Prove Quadruple Product Identity from Triple Product Identities

In summary, the homework statement asks for an equation that allows for the determination of the identity of a×b. The problem is that the equation is backwards, and the student is lost on how to fix it.
  • #1
jtleafs33
28
0

Homework Statement


I need to prove the identity:

(a×b)[itex]\cdot[/itex](c×d)= (a[itex]\cdot[/itex]c)(b[itex]\cdot[/itex]d)-(a[itex]\cdot[/itex]d)(b[itex]\cdot[/itex]c)

using the properties of the vector and triple products:

Homework Equations



a×(b×c)=b(a[itex]\cdot[/itex]c)-c(a[itex]\cdot[/itex]b)
a[itex]\cdot[/itex](b×c)=c[itex]\cdot[/itex](a×b)=b[itex]\cdot[/itex](c×a)

The Attempt at a Solution


I really don't know where to begin. I need to prove this identity simply so I can use it on a problem, and I know it CAN be proven using these identities from the triple products, but I'm lost on how to attempt such a proof.
 
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  • #2
jtleafs33 said:

Homework Statement


I need to prove the identity:

(a×b)[itex]\cdot[/itex](c×d)= (a[itex]\cdot[/itex]c)(b[itex]\cdot[/itex]d)-(a[itex]\cdot[/itex]d)(b[itex]\cdot[/itex]c)

using the properties of the vector and triple products:

Homework Equations



a×(b×c)=b(a[itex]\cdot[/itex]c)-c(a[itex]\cdot[/itex]b)
a[itex]\cdot[/itex](b×c)=c[itex]\cdot[/itex](a×b)=b[itex]\cdot[/itex](c×a)

The Attempt at a Solution


I really don't know where to begin. I need to prove this identity simply so I can use it on a problem, and I know it CAN be proven using these identities from the triple products, but I'm lost on how to attempt such a proof.
What have you tried?

Here's a hint:

Let u = c×d. Then use the scalar triple product, then substitute c×d back in for u, and see where that's leading you.

Added in Edit:

Putting in u, then applying the scalar triple product will simply let you switch a sclar product and a vector product, but that will allow you to get the desired result.
 
Last edited:
  • #3
Okay, I did that, but I still am not understanding.

Letting u=(c×d):

(a×b)[itex]\cdot[/itex](c×d) = u[itex]\cdot[/itex](a×b) = b[itex]\cdot[/itex](u×a)
=b[itex]\cdot[/itex]((c×da)
=-b[itex]\cdot[/itex](a×(c×d))
=b[itex]\cdot[/itex](c(a[itex]\cdot[/itex]d)-d(a[itex]\cdot[/itex]c))

And I don't know what to do with this either...
 
  • #4
jtleafs33 said:
Okay, I did that, but I still am not understanding.

Letting u=(c×d):

(a×b)[itex]\cdot[/itex](c×d) = u[itex]\cdot[/itex](a×b) = b[itex]\cdot[/itex](u×a)
=b[itex]\cdot[/itex]((c×da)
=-b[itex]\cdot[/itex](a×(c×d))
=b[itex]\cdot[/itex](c(a[itex]\cdot[/itex]d)-d(a[itex]\cdot[/itex]c))

And I don't know what to do with this either...
Use the distributive law to distribute the b vector. Don't forget, ad and ac are just scalars.
 
  • #5
Right, I've got that:

b[itex]\cdot[/itex]c(a[itex]\cdot[/itex]d)-b[itex]\cdot[/itex]d(a[itex]\cdot[/itex]c)

Can I go from here to:

(b[itex]\cdot[/itex]c)(a[itex]\cdot[/itex]d)-(b[itex]\cdot[/itex]d)(a[itex]\cdot[/itex]c) ?

But this is backwards of the identity I'm after...
 
  • #6
jtleafs33 said:
Okay, I did that, but I still am not understanding.

Letting u=(c×d):

(a×b)[itex]\cdot[/itex](c×d) = u[itex]\cdot[/itex](a×b) = b[itex]\cdot[/itex](u×a)
=b[itex]\cdot[/itex]((c×da)
=-b[itex]\cdot[/itex](a×(c×d))
The vector product is not commutative.

[itex]\textbf{r}\times\textbf{s}=-\textbf{s}\times\textbf{r}[/itex]
=b[itex]\cdot[/itex](c(a[itex]\cdot[/itex]d)-d(a[itex]\cdot[/itex]c))

And I don't know what to do with this either...

Fix that & you'll be OK .
 
  • #7
Thanks! I had the negative sign there but forgot to carry it through the last step. Thanks for the help!
 

1. How can I prove the quadruple product identity from triple product identities?

To prove the quadruple product identity from triple product identities, we can use the formula for the triple scalar product and the properties of vector operations such as distributivity and associativity.

2. Can you provide an example of the quadruple product identity from triple product identities?

One example of the quadruple product identity from triple product identities is (a x b) x (c x d) = (a x c) x (b x d) + (a x d) x (c x b) + (b x c) x (a x d).

3. What are the properties of vector operations used in proving the quadruple product identity?

The properties of vector operations used in proving the quadruple product identity include distributivity, associativity, and the triple scalar product formula.

4. Are there any other methods to prove the quadruple product identity from triple product identities?

Yes, there are other methods such as using the vector triple product, the Jacobi identity, or using geometric proofs.

5. Can the quadruple product identity from triple product identities be extended to higher order products?

Yes, the quadruple product identity from triple product identities can be extended to higher order products using the same principles and properties of vector operations.

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