Answer to Question #132571 in Analytic Geometry for Promise Omiponle

Question

2. Assume that A=<ax,0,0>, B=<bx,by,bz>, and C=<cx,cy,cz>, and demonstrate that
the following vector identity holds:
Ax(BxC) = (A.C)B-(A.B)C
Note that if you replace the subscript "x" by "i", "y" by j, and "z" by k, and then cycle
x->y->z->x appropriately, you've proven the identity is true in general.

Answer 1

$(1)\; B \times C = \vec i (b_xc_x-b_zc_y)+\vec j(b_zc_x-b_xc_z)+\vec k(b_xc_y-b_yc_x)\\ (2)\;A \times (B \times C) = \vec i(0-0)+\vec j (0-a_x(b_xc_y-b_yc_x)) + \\ +\vec k (a_x(b_zc_x-b_xc_z)-0)= \\ =\vec j (a_xb_yc_x-a_xb_xc_y)+\vec k (a_xb_zc_x-a_xb_xc_z)\\ (3)\;A*C=a_xc_x\\ (4)\;(A*C)*B=\vec i (a_xb_xc_x) + \vec j (a_xb_yc_x)+\vec k (a_xb_zc_x)\\ (5)\; A*B=a_xb_x\\ (6)\;(A*B)*C=\vec i (a_xb_xc_x)+\vec j (a_xb_xc_y)+\vec k (a_xb_xc_z)\\$

Then substract the expression obtained in (6) from expression obtained in (4).

$(A*C)*B - (A*B)*C=\vec j (a_xb_yc_x-a_xb_xc_y)+\vec k (a_xb_zc_x-a_xb_xc_z)$

This is equal to the expression obtained in (2). It means that$A\times (B\times C)=(A*C)*B-(A*B)*C$