Where $A$ and $B$ are vectors, and $\times$ is the cross product operator. I was able to get $A(A \cdot B) - B$ using the vector triple product, but it doesn't look like a simplified version to me.

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    $\begingroup$ This does not make sense to me. $\vec A \cdot(\vec A \cdot \vec B)= \vec A \cdot \text{scalar}$ which is ridiculous $\endgroup$ – user21436 Feb 14 '12 at 2:44
  • $\begingroup$ Do you want $\times$ to be the cross product? Or are you trying to get $A\cdot (A\times B)$, where $\cdot$ is dot product and $\times$ is cross product? $\endgroup$ – Joe Johnson 126 Feb 14 '12 at 2:50
  • $\begingroup$ I'm sorry, I meant x is the cross product. $\endgroup$ – Aeon Feb 14 '12 at 2:54

Firstly note that, the vector triple product you want to evaluate is actually: $$A \times (A \times B)=(A\cdot B)A-(|A|^2)B$$

And this is, in fact, the most simplified expression that we can arrive at with what little you have given us!

Try to prove a more general following result:

$$A \times (B \times C)=(A \cdot C)B-(A \cdot B)C$$

Hint: Take, $A=a_1\hat i+a_2 \hat j+a_3 \hat k$ and so on for other vectors and do brute force computation.


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