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 following formula:

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

Here, the equation of a plane is generally written as \(Ax + By + Cz + D = 0\), and \(D_1\) and \(D_2\) are the constants from the two plane equations. In our case, both planes have the same normal vector, given by \(x + 6y + z\), so \(A = 1\), \(B = 6\), and \(C = 1\).

The given planes are:
1. \(x + 6y + z = 73\) (so \(D_1 = -73\))
2. \(x + 6y + z = 44\) (so \(D_2 = -44\))

Now we apply the formula:

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

The distance between the two planes is:

\[
d = \frac{29}{\sqrt{38}} \approx 4.70
\]

Thus, the distance between the planes is approximately \(4.70\) units.

Would you like a further breakdown of this solution or have any questions?

Here are five related questions for practice:
1. Find the distance between the planes \(2x - y + 2z = 10\) and \(2x - y + 2z = 5\).
2. Are the planes \(x + 2y + 3z = 5\) and \(2x + 4y + 6z = 10\) parallel?
3. Determine if the planes \(x - 4y + z = 7\) and \(2x - 8y + 2z = 10\) are parallel.
4. What is the distance between the parallel lines \(3x - 2y = 5\) and \(3x - 2y = 1\)?
5. How do you find the angle between two planes?

**Tip:** When two planes are parallel, their normal vectors are proportional or identical.

Find the distance between the following parallel planes: x + 6y + z = 73 and x + 6y + z = 44

Learn how to find the distance between the parallel planes x + 6y + z = 73 and x + 6y + z = 44. This problem uses the distance formula for parallel planes and explores key geometric concepts in 3D space. Suitable for students in Grades 10-12.

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