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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