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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.

enter image description here enter image description here

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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The solution of which part is not given? The images you're including look like they're nothing but solution. (But: There has to be a better way to do it than that coordinate soup! I get dizzy just looking at it.) – Henning Makholm Jul 5 '12 at 22:26
on the other page ... it says that if you replace those A's and B's ... you will get the desired result. Well ... i would be happy to know some other nice method. – Santosh Linkha Jul 5 '12 at 22:31
So your idea of specifying what you're talking about is to say "on the other page" and point to an image with 37 lines of symbol-dense calculations? Good luck with that. – Henning Makholm Jul 5 '12 at 22:43
No ... i just meant to show the methodology used in my book. Since it's physics quite unconventional methods are used than in mathematics – Santosh Linkha Jul 5 '12 at 23:03
You still haven't explained what it is you need help with. "The solution of on the other page is not given" makes no sense. You have not explained what "the desired result" is. You have not pointed to any part of the book's solution that you have a particular problem with. But it looks vaguely like your images cut off in the middle of the solution. Do you mean to say that you're missing the next page of the book? – Henning Makholm Jul 5 '12 at 23:11
up vote 5 down vote accepted

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).

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thanks for answering ... :) – Santosh Linkha Jul 5 '12 at 23:21

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