1. **State the problem:** Find the angle between the surfaces defined by $$xy^2z = 3x + z^2$$ and $$3x^2 - y^2 + 2z = 1$$ at the point $(1, -2, 1)$. 2. **Formula and concept:** The angle between two surfaces at a point is the angle between their normal vectors at that point. 3. **Find the gradients (normal vectors) of each surface:** - For surface 1: Define $F(x,y,z) = xy^2z - 3x - z^2 = 0$ - For surface 2: Define $G(x,y,z) = 3x^2 - y^2 + 2z - 1 = 0$ 4. **Calculate the gradient vectors:** $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$$ $$\nabla G = \left(\frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z}\right)$$ 5. **Partial derivatives for $F$:** $$\frac{\partial F}{\partial x} = y^2 z - 3$$ $$\frac{\partial F}{\partial y} = 2 x y z$$ $$\frac{\partial F}{\partial z} = x y^2 - 2 z$$ 6. **Partial derivatives for $G$:** $$\frac{\partial G}{\partial x} = 6 x$$ $$\frac{\partial G}{\partial y} = -2 y$$ $$\frac{\partial G}{\partial z} = 2$$ 7. **Evaluate gradients at point $(1, -2, 1)$:** - For $F$: $$\frac{\partial F}{\partial x} = (-2)^2 \times 1 - 3 = 4 - 3 = 1$$ $$\frac{\partial F}{\partial y} = 2 \times 1 \times (-2) \times 1 = -4$$ $$\frac{\partial F}{\partial z} = 1 \times (-2)^2 - 2 \times 1 = 4 - 2 = 2$$ So, $\nabla F(1,-2,1) = (1, -4, 2)$. - For $G$: $$\frac{\partial G}{\partial x} = 6 \times 1 = 6$$ $$\frac{\partial G}{\partial y} = -2 \times (-2) = 4$$ $$\frac{\partial G}{\partial z} = 2$$ So, $\nabla G(1,-2,1) = (6, 4, 2)$. 8. **Find the angle $\theta$ between the two normal vectors using the dot product formula:** $$\cos \theta = \frac{\nabla F \cdot \nabla G}{|\nabla F| |\nabla G|}$$ 9. **Calculate dot product:** $$\nabla F \cdot \nabla G = 1 \times 6 + (-4) \times 4 + 2 \times 2 = 6 - 16 + 4 = -6$$ 10. **Calculate magnitudes:** $$|\nabla F| = \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$$ $$|\nabla G| = \sqrt{6^2 + 4^2 + 2^2} = \sqrt{36 + 16 + 4} = \sqrt{56}$$ 11. **Calculate cosine of angle:** $$\cos \theta = \frac{-6}{\sqrt{21} \times \sqrt{56}} = \frac{-6}{\sqrt{1176}} = \frac{-6}{34.29} \approx -0.175$$ 12. **Calculate angle:** $$\theta = \cos^{-1}(-0.175) \approx 100.1^\circ$$ 13. **Interpretation:** The angle between the surfaces at the point is approximately $100.1^\circ$. **Final answer:** The angle between the surfaces at $(1, -2, 1)$ is approximately $100.1^\circ$.