find the distance between the following parallel planes: x + 6y + z = 73 ; x+6y+z=44

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

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

where:
- \(Ax + By + Cz + D_1 = 0\) is the equation of the first plane,
- \(Ax + By + Cz + D_2 = 0\) is the equation of the second plane,
- \((A, B, C)\) are the coefficients of \(x\), \(y\), and \(z\) in both equations, which are the same for parallel planes.

In your case, the planes are:
- \(x + 6y + z = 73\)
- \(x + 6y + z = 44\)

The coefficients \(A = 1\), \(B = 6\), and \(C = 1\) are the same for both planes, so we can now apply the formula.

### Step-by-Step Calculation:

\[
d = \frac{|73 - 44|}{\sqrt{1^2 + 6^2 + 1^2}} = \frac{|29|}{\sqrt{1 + 36 + 1}} = \frac{29}{\sqrt{38}}
\]

Simplifying:

\[
d = \frac{29}{\sqrt{38}} \approx 4.7
\]

Thus, the distance between the two parallel planes is approximately **4.7 units**.

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the distance between two parallel lines in a 2D plane?
2. How do we find the angle between two planes?
3. How can the formula for the distance between planes be derived?
4. What are the conditions for two planes to be perpendicular?
5. Can we find the distance between non-parallel planes?

**Tip:** When finding distances between parallel planes, remember that their normal vectors are identical, simplifying the process!

Learn how to calculate the distance between two parallel planes with equations x + 6y + z = 73 and x + 6y + z = 44 using the distance formula. This problem covers key concepts in analytic geometry and is suitable for Grades 10-12.

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