The problem asks for a unit vector normal to the surface $x^2y + 2xz = 4$ at the point $(2, -2, 3)$.

To solve this, we need to:

1. Find the gradient vector of the given surface equation:
   The gradient of $f(x, y, z) = x^2y + 2xz - 4$ gives a normal vector to the surface at any point $(x, y, z)$.
   $\nabla f = \left(\frac{\partial f}{\partial x}\right)\mathbf{i} + \left(\frac{\partial f}{\partial y}\right)\mathbf{j} + \left(\frac{\partial f}{\partial z}\right)\mathbf{k}$

   Partial derivatives are:

   • $\frac{\partial f}{\partial x} = 2xy + 2z$
   • $\frac{\partial f}{\partial y} = x^2$
   • $\frac{\partial f}{\partial z} = 2x$

2. Evaluate the gradient at the point $(2, -2, 3)$:
   Substituting $x = 2$, $y = -2$, and $z = 3$:

   • $\frac{\partial f}{\partial x} = 2(2)(-2) + 2(3) = -8 + 6 = -2$
   • $\frac{\partial f}{\partial y} = (2)^2 = 4$
   • $\frac{\partial f}{\partial z} = 2(2) = 4$

   So, the gradient vector is: $\nabla f = -2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}$

3. Normalize the gradient vector to get the unit normal vector:
   The magnitude of $\nabla f$ is: $|\nabla f| = \sqrt{(-2)^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

   The unit vector is: $\mathbf{n} = \frac{\nabla f}{|\nabla f|} = \frac{-2}{6}\mathbf{i} + \frac{4}{6}\mathbf{j} + \frac{4}{6}\mathbf{k} = -\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}$

4. Check the options: The correct option is (b).

Would you like further details or explanations?

Related Questions:

1. How is the gradient vector related to the normal vector of a surface?
2. Why do we normalize vectors, and what does a unit vector signify?
3. Can you explain the significance of each partial derivative in this problem?
4. How does the choice of a point affect the calculation of the gradient?
5. What are some real-world applications of normal vectors in geometry?

Tip:

Always verify that the unit vector you calculate satisfies the original equation of the surface when plugged back into the gradient expression.