Question #132571
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.
1
Expert's answer
2020-09-15T13:51:36-0400

(1)  B×C=i(bxcxbzcy)+j(bzcxbxcz)+k(bxcybycx)(2)  A×(B×C)=i(00)+j(0ax(bxcybycx))++k(ax(bzcxbxcz)0)==j(axbycxaxbxcy)+k(axbzcxaxbxcz)(3)  AC=axcx(4)  (AC)B=i(axbxcx)+j(axbycx)+k(axbzcx)(5)  AB=axbx(6)  (AB)C=i(axbxcx)+j(axbxcy)+k(axbxcz)(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).

(AC)B(AB)C=j(axbycxaxbxcy)+k(axbzcxaxbxcz)(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 thatA×(B×C)=(AC)B(AB)CA\times (B\times C)=(A*C)*B-(A*B)*C


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS