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By definition of magnitude of cross product:

$\| \mathbf{F} \times \mathbf{r} \|= \| \mathbf{F} \|\ \| \mathbf{r} \| \sin (\mathbf{F},\mathbf{r}) \tag1$

$\| \mathbf{r} \times \mathbf{F} \|= \| \mathbf{r} \|\ \| \mathbf{F} \| \sin (\mathbf{r},\mathbf{F})\tag2$

By $(1)$ and $(2)$: $\| \mathbf{F} \times \mathbf{r} \|=-\| \mathbf{r} \times \mathbf{F} \|\tag3$

$\| \mathbf{F} \times \mathbf{r} \|$ and $\| \mathbf{r} \times \mathbf{F} \|$ are modes of vectors.

$\| \mathbf{F} \times \mathbf{r} \|$ and $\| \mathbf{r} \times \mathbf{F} \|$ are positive.

Then how can equation $3$ be true?

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    $\begingroup$ In (1) and (2) both, one takes the interior angle between $F$ and $r$ and hence the expressions are the same. Otherwise you'll have to take the absolute value. $\endgroup$ – Nephente Mar 17 '20 at 7:06
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    $\begingroup$ The magnitude of a vector cannot be negative so if (3) were true both sides would have to equal zero. $\endgroup$ – bemjanim Mar 17 '20 at 7:09
  • $\begingroup$ @Nephente: should we take the mode of $\sin θ$? $\endgroup$ – N.G.Tyson Mar 17 '20 at 7:12
  • $\begingroup$ I've never heard mode outside of statistics before, but yes, take the absolute value $|\sin|$. $\endgroup$ – Nephente Mar 17 '20 at 7:41
  • $\begingroup$ There is no need (to take the absolute value of the sin), because the angle between the vectors is (by definition) smaller than $\pi$. Moreover the two expressions are the same. $\endgroup$ – lcv Mar 17 '20 at 7:45
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We have \begin{align} \vec a \times \vec b &= \sum_{i,j=1}^3 a_i b_j \vec e_k \epsilon_{ijk} = -\sum_{i,j=1}^3 b_j a_i \vec e_k \epsilon_{jik} = - \vec b \times \vec a \end{align} where in the second equation we used $ \epsilon_{ijk} = - \epsilon_{jik} $. Taking the absolute value on both sides, we obtain $$ |\vec a \times \vec b| = |\vec b \times \vec a| $$ In conclusion: Yes, the magnitude of the cross product is commutative.

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