91Ó°ÊÓ

We need to expand the cross product \((\mathbf{F} \times \mathbf{G})\), which can be written as: $$(\mathbf{F} \times \mathbf{G})= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ F_{x} & F_{y} & F_{z} \\ G_{x} & G_{y} & G_{z} \end{vmatrix}$$ Now, find the components of the cross product: $$\mathbf{F} \times \mathbf{G} = (F_{y}G_{z} - F_{z}G_{y})\mathbf{i} - (F_{x}G_{z} - F_{z}G_{x})\mathbf{j} + (F_{x}G_{y} - F_{y}G_{x})\mathbf{k}$$

Now compute the divergence of the cross product, which can be written as: $$\nabla \cdot (\mathbf{F} \times \mathbf{G}) = \frac{\partial}{\partial x}\left(F_{y}G_{z} - F_{z}G_{y}\right) + \frac{\partial}{\partial y}\left(G_{x}F_{z} - F_{x}G_{z}\right) + \frac{\partial}{\partial z}\left(F_{x}G_{y} - F_{y}G_{x}\right)$$

Compute the curls of both vector fields, which can be written as: $$\begin{aligned} \nabla \times \mathbf{F} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_{x} & F_{y} & F_{z} \end{vmatrix}, \\ \nabla \times \mathbf{G} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ G_{x} & G_{y} & G_{z} \end{vmatrix} \end{aligned}$$ Now find the components of the curls: $$\begin{aligned} \nabla \times \mathbf{F} &= \left(\frac{\partial F_{z}}{\partial y} - \frac{\partial F_{y}}{\partial z}\right)\mathbf{i} - \left(\frac{\partial F_{z}}{\partial x} - \frac{\partial F_{x}}{\partial z}\right)\mathbf{j} + \left(\frac{\partial F_{y}}{\partial x} - \frac{\partial F_{x}}{\partial y}\right)\mathbf{k}, \\ \nabla \times \mathbf{G} &= \left(\frac{\partial G_{z}}{\partial y} - \frac{\partial G_{y}}{\partial z}\right)\mathbf{i} - \left(\frac{\partial G_{z}}{\partial x} - \frac{\partial G_{x}}{\partial z}\right)\mathbf{j} + \left(\frac{\partial G_{y}}{\partial x} - \frac{\partial G_{x}}{\partial y}\right)\mathbf{k} \end{aligned}$$

Compute the scalar products of the vector fields with the curls: $$\begin{aligned} \mathbf{G} \cdot (\nabla \times \mathbf{F}) &= G_{x}\left(\frac{\partial F_{z}}{\partial y} - \frac{\partial F_{y}}{\partial z}\right) - G_{y}\left(\frac{\partial F_{z}}{\partial x} - \frac{\partial F_{x}}{\partial z}\right) + G_{z}\left(\frac{\partial F_{y}}{\partial x} - \frac{\partial F_{x}}{\partial y}\right), \\ \mathbf{F} \cdot (\nabla \times \mathbf{G}) &= F_{x}\left(\frac{\partial G_{z}}{\partial y} - \frac{\partial G_{y}}{\partial z}\right) - F_{y}\left(\frac{\partial G_{z}}{\partial x} - \frac{\partial G_{x}}{\partial z}\right) + F_{z}\left(\frac{\partial G_{y}}{\partial x} - \frac{\partial G_{x}}{\partial y}\right) \end{aligned}$$

Finally, compute the difference of the scalar products: $$\mathbf{G} \cdot(\nabla \times \mathbf{F})-\mathbf{F} \cdot(\nabla \times \mathbf{G}) = \frac{\partial}{\partial x}\left(F_{y}G_{z} - F_{z}G_{y}\right) + \frac{\partial}{\partial y}\left(G_{x}F_{z} - F_{x}G_{z}\right) + \frac{\partial}{\partial z}\left(F_{x}G_{y} - F_{y}G_{x}\right)$$ From steps 2 and 5, we find that both expressions are the same, thus proving the identity: $$\nabla \cdot(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot(\nabla \times \mathbf{F})-\mathbf{F} \cdot(\nabla \times \mathbf{G})$$