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Prove Quadruple Product Identity from Triple Product Identities

  • Thread starter jtleafs33
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  • #1
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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.
 

Answers and Replies

  • #2
SammyS
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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
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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
SammyS
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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
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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
SammyS
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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
28
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Thanks! I had the negative sign there but forgot to carry it through the last step. Thanks for the help!!
 

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