If `\vec A = xz^2\hat i + 2xy\hat j + 3xz\hat k`
Then find `\nabla  \times \vec A(or\,curl\,\vec A)` at the point (1, 2, 1)

Solution:

We know that 

     `\vec A = xz^2\hat i + 2xy\hat j + 3xz\hat k`

Than

`\nabla  \times \vec A = (\partial/{\partialy} 3xz - \partial/{\partialz} 2xy)\hat i - (\partial/{\partialx} 3xz - \partial/{\partialz} xz^2)\hat j + (\partial/{\partialx} 2xy - \partial/{\partialy} xz^2)\hat k`

`\nabla  \times \vec A = (0)\hat i - (3z - 2xz)\hat j + (2y - 0)\hat k`

Now, we calculate the values of `\nabla \times \vec A` at  (1,2,1)

`\nabla  \times \vec A = (0)\hat i - (3(1) - 2(1)(1))\hat j + (2(2) - 0)\hat k`

`\nabla  \times \vec A = (0)\hat i - (3 - 2)\hat j + (4 - 0)\hat k`

`\nabla  \times \vec A = (0)\hat i - (1)\hat j + (4)\hat k`

`\nabla  \times \vec A =  -\hat j + 4\hat k`