评分及理由

（1）得分及理由（满分4分）

学生作答：-i-k

标准答案：\(\vec{i} - \vec{k}\)

理由：旋度\(\text{rot}\vec{F}\)（即\(\nabla \times \vec{F}\)）的计算公式为： \[ \nabla \times \vec{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \vec{i} - \left( \frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z} \right) \vec{j} + \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \vec{k} \] 给定\(\vec{F}(x, y, z) = xy\vec{i} - yz\vec{j} + xz\vec{k}\)，计算各分量： \[ \frac{\partial F_z}{\partial y} = \frac{\partial (xz)}{\partial y} = 0, \quad \frac{\partial F_y}{\partial z} = \frac{\partial (-yz)}{\partial z} = -y \] \[ \frac{\partial F_z}{\partial x} = \frac{\partial (xz)}{\partial x} = z, \quad \frac{\partial F_x}{\partial z} = \frac{\partial (xy)}{\partial z} = 0 \] \[ \frac{\partial F_y}{\partial x} = \frac{\partial (-yz)}{\partial x} = 0, \quad \frac{\partial F_x}{\partial y} = \frac{\partial (xy)}{\partial y} = x \] 因此， \[ \nabla \t...