Prove Quadruple Product Identity from Triple Product Identities

[jtleafs33]

Homework Statement

I need to prove the identity:

(a×b)\cdot(c×d)= (a\cdotc)(b\cdotd)-(a\cdotd)(b\cdotc)

using the properties of the vector and triple products:

Homework Equations

a×(b×c)=b(a\cdotc)-c(a\cdotb)
a\cdot(b×c)=c\cdot(a×b)=b\cdot(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.

 

[SammyS]

Homework Statement

I need to prove the identity:

(a×b)\cdot(c×d)= (a\cdotc)(b\cdotd)-(a\cdotd)(b\cdotc)

using the properties of the vector and triple products:

Homework Equations

a×(b×c)=b(a\cdotc)-c(a\cdotb)
a\cdot(b×c)=c\cdot(a×b)=b\cdot(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.

 

[jtleafs33]

Okay, I did that, but I still am not understanding.

Letting u=(c×d):

(a×b)\cdot(c×d) = u\cdot(a×b) = b\cdot(u×a)
=b\cdot((c×d)×a)
=-b\cdot(a×(c×d))
=b\cdot(c(a\cdotd)-d(a\cdotc))

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

 

[SammyS]

Okay, I did that, but I still am not understanding.

Letting u=(c×d):

(a×b)\cdot(c×d) = u\cdot(a×b) = b\cdot(u×a)
=b\cdot((c×d)×a)
=-b\cdot(a×(c×d))
=b\cdot(c(a\cdotd)-d(a\cdotc))

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

Use the distributive law to distribute the b vector. Don't forget, a∙d and a∙c are just scalars.

 

[jtleafs33]

Right, I've got that:

b\cdotc(a\cdotd)-b\cdotd(a\cdotc)

Can I go from here to:

(b\cdotc)(a\cdotd)-(b\cdotd)(a\cdotc) ?

But this is backwards of the identity I'm after...

 

[SammyS]

Okay, I did that, but I still am not understanding.

Letting u=(c×d):

(a×b)\cdot(c×d) = u\cdot(a×b) = b\cdot(u×a)
=b\cdot((c×d)×a)
=-b\cdot(a×(c×d))

The vector product is not commutative.

\textbf{r}\times\textbf{s}=-\textbf{s}\times\textbf{r}

=b\cdot(c(a\cdotd)-d(a\cdotc))

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

Fix that & you'll be OK .

 

[jtleafs33]

Thanks! I had the negative sign there but forgot to carry it through the last step. Thanks for the help!