Question

If \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) is the position vector of a point, then \( \text{curl}(\vec{r}) \) is:

• A. null vector
• B. \( 0 \)
• C. \( \hat{i} + \hat{j} + \hat{k} \)
• D. \( \hat{i} - \hat{j} \)

Hint:

The curl of the position vector \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) is always the null vector because its components are linear and partial cross derivatives vanish.

Answer 1

The Correct Option is A

We are given the position vector: \[ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \] We are asked to compute: \[ \text{curl}(\vec{r}) = \nabla \times \vec{r} \] Recall the formula for the curl of a vector field: \[ \nabla \times \vec{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}
x & y & z \end{vmatrix} \] Now expand the determinant: \[ = \hat{i} \left( \frac{\partial z}{\partial y} - \frac{\partial y}{\partial z} \right) - \hat{j} \left( \frac{\partial z}{\partial x} - \frac{\partial x}{\partial z} \right) + \hat{k} \left( \frac{\partial y}{\partial x} - \frac{\partial x}{\partial y} \right) \] Each of these partial derivatives is zero: \[ \frac{\partial z}{\partial y} = \frac{\partial y}{\partial z} = \frac{\partial z}{\partial x} = \frac{\partial x}{\partial z} = \frac{\partial y}{\partial x} = \frac{\partial x}{\partial y} = 0 \] Therefore: \[ \nabla \times \vec{r} = \vec{0} \] Hence, the curl is a null vector.