Posted by Wellness on Friday, October 30, 2009 at 10:38am.
A Ã— B = (Ax Ã® + Ay Äµ + Az Ë†k) Ã— (Bx Ã® + By Äµ + Bz Ë†k)
A Ã— B = Ax Ã® Ã— Bx Ã® + Ax Ã® Ã— By Äµ + Ax Ã® Ã— Bz Ë†k
+ Ay Äµ Ã— Bx Ã® + Ay Äµ Ã— By Äµ + Ay Äµ Ã— Bz Ë†k
+ Az Ë†k Ã— Bx Ã® + Az Ë†k Ã— By Äµ + Az Ë†k Ã— Bz Ë†k
A Ã— B = AxBy Ë†k âˆ’ AxBz Äµ
âˆ’ AyBx Ë†k + AyBz Ã®
+ AzBx Äµ âˆ’ AzBy Ã®
A Ã— B = (AyBz âˆ’ AzBy) Ã® + (AzBx âˆ’ AxBz) Äµ + (AxBy âˆ’ AyBx) Ë†k
When you dot this with C= Cxi + Cyj + Czk the dot products ii=jj=kk=1; while the dot products ij=ik=jk=kj=0.
B Ã— C =(ByCz âˆ’ BzCy) Ã® + (BzCx âˆ’ BxCz) Äµ + (BxCy âˆ’ ByCx) Ë†k
When you dot this with A= Axi + Ayj + Azk the dot products ii=jj=kk=1; while the dot products ij=ik=jk=kj=0.
NOTICE WHEN THIS IS DONE :
Ax(ByCz âˆ’ BzCy) +Ay(BzCx âˆ’ BxCz)+ Az(BxCy âˆ’ ByCx) = (AyBz âˆ’ AzBy) Cx + (AzBx âˆ’ AxBz)Cy +(AxBy âˆ’ AyBx)Cz
AxByCz - AxBzCy + AyBzCx - AyBxCz + AzBxCy - AzByCx =
AyBzCx âˆ’ AzByCx + AzBxCy âˆ’ AxBzCy + AxByCzâˆ’ AyBxCz
So each term on the left side has an identical term on the right side of the equal sign and the identity is proven
QED