Question

Show that (ax(bxc))xc=(a.c)(bxc) and (axb).(axc)+(a.b)(a.c)=(a.a)(b.c)​

Answer 1

Answer:

A= Ax i+Ay j+Az k

B= Bx i+By j+Bz k

C= Cx i+Cy j+Cz k

Now start with either L HS or R HS

Lets start with L.H.S

L.H.S = (A x(B x C))

First we will solve for (B x C)

(B x C) = ( ByCz - BzCy)i +(BzCx - BxCz)j+(BxCy - ByCx)k

Now we will solve for (A x(B x C))

(A x(B x C)) = (AyBxCy-AyByCx-AzBzCx+AzBxCz)i+(AzByCz-AzBzCy-AxBxCy+AxByCx)j+(AxBzCx-AxBxCz-AyByCz+AyBzCy)k ——-(1)

Now we will solve for R.H.S

R.H.S =B(A.C)-C(A.B)

=B.( AxCx +AyCy+AzCz )-C.(AxBx+AyBy+AzBz)

=(AyBxCy-AyByCx-AzBzCx+AzBxCz)i+(AzByCz-AzBzCy-AxBxCy+AxByCx)j+(AxBzCx-AxBxCz-AyByCz+AyBzCy)k —-(2)

Since equation 1 = equation 2

hence L.H.S =R.H.S

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