Help-vector proving?

given


A = (a1,a2,a3),B = (b1,b2,b3) , C =(c1,c2,c3)





how to prove


A x ( B x C ) = ( A .C) B - ( A . B )C





*A,B,C are vectors

Help-vector proving?
A x ( B x C ) =





= (a₁e₁+ a₂e₂+ a₃e₃) x ((b₂c₃- b₃c₂)e₁+ (-b₁c₃+ b₃c₁)e₂+ (b₁c₂- b₂c₁)e₃) =





= (a₂(b₁c₂- b₂c₁) - a₃(-b₁c₃+ b₃c₁))e₁+ (-a₁(b₁c₂- b₂c₁) + a₃(b₂c₃- b₃c₂)e₂+ (a₁(-b₁c₃+ b₃c₁) - a₂(b₂c₃- b₃c₂))e₃=





= (a₂b₁c₂- a₂b₂c₁+ a₃b₁c₃- a₃b₃c₁)e₁+ (-a₁b₁c₂+ a₁b₂c₁+ a₃b₂c₃- a₃b₃c₂)e₂+ (-a₁b₁c₃+ a₁b₃c₁- a₂b₂c₃+ a₂b₃c₂)e₃=





= (a₂b₁c₂+ a₃b₁c₃- a₂b₂c₁- a₃b₃c₁)e₁+ (a₁b₂c₁+ a₃b₂c₃- a₁b₁c₂- a₃b₃c₂)e₂+ (a₁b₃c₁+ a₂b₃c₂- a₁b₁c₃- a₂b₂c₃)e₃=





= (a₁b₁c₁+ a₂b₁c₂+ a₃b₁c₃– a₁b₁c₁- a₂b₂c₁- a₃b₃c₁)e₁+ (a₁b₂c₁+ a₂b₂c₂+ a₃b₂c₃- a₁b₁c₂- a₂b₂c₂- a₃b₃c₂)e₂+ (a₁b₃c₁+ a₂b₃c₂+ a₃b₃c₃- a₁b₁c₃- a₂b₂c₃- a₃b₃c₃)e₃=





= (a₁b₁c₁+ a₂b₁c₂+ a₃b₁c₃)e₁+ (a₁b₂c₁+ a₂b₂c₂+ a₃b₂c₃)e₂+ (a₁b₃c₁+ a₂b₃c₂+ a₃b₃c₃)e₃– (a₁b₁c₁+ a₂b₂c₁+ a₃b₃c₁)e₁– (a₁b₁c₂+ a₂b₂c₂+ a₃b₃c₂)e₂– (a₁b₁c₃+ a₂b₂c₃+ a₃b₃c₃)e₃=





= (a₁c₁+ a₂c₂+ a₃c₃)(b₁e₁) + (a₁c₁+ a₂c₂+ a₃c₃)(b₂e₂) + (a₁c₁+ a₂c₂+ a₃c₃)(b₃e₃) – (a₁b₁+ a₂b₂+ a₃b₃)(c₁e₁) – (a₁b₁+ a₂b₂+ a₃b₃) (c₂e₂) – (a₁b₁+ a₂b₂+ a₃b₃) (c₃e₃) =





= (a₁c₁+ a₂c₂+ a₃c₃)(b₁e₁+ b₂e₂+ b₃e₃) – (a₁b₁+ a₂b₂+ a₃b₃)( c₁e₁+ c₂e₂+ c₃e₃) =





= (A .C) B - (A . B) C





Q.E.D.





-
Reply:♠ [ B x C ] = (d1, d2, d3) =


│`i````j````k`│


│b1``b2``b3│ = (b2*c3-b3*c2, c1*b3-b1*c3, b1*c2-b2*c1);


│c1``c2``c3│


♣ [A x [ B x C ]] =[AxD] =


│`i````j````k`│


│a1``a2``a3│ = (a2*d3-a3*d2, d1*a3-a1*d3, a1*d2-a2*d1);


│d1``d2``d3│


♦ a2*d3 -a3*d2 =


=a2*(b1*c2 -b2*c1) -a3*(c1*b3 -b1*c3) =


= (a2*c2 +a3*c3)*b1 –(a2*b2 +a3*b3)*c1 =


= (a1*c1+a2*c2 +a3*c3)*b1 - a1*c1*b1 –


–(a1*b1+a2*b2 +a3*b3)*c1 +a1*b1*c1 =


= (a1*c1+a2*c2 +a3*c3)*b1 –(a1*b1+a2*b2 +a3*b3)*c1;


♦ d1*a3 –a1*d3 =


= (b2*c3 -b3*c2)*a3 –a1*(b1*c2 -b2*c1) =


= (a3*c3 + a1*c1)*b2 –(a1*b1+a3*b3)*c2 =


= (a1*c1+a2*c2 +a3*c3)*b2 –(a1*b1+a2*b2 +a3*b3)*c2;


♦ a1*d2 -a2*d1 =


= a1*(c1*b3-b1*c3) –a2*(b2*c3 -b3*c2) =


= (a1*c1 +a2*c2)*b3 –(a1*b1+a2*b2)*c3 =


= (a1*c1+a2*c2 +a3*c3)*b3 –(a1*b1+a2*b2 +a3*b3)*c3;


♥ [A x [ B x C ]] = [AxD] =(A·C)*B – (A·B)*C; QED;