Find the minimum distance between the two planes, a. P1: 2𝑥 + 3𝑦 − 6𝑧 = −10 and P2: 2𝑥 + 3𝑦 − 6𝑧 = 88. b. The point (-2, 0, 1) resides on P1. What is the nearest point on P2?

Let's solve the two parts of the problem step by step:

### Part (a): Minimum Distance Between Two Planes

Given planes:
- \( P_1: 2x + 3y - 6z = -10 \)
- \( P_2: 2x + 3y - 6z = 88 \)

Both planes have the same normal vector \( \mathbf{n} = \langle 2, 3, -6 \rangle \). Since the planes are parallel (same normal vector), the minimum distance between them can be calculated using the formula:

\[
\text{Distance} = \frac{|d_2 - d_1|}{\|\mathbf{n}\|}
\]

where:
- \( d_1 = -10 \) (from the equation of \( P_1 \))
- \( d_2 = 88 \) (from the equation of \( P_2 \))
- \( \|\mathbf{n}\| \) is the magnitude of the normal vector \( \mathbf{n} \).

First, calculate the magnitude of \( \mathbf{n} \):

\[
\|\mathbf{n}\| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7
\]

Now, calculate the distance:

\[
\text{Distance} = \frac{|88 - (-10)|}{7} = \frac{98}{7} = 14
\]

So, the minimum distance between the two planes is **14 units**.

### Part (b): Nearest Point on \( P_2 \) from Point \((-2, 0, 1)\) on \( P_1 \)

To find the nearest point on \( P_2 \), we project the point \((-2, 0, 1)\) onto the plane \( P_2 \).

1. **Find the normal vector**: \( \mathbf{n} = \langle 2, 3, -6 \rangle \).

2. **Equation of the line** passing through \((-2, 0, 1)\) and along the direction of the normal vector is:

\[
\mathbf{r}(t) = (-2, 0, 1) + t \langle 2, 3, -6 \rangle = (-2 + 2t, 3t, 1 - 6t)
\]

3. **Substitute these coordinates** into the equation of plane \( P_2 \):

\[
2(-2 + 2t) + 3(3t) - 6(1 - 6t) = 88
\]

4. Simplify:

\[
-4 + 4t + 9t - 6 + 36t = 88
\]

\[
49t - 10 = 88
\]

\[
49t = 98 \quad \Rightarrow \quad t = 2
\]

5. **Substitute \( t = 2 \)** back into the equation of the line to find the coordinates of the nearest point on \( P_2 \):

\[
x = -2 + 2(2) = 2
\]
\[
y = 0 + 3(2) = 6
\]
\[
z = 1 - 6(2) = -11
\]

So, the nearest point on \( P_2 \) is \((2, 6, -11)\).

### Summary:
- The minimum distance between the two planes is **14 units**.
- The nearest point on \( P_2 \) to the point \((-2, 0, 1)\) on \( P_1 \) is \((2, 6, -11)\).

Do you want further details or have any questions?

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**Related Questions:**
1. How do you determine if two planes are parallel?
2. What is the formula for the distance between a point and a plane?
3. How can you find the distance between two skew lines?
4. What is the geometrical significance of the normal vector of a plane?
5. How do you find the angle between two intersecting planes?

**Tip:** Always ensure to calculate the magnitude of the normal vector carefully, as it is crucial for determining distances between planes.

Learn how to calculate the minimum distance between two parallel planes and find the nearest point on one plane to a given point on another. This problem involves advanced concepts in 3D geometry, including the formula for distance between parallel planes and projecting points onto planes.

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