Geometric Proof of Vector Triple Product: Find Coefficients b and c

In summary, the conversation discusses a proof for the vector triple product and the definition of the curl of a function in ℝn. The individual attempting the proof shares their own attempt and asks for help on how to continue. The conversation then delves into the definition of the curl of a function in ℝn and the possibility of generalizing it to higher dimensions. The attempt at a solution involves using the operations of gradient and divergence, but it is eventually shown that there is a contradiction in this approach.
  • #1

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] [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:


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)


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.
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  • #2
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.


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:


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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  • #3
I don;t understand your dilemma. The curl is well defined in R^n, likewise the divergence and the gradient and also the laplacian.
  • #4
What terms had to vanish in this step:

∇×C → (∇×)2C

such that (∇×)2C is expressible in only ∇ and (∇.) alone?
  • #5
This paper might help: [Broken]
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  • #6
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


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

1. What is the purpose of a geometric proof of vector triple product?

A geometric proof of vector triple product helps to understand the relationship between three vectors in 3-dimensional space and how they form a new vector using their cross product. It also helps in determining the coefficients b and c in the resulting vector.

2. How is the geometric proof of vector triple product different from the algebraic proof?

The geometric proof involves visualizing the vectors in 3-dimensional space and using geometric principles to prove the relationship, while the algebraic proof uses mathematical equations and properties to show the relationship.

3. What are the steps involved in the geometric proof of vector triple product?

The steps involve determining the cross product of two of the given vectors, finding the angle between the resulting vector and the third vector, and then using trigonometric functions to solve for the coefficients b and c.

4. Can the geometric proof of vector triple product be used for any three vectors?

Yes, the geometric proof can be used for any three non-collinear vectors in 3-dimensional space, as long as the cross product of any two of the vectors is non-zero.

5. How is the geometric proof of vector triple product useful in real-world applications?

The geometric proof of vector triple product is useful in many engineering and physics applications, such as determining torque and angular momentum, calculating work done by a force, and analyzing forces in 3-dimensional structures.

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