VC3.8

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Given F=[y{\frac  {\partial f}{\partial z}}-z{\frac  {\partial f}{\partial y}}]i+[z{\frac  {\partial f}{\partial x}}-x{\frac  {\partial f}{\partial z}}]j+[x{\frac  {\partial f}{\partial y}}-y{\frac  {\partial f}{\partial x}}]k\, [Equation 1]

Also,here \nabla f={\frac  {\partial f}{\partial x}}i+{\frac  {\partial f}{\partial y}}j+{\frac  {\partial f}{\partial z}}k\, and r=xi+yj+zk\, Let this be Equation 2.

i). r\times \nabla f={\begin{vmatrix}i&j&k\\x&y&z\\{\frac  {\partial f}{\partial x}}&{\frac  {\partial f}{\partial y}}&{\frac  {\partial f}{\partial z}}\end{vmatrix}}\,

=F=[y{\frac  {\partial f}{\partial z}}-z{\frac  {\partial f}{\partial y}}]i+[z{\frac  {\partial f}{\partial x}}-x{\frac  {\partial f}{\partial z}}]j+[x{\frac  {\partial f}{\partial y}}-y{\frac  {\partial f}{\partial x}}]k\, which is equal to

=F\, by (1),Hence the first equality was proved.

ii).F\cdot r=(r\times \nabla f)\cdot r\, by the above proof (i).

=0\, [Since scalar tripple product vanishes if it contain two equal vectors]

iii). F\cdot \nabla f=(r\times \nabla f)\cdot \nabla f=0\,

[Since scalar tripple product vanishes if it contain two equal vectors]


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