1. **State the problem:** Given the vector field $v = (3xyz, 2xy, -xyz)$ and scalar field $\phi = 3x^2 - yz$, find (i) $\text{div } v$, (ii) $v \cdot \nabla \phi$, and (iii) $\text{div} (\phi v)$ at the point $(1, -1, 1)$. 2. **Calculate (i) $\text{div } v$:** The divergence of $v$ is $$\text{div } v = \frac{\partial}{\partial x}(3xyz) + \frac{\partial}{\partial y}(2xy) + \frac{\partial}{\partial z}(-xyz).$$ Calculate each partial derivative: - $\frac{\partial}{\partial x}(3xyz) = 3yz$ - $\frac{\partial}{\partial y}(2xy) = 2x$ - $\frac{\partial}{\partial z}(-xyz) = -xy$ Thus, $$\text{div } v = 3yz + 2x - xy.$$ Evaluate at $(1, -1, 1)$: $$3 \cdot (-1) \cdot 1 + 2 \cdot 1 - 1 \cdot (-1) = -3 + 2 + 1 = 0.$$ 3. **Calculate (ii) $v \cdot \nabla \phi$:** First compute $\nabla \phi$: $$\frac{\partial \phi}{\partial x} = 6x, \quad \frac{\partial \phi}{\partial y} = -z, \quad \frac{\partial \phi}{\partial z} = -y.$$ At $(1, -1, 1)$: $$\nabla \phi = (6 \cdot 1, -1, -(-1)) = (6, -1, 1).$$ Vector $v$ at $(1, -1, 1)$: $$v = (3 \cdot 1 \cdot (-1) \cdot 1, 2 \cdot 1 \cdot (-1), -1 \cdot (-1) \cdot 1) = (-3, -2, 1).$$ Now compute dot product: $$v \cdot \nabla \phi = (-3)(6) + (-2)(-1) + (1)(1) = -18 + 2 + 1 = -15.$$ 4. **Calculate (iii) $\text{div} (\phi v)$:** Use the product rule for divergence: $$\text{div}(\phi v) = (\nabla \phi) \cdot v + \phi \, \text{div } v.$$ Using values already found: $$ (\nabla \phi) \cdot v = -15, \quad \text{div } v = 0, \quad \phi(1, -1, 1) = 3(1)^2 - (-1)(1) = 3 + 1 = 4.$$ Thus, $$\text{div}(\phi v) = -15 + 4 \times 0 = -15.$$ **Final answers:** - (i) $\text{div } v = 0$ - (ii) $v \cdot \nabla \phi = -15$ - (iii) $\text{div} (\phi v) = -15$