prove that grad [r a b]=a×b
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I assume that by ar the OP means a⋅r, where r is the position vector. Then with a=(ax,ay,az)T and r=(x,y,z)T, we have
a⋅r=axx+ayy+azz;(1)
and since
∇(a⋅r)=(∂a⋅r∂x,∂a⋅r∂y,∂a⋅r∂z)T,(2)
we merely need observe that
∂a⋅r∂x=∂(axx+ayy+azz)∂x=ax,(3)
with the analogous results holding for partial differentiation with respect to y and z. Thus
∇(a⋅r)=(ax,ay,az)T,(4)
as was to be shown. QED.
Hope this helps
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