# Vectors product proof

1. Mar 6, 2014

### JasonHathaway

1. The problem statement, all variables and given/known data

Proof that (A×B) . (B×A) + (A . B)^2= A^2 . B^2

2. Relevant equations

A×(B×C)=(A . C)B - (A . B)C

3. 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

2. Mar 6, 2014

### Staff: Mentor

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.

3. Mar 6, 2014

### JasonHathaway

Isn't the dot product of a vector (A for example) with itself is equal to A^2?

I see that the idea behind this proof is to eliminate the (A . B)(B . A) and (A . B)(A . B)

4. Mar 6, 2014

### Staff: Mentor

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.)