Under what condition does $A^T(B \times C) + (B\times C)^T A = 2A^T(B \times C)$, A,B,C vectors

Question

In my classical mechanics text book there is a formula that states

$(\dot r_c + \omega_i \times d_i)^T (\dot r_c + \omega_i \times d_i)$

give rise to

$\dot r_c^T \dot r_c + 2\dot r_c^T(\omega_i \times d_i) + (\omega_i \times d_i)^T(\omega_i \times d_i)$

I don't understand how they were able to collapse the middle express like that

Can someone give a condition for which

$A^T(B \times C) + (B\times C)^T A = 2A^T(B \times C)$

Answer 1

It holds without condition for any three vectors $A$, $B$, $C$.

Note that $(B \times C)^T A$ is a scalar quantity, as is $A^T(B \times C)$. Therefore each is symmetric in the sense of being equal to its own transpose. We thus have

$(B \times C)^T A = ((B \times C)^T A)^T$ $ = A^T ((B \times C)^T)^T = A^T(B \times C), \tag{1}$

whence

$A^T(B \times C) + (B \times C)^TA = 2A^T(B \times C), \tag{2}$

and that does it!