Proving the Mixed Product Formula for Vectors in R3

[JasonHathaway]

Homework Statement

Prove that (A×B) . [(B×C)×(C×A)]=(A,B,C)^2
where A, B, C are vectors in R3.

Homework Equations

W×(U×V)=(W . V) U - (W × U) V

The Attempt at a Solution

Assuming K=(A×B), M=(B×C):
K . [M×(C×A)]
K . [(M . A) C - (M . C) A]
[(M . A)(K . C) - (M . C)(K . A)]

Then:
[(B×C) . A] [(A×B) . C] - [(B×C) . C] [(A×B) . A]

 

[SammyS]

Homework Statement

Prove that (A×B) . [(B×C)×(C×A)]=(A,B,C)^2
where A, B, C are vectors in R3.

Homework Equations

W×(U×V)=(W . V) U - (W × U) V

The Attempt at a Solution

Assuming K=(A×B), M=(B×C):
K . [M×(C×A)]
K . [(M . A) C - (M . C) A]
[(M . A)(K . C) - (M . C)(K . A)]

Then:
[(B×C) . A] [(A×B) . C] - [(B×C) . C] [(A×B) . A]

You're almost there!

What is $\displaystyle \left(\vec{B}\times\vec{C}\right)\cdot\vec{C} \ ?$

 

[JasonHathaway]

(b×c) . C = (c×b) . C = (c×c) . B = (0) . B = 0

 

[SammyS]

(b×c) . C = (c×b) . C = (c×c) . B = (0) . B = 0

Yes.

Does that get you to the result?

 

[JasonHathaway]

(A×B) . A will vanish as well, and I'll end up with [(B×C) . A] and [(A×B) . C] which are equal.

But how - by algebra - can they be equal to (A,B,C)^2?

 

[SammyS]

(A×B) . A will vanish as well, and I'll end up with [(B×C) . A] and [(A×B) . C] which are equal.

But how - by algebra - can they be equal to (A,B,C)^2?

What is (A,B,C) ?

Isn't it the triple scalar product ?

 

[JasonHathaway]

I understand now.

Thank you very much.