc anymore.)

Hence, the curl of the curl of a function exists in ℝn, since the gradient is defined as

∇ = (d/dx1, d/dx2, ... , d/dxn)

and the divergence is defined as the dot product of ∇ and a vector of the same size.

As you can see, one question leads right into another. This still isn't a coursework question (or even remotely related to anything I'm doing in class), but that doesn't mean it's inferior to coursework questions. I hope you people of PF can actually try (gasp!) -responding- this time instead of completely ignoring my threads. :(

## Homework Statement

Why is the curl of the curl of a function C in ℝn defined under the pre-existing definition of the gradient, while the curl of C is only defined when the function is in ℝ3? In symbols, let the curl of the curl of C be written as:

(∇×)2C

How come (∇×)2nC (where n∈Z) is defined in ℝn, but not (∇×)2n+1C? Does there exist a process ((∇×)2)1/2 such that after two iterations on C, ∇(∇.C)-∇.(∇C) is obtained? That's the equivalent of asking if there exist operations (∇×) and (∇×)-1. If so, how would I go about finding out what they are, and are they generalizable to ℝn such that the curl is defined for ℝn?

## Homework Equations

I want to avoid all tensor use.

## The Attempt at a Solution

Let C = (x1, x2, x3)
Let ∇×C in ℝ3 be (v1, v2, v3), where
v1 = (d/dx2)x3-(d/dx3)x2
v2 = (d/dx3)x1-(d/dx1)x3
v3 = (d/dx1)x2-(d/dx2)x1

Then it makes sense to generalize the curl by:

∇×C = (v1, v2, ... , vn)

v1 = (d/dx2)x3-(d/dx3)x2
v2 = (d/dx3)x4-(d/dx4)x3
.
.
.
vn-1 = (d/dxn)x1-(d/dx1)xn
vn = (d/dx1)x2-(d/dx2)x1

Is there a way to write this using ∇ and (∇.) only?

Since this is a vector function that is arrived at from a vector function, both operations must be used at least once. The divergence produces a scalar, and the gradient produces a vector. Therefore, C = ∇F for some function F. But it's known that ∇×(∇F) = 0. So there's obviously a contradiction.

-Is super confused-

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dextercioby
Homework Helper
I don;t understand your dilemma. The curl is well defined in R^n, likewise the divergence and the gradient and also the laplacian.

What terms had to vanish in this step:

∇×C → (∇×)2C

such that (∇×)2C is expressible in only ∇ and (∇.) alone?

This paper might help:

Last edited by a moderator:
lurflurf
Homework Helper
You cannot avoid all tensor use, you can only hide it. The tensors are still there.
I think all those planes obscure the point

Ax(BxC) = aA(B.C)+bB(A.C) + cC(B.A)

for suitable constants a,b,c which are obviously a=0,b=1,c=-1

(∇×)2C is analogous

The generalization you suggest is not the useful accepted one which is

Δ=dδ+δd

curl is only defined for R3 and in R7 we have something very curl like