Question Number 4 by user1 last updated on 25/Jan/15
$$\mathrm{If} \\ $$$$\mathrm{F}\left({y}\frac{\partial{f}}{\partial{z}}−{z}\frac{\partial{f}}{\partial{y}}\right)\boldsymbol{\mathrm{i}}+\left({z}\frac{\partial{f}}{\partial{x}}−{x}\frac{\partial{f}}{\partial{z}}\right)\boldsymbol{\mathrm{j}}+\left({x}\frac{\partial{f}}{\partial{y}}−{y}\frac{\partial{f}}{\partial{x}}\right)\boldsymbol{\mathrm{k}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\:\:\boldsymbol{\mathrm{F}}=\boldsymbol{\mathrm{r}}×\bigtriangledown{f}. \\ $$
Answered by user1 last updated on 29/Oct/14
$$\:\:\boldsymbol{\mathrm{r}}×\bigtriangledown\boldsymbol{\mathrm{f}}=\begin{vmatrix}{\boldsymbol{\mathrm{i}}}&{\boldsymbol{\mathrm{j}}}&{\boldsymbol{\mathrm{k}}}\\{{x}}&{{y}}&{{z}}\\{\frac{\partial{f}}{\partial{x}}}&{\frac{\partial{f}}{\partial{y}}}&{\frac{\partial{f}}{\partial{z}}}\end{vmatrix} \\ $$$$=\left({y}\frac{\partial{f}}{\partial{z}}−{z}\frac{\partial{f}}{\partial{y}}\right)\boldsymbol{\mathrm{i}}+\left({z}\frac{\partial{f}}{\partial{x}}−{x}\frac{\partial{f}}{\partial{z}}\right)\boldsymbol{\mathrm{j}}+\left({x}\frac{\partial{f}}{\partial{y}}−{y}\frac{\partial{f}}{\partial{x}}\right)\boldsymbol{\mathrm{k}} \\ $$$$=\:\:\boldsymbol{\mathrm{F}} \\ $$