To solve the problem, we need to find the point $P(x_0, y_0, z_0)$ on the surface $z = 2y^2 + 2x^2$ that is closest to the plane $z = y + x - 3$. We then calculate the distance $d$ between this point and the plane.

Step 1: Point on the Surface

Given that $[x_0, y_0, z_0] = \left[\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right]$, we can confirm that this point lies on the surface $z = 2y^2 + 2x^2$.

Plugging in the coordinates: $z_0 = 2\left(\frac{1}{4}\right)^2 + 2\left(\frac{1}{4}\right)^2 = 2\left(\frac{1}{16}\right) + 2\left(\frac{1}{16}\right) = 2\left(\frac{2}{16}\right) = \frac{4}{16} = \frac{1}{4}$ Since $z_0 = \frac{1}{4}$, the point $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$ is indeed on the surface.

Step 2: Distance to the Plane

The formula for the distance from a point $(x_0, y_0, z_0)$ to a plane $ax + by + cz + d = 0$ is: $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ For the plane $z = y + x - 3$, we can rewrite it as: $x + y - z - 3 = 0$ Here, $a = 1$, $b = 1$, $c = -1$, and $d = -3$.

Plugging in the coordinates $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$ into the distance formula: $d = \frac{|1\cdot\frac{1}{4} + 1\cdot\frac{1}{4} - 1\cdot\frac{1}{4} - 3|}{\sqrt{1^2 + 1^2 + (-1)^2}} = \frac{\left|\frac{1}{4} + \frac{1}{4} - \frac{1}{4} - 3\right|}{\sqrt{1 + 1 + 1}} = \frac{\left|\frac{1}{4} - 3\right|}{\sqrt{3}}$ Simplifying further: $d = \frac{\left| \frac{1}{4} - \frac{12}{4} \right|}{\sqrt{3}} = \frac{\left| \frac{-11}{4} \right|}{\sqrt{3}} = \frac{\frac{11}{4}}{\sqrt{3}} = \frac{11}{4\sqrt{3}} = \frac{11\sqrt{3}}{12}$ Thus, the distance $d$ from the point $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$ to the plane is: $d = \frac{11\sqrt{3}}{12}$

Final Answer

• The point $P(x_0, y_0, z_0) = \left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$.
• The distance $d = \frac{11\sqrt{3}}{12}$.

Would you like further details or have any questions? Here are some related questions to consider:

1. What is the general method to find the distance between a point and a plane?
2. How do you derive the equation for the distance between a point and a plane?
3. Can you calculate the gradient of the surface $z = 2y^2 + 2x^2$?
4. How do you determine the closest point on a surface to a plane using Lagrange multipliers?
5. What is the significance of the Hessian matrix in optimizing distances?
6. How would you find the closest point on a surface to a point in space, not a plane?
7. Can the distance formula be used for non-Euclidean geometries? If so, how?
8. How does the orientation of a plane affect the distance to a point?

Tip: Always verify that a given point satisfies the equation of the surface or plane before proceeding with calculations.