91Ó°ÊÓ

We are given that \(\mathbf{F} = \mathbf{a} \times \mathbf{r}\), where \(\mathbf{a}\) is a nonzero constant vector and \(\mathbf{r} = \langle x, y, z \rangle\). Let \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\). Now we will compute the cross product of \(\mathbf{a}\) and \(\mathbf{r}\): $$ \mathbf{F} = \mathbf{a} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ x & y & z \end{vmatrix} = \left\langle \left|\begin{matrix} a_2 & a_3 \\ y & z \end{matrix}\right|, -\left|\begin{matrix} a_1 & a_3 \\ x & z \end{matrix}\right|, \left|\begin{matrix} a_1 & a_2 \\ x & y \end{matrix}\right| \right\rangle = \left\langle a_2z-a_3y, a_3x-a_1z, a_1y-a_2x \right\rangle $$

Now let's compute the divergence of the vector field \(\mathbf{F}\). The divergence of a vector field \(\mathbf{F}=\langle F_x, F_y, F_z \rangle\) is given by: $$ \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} $$ In our case, the divergence of \(\mathbf{F}\) is: $$ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(a_2z-a_3y) + \frac{\partial}{\partial y}(a_3x-a_1z) + \frac{\partial}{\partial z}(a_1y-a_2x) $$

Now, we will compute these partial derivatives and simplify the resulting expression: $$ \nabla \cdot \mathbf{F} = \left(-a_3\frac{\partial y}{\partial x} + a_2\frac{\partial z}{\partial x}\right) + \left(a_3\frac{\partial x}{\partial y} - a_1\frac{\partial z}{\partial y}\right) + \left(a_1\frac{\partial y}{\partial z} - a_2\frac{\partial x}{\partial z}\right) $$ However, we know that: $$ \frac{\partial y}{\partial x} = \frac{\partial z}{\partial x} = \frac{\partial x}{\partial y} = \frac{\partial z}{\partial y} = \frac{\partial x}{\partial z} = \frac{\partial y}{\partial z} = 0 $$ This is because the partial derivatives of a variable with respect to other variables are zero. Thus, upon substitution, we get: $$ \nabla \cdot \mathbf{F} = 0 $$ Finally, we have shown that the general rotation field \(\mathbf{F} = \mathbf{a} \times \mathbf{r}\) has zero divergence.