Answer
See proof
Work Step by Step
Let \[ \mathbf{F} = f\vec{i} + g\vec{j} + h\vec{k} \] If \[ \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \] and \[ \mathbf{b} = b_1\mathbf{i} + jb_2\mathbf{j} + kb_3\mathbf{k} \] \[ \mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{array} \right| = (\mathbf{i}(a_2b_3 - a_3b_2) + \mathbf{j}(a_3b_1 - a_1b_3) + \mathbf{k}(a_1b_2 - b_1a_2)) \] \[ \nabla \cdot \mathbf{a} = \left(\frac{\partial a_1}{\partial x} + \frac{\partial a_2}{\partial y} + \frac{\partial a_3}{\partial z}\right) \] \[ curl(kF) = \left(\mathbf{i}(k)\left(\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}\right) + \mathbf{j}(k)\left(\frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}\right) +(k)\mathbf{k}\left(\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right)\right) \] \[ = k\operatorname{curl}\vec{F} \] Result : \[ \operatorname{curl}(k\vec{F}) = k\operatorname{curl}\vec{F} \]
