Suppose one try to introduce a new product of two vectors as $C = A\operatorname{XX}B$, where $A,B,C$ are all vectors. Now it is defined as \begin{align} C_x &=A_y B_z + A_z B_y\\ C_y &= A_z B_x+A_x B_z\\ C_z &=A_x B_y+A_y B_x \end{align} choose the simple case where vector $A=(A_x,A_y,0)$ and vector $B=(B_x,B_y,0)$. Can $C$ be considered as a vector under orthogonal transformation? ($A_x$ means $A$ suffix $x$ and applies for the rest).
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Have you tried performing the requested computation (using the simple values, or even simpler ones like $(1,0,0)$ and $(0,1,0)$) ? what does the result tell you when you perform some rotation around the $z$-axis ? – lmsteffan Oct 14 '15 at 07:27
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i could not undertand – rickriordan Oct 14 '15 at 08:11
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What do you recall about orthogonal transformations with respect to vectors? and what is the simplest kind of orthogonal transformation ?... – lmsteffan Oct 14 '15 at 08:39
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Let $A$ and $B$ be two (unit) vectors along the $x$- and $y$-axes, and $C$ their "product" according to your definition : then
$$C = (1,0,0) \operatorname{XX}\, (0,1,0) = (0, 0, 1)$$
which is a unit vector lying along the $z$-axis.
Now rotate the vectors $A$ and $B$ by $\frac{\pi}{2}$ (i.e. $90°$) around the $z$-axis (or equivalently, perform a change of basis) : the equation above becomes
$$C = (0,1,0) \operatorname{XX}\, (-1,0,0) = (0, 0, -1)$$
which means that the "product vector" has reversed direction along the $z$-axis even though that axis was left untouched by the rotation.
The result is even more striking if you perform instead a $\frac{\pi}{4}$ rotation : in that case $C$ becomes the null vector $(0,0,0)$ so not only its direction but also its length has changed.
These changes which are not induced by the rotation (an orthogonal transformation) prove that $\operatorname{XX}$ is not a true "vector product".
lmsteffan
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