( 1 ) B × C = i ⃗ ( b x c x − b z c y ) + j ⃗ ( b z c x − b x c z ) + k ⃗ ( b x c y − b y c x ) ( 2 ) A × ( B × C ) = i ⃗ ( 0 − 0 ) + j ⃗ ( 0 − a x ( b x c y − b y c x ) ) + + k ⃗ ( a x ( b z c x − b x c z ) − 0 ) = = j ⃗ ( a x b y c x − a x b x c y ) + k ⃗ ( a x b z c x − a x b x c z ) ( 3 ) A ∗ C = a x c x ( 4 ) ( A ∗ C ) ∗ B = i ⃗ ( a x b x c x ) + j ⃗ ( a x b y c x ) + k ⃗ ( a x b z c x ) ( 5 ) A ∗ B = a x b x ( 6 ) ( A ∗ B ) ∗ C = i ⃗ ( a x b x c x ) + j ⃗ ( a x b x c y ) + k ⃗ ( a x b x c z ) (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)\\ ( 1 ) B × C = i ( b x c x − b z c y ) + j ( b z c x − b x c z ) + k ( b x c y − b y c x ) ( 2 ) A × ( B × C ) = i ( 0 − 0 ) + j ( 0 − a x ( b x c y − b y c x )) + + k ( a x ( b z c x − b x c z ) − 0 ) = = j ( a x b y c x − a x b x c y ) + k ( a x b z c x − a x b x c z ) ( 3 ) A ∗ C = a x c x ( 4 ) ( A ∗ C ) ∗ B = i ( a x b x c x ) + j ( a x b y c x ) + k ( a x b z c x ) ( 5 ) A ∗ B = a x b x ( 6 ) ( A ∗ B ) ∗ C = i ( a x b x c x ) + j ( a x b x c y ) + k ( a x b x c z )
Then substract the expression obtained in (6) from expression obtained in (4).
( A ∗ C ) ∗ B − ( A ∗ B ) ∗ C = j ⃗ ( a x b y c x − a x b x c y ) + k ⃗ ( a x b z c x − a x b x c z ) (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) ( A ∗ C ) ∗ B − ( A ∗ B ) ∗ C = j ( a x b y c x − a x b x c y ) + k ( a x b z c x − a x b x c z )
This is equal to the expression obtained in (2). It means thatA × ( B × C ) = ( A ∗ C ) ∗ B − ( A ∗ B ) ∗ C A\times (B\times C)=(A*C)*B-(A*B)*C A × ( B × C ) = ( A ∗ C ) ∗ B − ( A ∗ B ) ∗ C
Comments