91Ó°ÊÓ

The cross product between two vectors is given by the determinant of a special matrix defined by the vectors. For vectors \(\mathbf{A}=a_{1}\mathbf{i}+a_{2}\mathbf{j}+a_{3}\mathbf{k}\) and \(\mathbf{B}=b_{1}\mathbf{i}+b_{2}\mathbf{j}+b_{3}\mathbf{k}\) the cross product is given as \(\mathbf{A}\times\mathbf{B}=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \end{vmatrix}\). Applying this to \(\mathbf{F}\) and \(\mathbf{G}\), we get \(\mathbf{F} \times \mathbf{G} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & -z \ x^{2} & y & z^{2} \end{vmatrix}\). Thus, \(\mathbf{F} \times \mathbf{G}=(-z y) \mathbf{i} - (-z x^{2}) \mathbf{j} + (x y - 0) \mathbf{k}= - z y \mathbf{i} + z x^2 \mathbf{j} + x y \mathbf{k}\).

The curl of a vector field \(\mathbf{A}(x, y, z)=A_{1}\mathbf{i}+A_{2}\mathbf{j}+A_{3}\mathbf{k}\) is given by the determinant of a special matrix defined by the unit vectors, the gradient operator and the components of \(\mathbf{A}\) i.e. \(\nabla \times \mathbf{A}= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ A_{1} & A_{2} & A_{3} \end{vmatrix}\). Applying this to \(\mathbf{F}\times\mathbf{G}\) from Step 1, we get \(\operatorname{curl}(\mathbf{F} \times \mathbf{G})= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ -z y & z x^{2} & x y \end{vmatrix} = \left(\frac{\partial}{\partial y} (x y) - \frac{\partial}{\partial z}(-z x^{2})\right) \mathbf{i} - \left(\frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial z} (- z y)\right) \mathbf{j} + \left(\frac{\partial}{\partial x} (- z x^{2}) - \frac{\partial}{\partial y} (- z y)\right) \mathbf{k} = (y + 2 z x) \mathbf{i} - (y + x + z) \mathbf{j} + 2 x z \mathbf{k}\).