VC3.37

From Exampleproblems

Since $v=v_1i+v_2j+v_3k\,$ ,we have

$\nabla\times v=[\frac{\partial v_3}{\partial y}-\frac{\partial v_2}{\partial z}]i+[\frac{\partial v_1}{\partial z}-\frac{\partial v_3}{\partial x}]j+[\frac{\partial v_2}{\partial x}-\frac{\partial v_1}{\partial y}]k\,$ --(1)

Now, $\nabla v_1\times i=[i\frac{\partial v_1}{\partial x}+j\frac{\partial v_2}{\partial y}+k\frac{\partial v_3}{\partial z}]\times i=\frac{\partial v_1}{\partial y}(j\times i)+\frac{\partial v_1}{\partial z}(k\times i)=-\frac{\partial v_1}{\partial y}k+\frac{\partial v_1}{\partial z}j\,$ --(2)

$\nabla v_2\times j=[i\frac{\partial v_2}{\partial x}+j\frac{\partial v_2}{\partial y}+k\frac{\partial v_2}{\partial z}]\times j=\frac{\partial v_2}{\partial x}(i\times j)+\frac{\partial v_2}{\partial z}(k\times i)=\frac{\partial v_2}{\partial x}k-\frac{\partial v_2}{\partial z}i\,$ --(3)

and $\nabla v_3\times k=[i\frac{\partial v_3}{\partial x}+j\frac{\partial v_3}{\partial y}+k\frac{\partial v_3}{\partial z}]\times k=\frac{\partial v_3}{\partial x}(i\times k)+\frac{\partial v_3}{\partial y}(j\times k)=-\frac{\partial v_3}{\partial x}j+\frac{\partial v_3}{\partial y}i\,$ --(4)

Adding (2),(3),(4), we have

$\nabla v_1\times i+\nabla v_2\times j+\nabla v_3\times k=[\frac{\partial v_3}{\partial y}-\frac{\partial v_2}{\partial z}]i+[\frac{\partial v_1}{\partial z}-\frac{\partial v_3}{\partial x}]j+[\frac{\partial v_2}{\partial x}-\frac{\partial v_3}{\partial y}]k\,$ implies

$\nabla v_1\times i+\nabla v_2\times j+\nabla v_3\times k=\nabla\times v\,$ by (1)

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