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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is \[\nabla.\vec A \neq \vec A.\nabla\] Consider \[\vec A=A_{1}i+A_{2}j+A_{3}k\] then \[\nabla . \vec A=(\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}).(A_{1}i+A_{2}j+A_{3}k)\]\[\nabla . \vec A=\frac{\partial A_{1}}{\partial x}+\frac{\partial A_{2}}{\partial y}+\frac{\partial A_{3}}{\partial z}\] Why wouldn't \[\vec A . \nabla\] give the same result?Isn't it a dot product
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