# Prove Quadruple Product Identity from Triple Product Identities

1. Sep 28, 2012

### jtleafs33

1. The problem statement, all variables and given/known data
I need to prove the identity:

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

using the properties of the vector and triple products:

2. Relevant equations

a×(b×c)=b(a$\cdot$c)-c(a$\cdot$b)
a$\cdot$(b×c)=c$\cdot$(a×b)=b$\cdot$(c×a)

3. 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.

2. Sep 28, 2012

### SammyS

Staff Emeritus
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.

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: Sep 28, 2012
3. Sep 28, 2012

### 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×da)
=-b$\cdot$(a×(c×d))
=b$\cdot$(c(a$\cdot$d)-d(a$\cdot$c))

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

4. Sep 28, 2012

### SammyS

Staff Emeritus
Use the distributive law to distribute the b vector. Don't forget, ad and ac are just scalars.

5. Sep 28, 2012

### jtleafs33

Right, I've got that:

b$\cdot$c(a$\cdot$d)-b$\cdot$d(a$\cdot$c)

Can I go from here to:

(b$\cdot$c)(a$\cdot$d)-(b$\cdot$d)(a$\cdot$c) ?

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

6. Sep 28, 2012

### SammyS

Staff Emeritus
The vector product is not commutative.

$\textbf{r}\times\textbf{s}=-\textbf{s}\times\textbf{r}$
Fix that & you'll be OK .

7. Sep 28, 2012

### jtleafs33

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