Find the distance between the following parallel planes:
4x+ 2y+2z = 62 and 4x+ 2y+2z = 45

To find the distance between two parallel planes, we can use the formula:

\[
\text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}
\]

Where the general equation of a plane is \( Ax + By + Cz = D \), and \( A, B, C \) are the coefficients of \( x, y, z \) respectively, and \( D_1 \) and \( D_2 \) are the constant terms from the two plane equations.

For the given planes:

1. \( 4x + 2y + 2z = 62 \) (First plane)
2. \( 4x + 2y + 2z = 45 \) (Second plane)

Here, \( A = 4 \), \( B = 2 \), and \( C = 2 \), and \( D_1 = 62 \), \( D_2 = 45 \).

Now, applying the formula:

\[
\text{Distance} = \frac{|62 - 45|}{\sqrt{4^2 + 2^2 + 2^2}} = \frac{|17|}{\sqrt{16 + 4 + 4}} = \frac{17}{\sqrt{24}} = \frac{17}{2\sqrt{6}} = \frac{17\sqrt{6}}{12}
\]

So, the distance between the planes is:

\[
\frac{17\sqrt{6}}{12}
\]

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**Related Questions:**
1. How do you find the distance between a point and a plane?
2. Can two non-parallel planes have a distance between them?
3. What is the geometric meaning of the normal vector of a plane?
4. How do you find the angle between two planes?
5. What are the conditions for two planes to be parallel?

**Tip:** When two planes are parallel, they share the same normal vector.

Find the distance between the following parallel planes: 4x + 2y + 2z = 62 and 4x + 2y + 2z = 45

Discover how to calculate the distance between two parallel planes with the equations 4x + 2y + 2z = 62 and 4x + 2y + 2z = 45. This detailed solution walks through the key concepts of 3D geometry and the formula for the distance between planes. Suitable for high school students in grades 11-12.

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