If v = yz î + 3zx ĵ + z k̂, then curl v is
-3xî + yĵ + 2zk̂
The question asks us to find the curl of the given vector field $ \mathbf{v} = yz \hat{i} + 3zx \hat{j} + z \hat{k} $. The curl is a vector operation that describes the infinitesimal rotation of the vector field. It is often denoted as $ \nabla \times \mathbf{v} $.
The curl of a vector field $ \mathbf{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} $ is calculated using the determinant of a special matrix:
$$ \nabla \times \mathbf{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_x & v_y & v_z \end{vmatrix} $$For the given vector field $ \mathbf{v} = yz \hat{i} + 3zx \hat{j} + z \hat{k} $, we have:
Let's substitute the components into the determinant formula:
| $ \nabla \times \mathbf{v} $ = | $ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz & 3zx & z \end{vmatrix} $ |
Now, we expand the determinant:
Let's calculate the partial derivatives:
Substitute these back into the expansion:
Combining these components, the curl of the vector field is:
$$ \nabla \times \mathbf{v} = -3x \hat{i} + y \hat{j} + 2z \hat{k} $$Comparing our result $ -3x \hat{i} + y \hat{j} + 2z \hat{k} $ with the given options:
The calculated curl matches the first option.
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