Proving the Mixed Product Formula for Vectors in R3

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Homework Help Overview

The problem involves proving the mixed product formula for vectors in R3, specifically showing that (A×B) . [(B×C)×(C×A)] equals (A,B,C)^2, where A, B, and C are vectors. The discussion centers around vector identities and properties of the cross and dot products.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to manipulate the expression using vector identities, particularly the vector triple product. Questions arise regarding specific terms vanishing and the implications of those results on the overall proof.

Discussion Status

The discussion is ongoing, with participants exploring various aspects of the problem. Some have identified terms that vanish, while others are questioning how these relate to the final expression. There is no explicit consensus yet, but productive lines of reasoning are being developed.

Contextual Notes

Participants are considering the implications of the properties of the dot and cross products, particularly in relation to the triple scalar product. There is a focus on ensuring all assumptions and definitions are clear as they work through the proof.

JasonHathaway
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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]
 
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JasonHathaway said:

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} \ ?##
 
(b×c) . C = (c×b) . C = (c×c) . B = (0) . B = 0
 
JasonHathaway said:
(b×c) . C = (c×b) . C = (c×c) . B = (0) . B = 0

Yes.

Does that get you to the result?
 
(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?
 
JasonHathaway said:
(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 ?
 
I understand now.

Thank you very much.
 

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