Question

Let $\vec{f} = x^2 y z \hat{i} + x y^2 z \hat{j} + x y z^2 \hat{k}$ be a vector field. If $\vec{F}(x,y,z) = \text{curl }\vec{f}$, then the vector $\vec{F}(1,-2,1)$ is equal to _______}

• A. $-3\hat{i} + 4\hat{j} + 3\hat{k}$
• B. $-3\hat{i} + 3\hat{k}$
• C. $3\hat{i} - 3\hat{k}$
• D. $-3\hat{i} - 4\hat{j} + 3\hat{k}$

Hint:

Double check signs when expanding the cross product determinant.
The middle term ($\hat{j}$) has a negative sign in the expansion: $-\hat{j}(\frac{\partial f_z}{\partial x} - \frac{\partial f_x}{\partial z})$.

Answer 1

The Correct Option is B