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Vectors product proof

  1. Mar 6, 2014 #1
    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. jcsd
  3. Mar 6, 2014 #2

    Mark44

    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.
     
  4. Mar 6, 2014 #3
    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)
     
  5. Mar 6, 2014 #4

    Mark44

    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.)
     
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