The discussion revolves around proving the equation (A×B) . (B×A) + (A . B)^2 = A^2 . B^2, which involves vector operations including the cross product and dot product. The subject area is vector algebra.
The discussion is active, with participants providing insights into the properties of vector products and questioning the assumptions made in the proof. There is no explicit consensus, but various interpretations and clarifications are being explored.
Participants note potential issues with notation and definitions, particularly regarding the representation of the dot product and the implications of using A^2 versus A . A.
Shouldn't the right side be |A|2 |B|2? The left side is a scalar (i.e., a number), so the right side needs to be a scalar as well.JasonHathaway said:Homework Statement
Proof that (A×B) . (B×A) + (A . B)^2= A^2 . B^2
JasonHathaway said:Homework Equations
A×(B×C)=(A . C)B - (A . B)C
The Attempt at a Solution
Assuming K=(A×B)
K . (B×A) + (A . B)^2 = A^2 . B^2
B . (A×K) + (A . B)^2 = A^2 . B^2
B . [A×(A×B)] + (A . B)^2 = A^2 . B^2
B . [(A . B)A - (A . A)B] + (A . B)(A . B) = A^2 . B^2
(A . B)(B . A) - (A . A)(B . B) + (A . B)(A . B) = A^2 . B^2
Strictly speaking, no, but I understand what you're trying to say. The product of a vector with itself (which you write as A2) is normally written as ##A \cdot A## or ##A \times A##, depending on which kind of product you mean. Another argument against A2 is that it can't be extended to, say, A3, because ##A \cdot A \cdot A## isn't defined. (The first dot product produces a scalar, which can't be dotted with a vector.)JasonHathaway said:Isn't the dot product of a vector (A for example) with itself is equal to A^2?
JasonHathaway said:I see that the idea behind this proof is to eliminate the (A . B)(B . A) and (A . B)(A . B)