Question

Curl of \[ \vec{V}=e^{xyz}(\hat{i}+\hat{j}+\hat{k}) \] at the point \((1,1,1)\) is

• A. \(e\)
• B. \(1\)
• C. \(0\)
• D. \(\hat{0}\)

Hint:

If all three components of a vector field are identical, \[ \boxed{ P=Q=R, } \] then the mixed partial derivatives cancel, giving \[ \boxed{\nabla\times\vec{V}=\hat{0}.} \]

Answer 1

The Correct Option is D

Step 1: Write the vector components. Given, \[ \vec{V} = e^{xyz}\hat{i} + e^{xyz}\hat{j} + e^{xyz}\hat{k}. \] Thus, \[ P=Q=R=e^{xyz}. \]

Step 2: Compute the curl. The curl is \[ \nabla\times\vec{V} = \begin{vmatrix} \hat{i}&\hat{j}&\hat{k} \dfrac{\partial}{\partial x}& \dfrac{\partial}{\partial y}& \dfrac{\partial}{\partial z} P& Q& R \end{vmatrix}. \] Since \[ P=Q=R=e^{xyz}, \] we have \[ \frac{\partial R}{\partial y} = \frac{\partial Q}{\partial z}, \qquad \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}, \qquad \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}. \] Hence, \[ \nabla\times\vec{V} = \hat{0}. \] Therefore, at \((1,1,1)\), \[ \boxed{\nabla\times\vec{V}=\hat{0}.} \] Therefore, \[ \boxed{(D)} \] is the correct answer.