Vector Algebra - Vector Triple Product Proof

[H2instinct]

Homework Statement

Prove, by writing out in component form, that
\left(a \times b \right) \times c \equiv \left(a \bullet c\right) b - \left(b \bullet c\right) aand deduce the result, \left(a \times b\right) \times c \neq a \times \left(b \times c\right), that the operation of forming the vector product is non-associative.

The Attempt at a Solution

So I took the LHS and made it into component form:

LHS = \left(\left(\left(a_{z} \cdot b_{x} \cdot c_{z}\right) - \left(a_{x} \cdot b_{z} \cdot c_{z}\right)\right) - \left(\left(a_{x} \cdot b_{y} \cdot c_{y}\right) + \left(a_{y} \cdot b_{x} \cdot c_{y}\right)\right)\right) \hat{i} + \left(\left(\left(a_{x} \cdot b_{y} \cdot c_{x}\right) - \left(a_{y} \cdot b_{x} \cdot c_{x}\right)\right) - \left(\left(a_{y} \cdot b_{z} \cdot c_{z}\right) + \left(a_{z} \cdot b_{y} \cdot c_{z}\right)\right)\right) \hat{j}
+ \left(\left(\left(a_{y} \cdot b_{z} \cdot c_{y}\right) - \left(a_{z} \cdot b_{y} \cdot c_{y}\right)\right) - \left(\left(a_{z} \cdot b_{x} \cdot c_{x}\right) + \left(a_{x} \cdot b_{z} \cdot c_{x}\right)\right)\right) \hat{k}

RHS = \left(a_{x} \cdot b_{x} \cdot c_{x}\right) \hat{i} +\left(a_{y} \cdot b_{y} \cdot c_{y}\right) \hat{j} + \left(a_{z} \cdot b_{z} \cdot c_{z}\right) \hat{k} - \left(a_{x} \cdot b_{x} \cdot c_{x}\right) \hat{i} + \left(a_{y} \cdot b_{y} \cdot c_{y}\right) \hat{j} + \left(a_{z} \cdot b_{z} \cdot c_{z}\right) \hat{k}

This doesn't seem very difficult to me, however, the algebra of it is all over the place. I believe I found the component form of each side properly, but I could have screwed up somewhere. I am definitely not getting the RHS = LHS, so I am either screwing up somewhere or am totally on the wrong path. Any help, either noticing what I have done wrong, or a fresh start is much appreciated. Thanks.

 

[rickz02]

Recheck your RHS, all important terms are missing. Notice that a, b, and c have 3 components each, be careful in the multiplication. Also, you have problems on the signs in your LHS.

 

[lanedance]

one thing i find useful for thrse problems is working in subcript notation, where repeated indicies mean a sum is performed over that index. The dot products & cross products then become:

a \bullet b = a_i b_j \delta_{ij} = a_i b_i
(a \times b)_k = a_i b_j \epsilon_{ijk}

where \delta_{ij} is the kronecker delat & \epsilon_{ijk} is the levi cevita
http://en.wikipedia.org/wiki/Kronecker_delta
http://en.wikipedia.org/wiki/Levi-Civita_symbol

takes a little time to learn this notation to start, but i find they're an extra tool in the toolbox & can save a lot of time

 

[H2instinct]

Recheck your RHS, all important terms are missing. Notice that a, b, and c have 3 components each, be careful in the multiplication. Also, you have problems on the signs in your LHS.

I am not fully understanding the dot product multiplication rule because how I think it is supposed to be multiplied to show component form just keeps coming out as I have it shown above.

Also, I fixed the signs of the LHS, I believe that is now correct.

This is where I get stuck:

RHS = \left(a_{x} \cdot c_{x} + a_{y} \cdot c_{y} + a_{z} \cdot c_{z}\right) b - \left(b_{x} \cdot c_{x} + b_{y} \cdot c_{y} + b_{z} \cdot c_{z}\right) a

 

[rickz02]

In dot product multiplication you just have to multiply the components of the vectors with the same unit vectors (i.e. i to i, j to j, k to k).

I can see in your solution above that you don't have a problem with the dot product multiplication. You just confused it (indicated by a dot) with the ordinary multiplication (indicated by the the parentheses).

 

[H2instinct]

In dot product multiplication you just have to multiply the components of the vectors with the same unit vectors (i.e. i to i, j to j, k to k).

I can see in your solution above that you don't have a problem with the dot product multiplication. You just confused it (indicated by a dot) with the ordinary multiplication (indicated by the the parentheses).

You are correct... I confused regular multiplication with dot product. But what exactly is the parentheses multiplication of vectors? I though that was the dot product. Still confused...

 

[rickz02]

Another thing, you have sign problems in your RHS as well, all the terms you have there should cancel out.

 

[H2instinct]

Another thing, you have sign problems in your RHS as well, all the terms you have there should cancel out.

Ya. Just don't understand from here...

<br /> RHS = \left(a_{x} \cdot c_{x} + a_{y} \cdot c_{y} + a_{z} \cdot c_{z}\right) b - \left(b_{x} \cdot c_{x} + b_{y} \cdot c_{y} + b_{z} \cdot c_{z}\right) a<br />

to the solution.

 

[rickz02]

Just do the normal multiplication like (a+x)(b+y) = ab + bx + ay + xy.

 

[rickz02]

Remember that b has three components; bx, by and bz. The same goes with a.

 

[H2instinct]

Remember that b has three components; bx, by and bz. The same goes with a.

If you are saying to multiply them like this...


<br /> <br /> RHS = \left((a_{x} \cdot c_{x} \cdot b) + (a_{y} \cdot c_{y} \cdot b) + (a_{z} \cdot c_{z} \cdot b) \right) - \left((b_{x} \cdot c_{x} \cdot a) + (b_{y} \cdot c_{y} \cdot a) + (b_{z} \cdot c_{z} \cdot a)\right)<br /> <br />

That really only makes me consider the same questions I was considering before -.-

 

[rickz02]

Exactly, now your b = bx i + by j + bz k and your a = ax i + ay j + az k. Substitute and simplify again. Remove the dots so you won't be confused it with the dot product.

 

[rickz02]

Actually an easy way to do this is suggested by lanedance, but it needs some more understanding of the tensor notations. Using the notations you're actually left working with indices.

 

[H2instinct]

Luckily the easy way to do this also involves more learning. It's still not making sense btw.

 

[rickz02]

<br /> <br /> RHS = \left((a_{x} \cdot c_{x} \cdot b) + (a_{y} \cdot c_{y} \cdot b) + (a_{z} \cdot c_{z} \cdot b) \right) - \left((b_{x} \cdot c_{x} \cdot a) + (b_{y} \cdot c_{y} \cdot a) + (b_{z} \cdot c_{z} \cdot a)\right)<br /> <br />

This one is almost close to the proof. Just continue this.