# Find the curl of the vector field at the given point. F(x, y, z) = xyz i + xyz j + xyz k; (1, 3, 2).

## Question:

Find the curl of the vector field at the given point.

{eq}\vec{F}(x, y, z) = xyz \, i + xyz \, j + xyz \, k \, ; \; (1, \; 3, \; 2) {/eq}

## Answer and Explanation:

We are given:

{eq}\vec{F}(x, y, z) = xyz \, i + xyz \, j + xyz \, k {/eq}

{eq}Curl \vec{F} =\begin{vmatrix} i & j & k \\ \dfrac{\partial}{ \partial x} & \dfrac{\partial}{ \partial y} & \dfrac{\partial}{ \partial z} \\ xyz & xyz & xyz \end{vmatrix} {/eq}

{eq}= i \left(\dfrac{\partial}{ \partial y} (xyz) - \dfrac{\partial}{ \partial z} (xyz) \right) -j \left(\dfrac{\partial}{ \partial x} (xyz) - \dfrac{\partial}{ \partial z} (xyz) \right) +k \left(\dfrac{\partial}{ \partial x} (xyz) - \dfrac{\partial}{ \partial y} (xyz) \right) {/eq}

{eq}=x(z-y)i +y(z- x) j +z(y-x)k {/eq}

Plug in the given point {eq}(1,3,2) {/eq}

{eq}=-1 i +3 j +4k {/eq}

Therefore the solution is:

{eq}Curl \vec{F} =-1 i +3 j +4k {/eq}

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