on the surface

z=2⋅y2+2⋅x2

Find the point P(x0,y0,z0)

 closest to the plain

z=y+x−3

and calculate the distance d

 Including:

[x0,y0,z0]

=

[1/4,1/4,1/4]

Below is your answer, as understood by the system:

[14,14,14]

Correct answer, well done.

Score for this submission: 12.50 out of 12.50.

d=


Incorrect answer.

Score for this submission: 0.00 out of 12.50.

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.

Learn how to find the closest point on a given surface to a specified plane using analytic geometry. This problem involves determining a point (x0, y0, z0) on the surface z = 2y^2 + 2x^2 closest to the plane z = y + x - 3, and calculating the distance d between them.

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