VC3.36

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Let V=V_{1}i+V_{2}j+V_{3}k,V^{2}=V_{1}^{{2}}+V_{2}^{{2}}+V_{3}^{{2}}\, --(1)

Therefore {\mathrm  {curl}}V=[{\frac  {\partial V_{3}}{\partial y}}-{\frac  {\partial V_{2}}{\partial z}}]i+[{\frac  {\partial V_{1}}{\partial z}}-{\frac  {\partial V_{3}}{\partial x}}]j+[{\frac  {\partial V_{2}}{\partial x}}-{\frac  {\partial V_{1}}{\partial y}}]k\, --(2)

Using (1) and (2)

V\times {\mathrm  {curl}}V={\begin{vmatrix}i&j&k\\V_{1}&V_{2}&V_{3}\\{\frac  {\partial V_{3}}{\partial y}}-{\frac  {\partial V_{2}}{\partial z}}&{\frac  {\partial V_{1}}{\partial z}}-{\frac  {\partial V_{3}}{\partial x}}&{\frac  {\partial V_{2}}{\partial x}}-{\frac  {\partial V_{1}}{\partial y}}\end{vmatrix}}\,

=\sum i{V_{2}[{\frac  {\partial V_{2}}{\partial x}}-{\frac  {\partial V_{1}}{\partial y}}]-V_{3}[{\frac  {\partial V_{1}}{\partial z}}-{\frac  {\partial V_{3}}{\partial x}}]}\,

=\sum i{[V_{2}{\frac  {\partial V_{2}}{\partial x}}+V_{3}{\frac  {\partial V_{3}}{\partial x}}]-[V_{2}{\frac  {\partial V_{1}}{\partial y}}+V_{3}{\frac  {\partial V_{1}}{\partial z}}]}\,

Adding and subtracting V_{1}{\frac  {\partial V_{1}}{\partial x}}\, to the above, we get

=\sum i{[V_{1}{\frac  {\partial V_{1}}{\partial x}}+V_{2}{\frac  {\partial V_{2}}{\partial x}}+V_{3}{\frac  {\partial V_{3}}{\partial x}}]-[V_{1}{\frac  {\partial V_{1}}{\partial x}}+V_{2}{\frac  {\partial V_{1}}{\partial y}}+V_{3}{\frac  {\partial V_{1}}{\partial z}}]}\,

=\sum i{{\frac  {1}{2}}{\frac  {\partial }{\partial x}}(V_{1}^{{2}}+V_{2}^{{2}}+V_{3}^{{2}})}-\sum i[V_{1}{\frac  {\partial }{\partial x}}+V_{2}{\frac  {\partial }{\partial y}}+V_{3}{\frac  {\partial }{\partial z}}]V_{1}\,

={\frac  {1}{2}}\sum i{\frac  {\partial V^{2}}{\partial x}}-[V_{1}{\frac  {\partial }{\partial x}}+V_{2}{\frac  {\partial }{\partial y}}+V_{3}{\frac  {\partial }{\partial z}}]\sum V_{1}i\,, by (1)

={\frac  {1}{2}}[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}]V^{2}-(V_{1}i+V_{2}j+V_{3}k)\cdot [i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}]V\, [Since V=V_{1}i+V_{2}j+V_{3}k=\sum V_{1}i\,]

={\frac  {1}{2}}\nabla V^{2}-(V\cdot \nabla )V\,

Hence,the required is proved.

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