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Question goes as

If $\vec A$ and $\vec B$ are invariant under rotation, the prove that $ \vec A \times \vec B $ is also invariant.

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However solution of on the other page is not given. Says that if you replace A and B with A' and B' and i,j,k with i', j', k' you will get the desired result.
I tried to change $ (A_2 B_3 - A_3 B_2 ) \hat i$ into $ A' $ and $ B'$ but I am getting zero.I would like to have a hint on this.

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    I need to show that $ \vec A \times \vec B = \vec A' \times \vec B' $ for given transformation rule $ x = a_{11}x' + a_{12}y' + a_{13}z'\\ y = a_{21}x' + a_{22}y' + a_{23}z'\\ z = a_{31}x' + a_{32}y' + a_{33}z'$2012-07-05

1 Answers 1

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I'm not exactly sure what you are asking, however the following may be useful as a less verbose way of obtaining the same result.

The key fact is that the cross product of $A,B$ is the unique element $A\times B$ such that $\langle x, A\times B \rangle = \det \begin{bmatrix} A & B & x\end{bmatrix}$, $\forall x$.

Let $Q$ be a rotation (ie, $Q^TQ = I$), then using the properties of $\det$ we have $\det \begin{bmatrix} A & B & x\end{bmatrix} =\det Q^T Q \det \begin{bmatrix} A & B & x\end{bmatrix} = \det Q \det \begin{bmatrix} QA & QB & Qx\end{bmatrix},$ from which we obtain $\langle x, A\times B \rangle = \det Q \langle Qx, QA\times QB \rangle = \langle x, (\det Q) Q^T(QA\times QB) \rangle$. Since this is true for all $x$, we have $A\times B = (\det Q) Q^T(QA\times QB)$, or $Q(A\times B) = (\det Q) QA\times QB.$

Thus, if $Q$ is a proper rotation ($\det Q = +1$), you have $Q(A\times B) = QA\times QB$ (ie, the cross product is invariant under proper rotations).

Your question posits that both $A,B$ are invariant under rotation (which seems like a fairly restrictive condition), in which case $A = QA$, $B= QB$, from which it follows that $Q(A\times B) = A\times B$, ie, $A\times B$ is invariant under rotation too (assuming a proper rotation, of course).