Question

Q2) IF F= (x+y+1)i+j-(x+y)k then show that F.curl F=0
​

Answer 1

SOLUTION

GIVEN

$\vec{F} = (x + y + 1) \hat{i} + \hat{j} - (x + y) \hat{k}$

TO PROVE

$\vec{F}.(\nabla \times \vec{F})$

EVALUATION

Here the given vector is

$\vec{F} = (x + y + 1) \hat{i} + \hat{j} - (x + y) \hat{k}$

Now

$\nabla \times \vec{F}$

$= \displaystyle\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \\ \frac{ \partial}{ \partial x} & \frac{ \partial}{ \partial y} & \frac{ \partial}{ \partial z} \\ \\ x + y + 1 & 1 & - x - y \end{vmatrix}$

$= \hat{i}( - 1) - \hat{j} ( - 1) + \hat{k}(0 - 1)$

$= - \hat{i} + \hat{j} - \hat{k}$

Now

$\vec{F}.(\nabla \times \vec{F})$

$= \big[(x + y + 1) \hat{i} + \hat{j} - (x + y) \hat{k} \big]. ( - \hat{i} + \hat{j} - \hat{k})$

$= - x - y - 1 + 1 + x + y$

$= 0$

Hence proved

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