Reconcile Geometric Form of Cross Product with Algebric Form

[noir1993]

Basically, we have to prove that A X B is equal to ABsinθ

Now a crucial step towards the proof is proving that
(AXB).(AXB) is equal to (AB)² - (A.B)²

After that it is fairly simple. But unfortunately, I can not prove the identity. I've tried expanding it into components but things are becoming messy. Any idea on how to prove this identity (namely : (AXB).(AXB)= (AB)² - (A.B)² ) ?

Bold face indicates a Vector

 

[LCKurtz]

Basically, we have to prove that A X B is equal to ABsinθ

Now a crucial step towards the proof is proving that
(AXB).(AXB) is equal to (AB)² - (A.B)²

After that it is fairly simple. But unfortunately, I can not prove the identity. I've tried expanding it into components but things are becoming messy. Any idea on how to prove this identity (namely : (AXB).(AXB)= (AB)² - (A.B)² ) ?

Bold face indicates a Vector

I will assume you know the identities $A \cdot B \times C = A \times B \cdot C$ (interchange of dot and cross) and $A\times (B\times C) = (A\cdot C)B-(A\cdot C)B$.

$$\|A\times B\|^2=(A\times B)\cdot (A\times B) = A\cdot (B\times (A\times B))$$ $$
=A\cdot ((B\cdot B)A - (B\cdot A)B)= \|B\|^2\|A\|^2-(B\cdot A)(B\cdot A)$$

 

[noir1993]

I will assume you know the identities $A \cdot B \times C = A \times B \cdot C$ (interchange of dot and cross) and $A\times (B\times C) = (A\cdot C)B-(A\cdot C)B$.

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Hmm, thanks, :). That was short and sweet.

By the way, I guess you made a small typo(in the second term of the vector triple product.), it will be AX(BXC) = B(A.C)-C(A.B).

 

[LCKurtz]

Hmm, thanks, :). That was short and sweet.

By the way, I guess you made a small typo(in the second term of the vector triple product.), it will be AX(BXC) = B(A.C)-C(A.B).

Yes that was a typo. Unfortunately it's too late to correct it but that doesn't matter now anyway.