Prove the identity between vectors A, B, C

Question

2(A.B)(A.C)=(A.A)(B.C)+A^2.BC

A^2 is neither a dot or cross product

it's used in optics, to calculate the aberrations of a non-axially symmetric system

Answer 1

It appears to me from the context like perhaps these $A,B,C$ are two dimensional vector representations of complex numbers and that the dot product is the usual dot product while the other product occurring is the usual product of complex or real numbers... That is to say, if we have two vectors $[u,v]$ and $[x,y]$ we have $[u,v]\cdot [x,y]=ux+vy$ and we have $[u,v][x,y]=[ux-vy,uy+vx]$

Indeed, if we have $A=[a_r,a_i],B=[b_r,b_i],C=[c_r,c_i]$ on the LHS we would have:

$2(A\cdot B)(A\cdot C) = 2(a_rb_r+a_ib_i)(a_rc_r+a_ic_i) = 2a_r^2b_rc_r+2a_ra_ib_rc_i+2a_ra_ib_ic_r+2a_i^2b_ic_i$

On the RHS we have:

$(A\cdot A)(B\cdot C) + A^2\cdot BC$ $ = (a_r^2+a_i^2)(b_rc_r+b_ic_i) + [a_r^2-a_i^2,2a_ra_i]\cdot [b_rc_r-b_ic_i,b_rc_i+b_ic_r]$

$=(a_r^2+a_i^2)(b_rc_r+b_ic_i)+(a_r^2-a_i^2)(b_rc_r-b_ic_i) + (2a_ra_i)(b_rc_i+b_ic_r)$

Noting some nice cancellations, this simplifies as:

$2a_r^2b_rc_r+2a_i^2b_ic_i+2a_ra_ib_rc_i+2a_ra_ib_ic_r$ same as before.