(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Vector Triple Product

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**