Applying Vector Cross Product Properties for (AxB).(CxD) Calculation

[Saladsamurai]

Homework Statement

I am following along in a book and in one line the author asserts that

(\mathbf{A}\times\mathbf{B})\cdot(\mathbf{C}\times\mathbf{D}) = (\mathbf{A}\cdot\mathbf{C})(\mathbf{B}\cdot\mathbf{D}) - (\mathbf{A}\cdot\mathbf{D})(\mathbf{B}\cdot\mathbf{C})\qquad(1)

Homework Equations

I believe that he is somehow using the rule that

\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})\qquad(2)

The Attempt at a Solution

Is this the only rule he is using to arrive at (1) ?
I am having trouble see how to implement this to arrive at the same result. Am I missing something painfully obvious?

 

[radou]

Did you try to do this in terms of components, i.e. using the definition of the cross and dot product? (Didn't try it myself, only suggesting.)

Edit: although it might get a little messy...

 

[rock.freak667]

Homework Equations

I believe that he is somehow using the rule that

\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})\qquad(2)

I would think they'd use this formula as well as this one


A\cdot (B \times C) = B \cdot (C \times A) = C \cdot (A \times B)

 

[Saladsamurai]

I would think they'd use this formula as well as this one


A\cdot (B \times C) = B \cdot (C \times A) = C \cdot (A \times B)

Ah yes, totally useful . Seeing as I have, in essence, a scalar triple product I would be hard pressed to start this without that rule

I have solved it now.

Thanks again!