Prove that A.(BxC) = (AxB).C

Answer 1

A × B = (Ax î + Ay ĵ + Az ˆk) × (Bx î + By ĵ + Bz ˆk)

A × B = Ax î × Bx î + Ax î × By ĵ + Ax î × Bz ˆk
+ Ay ĵ × Bx î + Ay ĵ × By ĵ + Ay ĵ × Bz ˆk
+ Az ˆk × Bx î + Az ˆk × By ĵ + Az ˆk × Bz ˆk

A × B = AxBy ˆk − AxBz ĵ
− AyBx ˆk + AyBz î
+ AzBx ĵ − AzBy î

A × B = (AyBz − AzBy) î + (AzBx − AxBz) ĵ + (AxBy − AyBx) ˆk

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 × C =(ByCz − BzCy) î + (BzCx − BxCz) ĵ + (BxCy − ByCx) ˆk

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 − BzCy) +Ay(BzCx − BxCz)+ Az(BxCy − ByCx) = (AyBz − AzBy) Cx + (AzBx − AxBz)Cy +(AxBy − AyBx)Cz

AxByCz - AxBzCy + AyBzCx - AyBxCz + AzBxCy - AzByCx =
AyBzCx − AzByCx + AzBxCy − AxBzCy + AxByCz− 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

Answer 2

So, whats the property of it