Using index notation prove the following identities among ve

Using index notation prove the following identities among vectors A,B,C,D

a- (AXB)X(CXD)=[A.(CXD)]B-[B.(CXD]A

Solution

A has A1,A2,A3 components similarly others(B,C,D)

so A x B = |i j k;A1 A2 A3;B1 B2 B3| = i(A2B3 -A3B2) - j(A1B3 - A3B1) + k(A1B2 -A2B1)

similarly CxD = |i j k;C1 C2 C3;D1 D2 D3| = i(C2D3-C3D2) -j(C1D3 -C3D1) + k(C1D2 -C2D1)

L.H.S = (AxB) x(CxD) = |i j k;A2B3-A3B2 A1B3-A3B1 A1B2-A2B1;C2D3-C3D2 C1D3-C3D1 C1D2-C2D1|

=i((A1B3-A3B1)(C1D2 -C2D1) - (A1B2 -A2B1)(C1D3 -C3D1)) -j((A2B3-A3B2)(C1D2-C2D1)-(A1B2-A2B1)(C2D3-C3D2)) +k((A2B3-A3B2)(C1D3-C3D1) -(C1D3-C3D1)(A2B3-A3B2))

R.H.S:

B.D =B1D1+B2D2 +B3D3;B.C=B1C1 +B2C2 +B3C3

Ax(B.D) =|i j k;A1 A2 A3;B1D1 B2D2 B3D3| =i(A2B3D3 -A3B2D2) -j(A1B3D3-A3B1D1) +k(A1B2D2 -A2B1D1)

Ax(B.C) =|i j k;A1 A2 A3;B1C1 B2C2 B3C3| =i(A2B3C3 -A3B2C2) -j(A1B3C3-A3B1C1) +k(A1B2C2 -A2B1C1)

(Ax B.D)C - (Ax(B.C))D= (A2B3D3 -A3B2D2)C1 -(A1B3D3-A3B1D1)C2 +(A1B2D2 -A2B1D1)C3 + (A2B3C3 -A3B2C2)D1 -(A1B3C3-A3B1C1)D2 +(A1B2C2 -A2B1C1)D3

so this statement isnt true as clerified above