Is Lagrange's Theorem the Key to Solving This Vector Equation?

[rroy81]

Proof using lagrange!

Homework Statement

(A x B) . (C x D) = (A . B) (C . D) - (A . D) (B . C)

Homework Equations

This is all that's given..I am sort of lost on how to proof this. Spent 4hrs +

The Attempt at a Solution

Completely lost and don't know where to start

 

[I like Serena]

Homework Statement

(A x B) . (C x D) = (A . B) (C . D) - (A . D) (B . C)

Homework Equations

This is all that's given..I am sort of lost on how to proof this. Spent 4hrs +

The Attempt at a Solution

Completely lost and don't know where to start

Welcome to PF, rroy!

What did you try?
Which relevant equations do you have (and are you allowed to use)?

I presume you are allowed to use the definitions of the cross product and dot product.
Did you use those?
Any other identities?
Are you allowed to use identities that you can find with wikipedia?

 

[rroy81]

So turns out the instructor had typo in the problem. it should be
(A x B) . (C x D) = (A . C) (B . D) - (A . D) (B . C)

The way I started was let A = {a1, a2, a3 } and so on for the remaining (B, C, D). I assume it is a matrice and will proceed as a dot product. Than I am unsure if this right, if so than I stuck.

 

[I like Serena]

What do you think is representing a matrix?

Do you know what the definition for the cross product for vectors is?

 

[I like Serena]

Actually, what you are supposed to proof is Lagrange's identity.
See for instance here.

 

[rroy81]

I like Serena,

My understanding to cross product is using the matrix, am I right? (ith, jth and kth terms)
and as I have given the terms A=(a1, a2, a3) B=(b1, b2, b3) and so on..than to solve the cross product. I simply took the crossed out the first column and row and than did the cross product of remaining 4 and did the same with 2 column and row and so on

 

[I like Serena]

Ah okay.
Yes, that is the standard way to go.
If you multiply out the cross products and dot products, you should find the equality.
It is a bit of work though.