find the distance between the following parallel lines 9x+5y+3z=69 , 9x+5y+3z=19

The distance between two parallel planes can be found using the following formula:

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

Where:
- \(A\), \(B\), and \(C\) are the coefficients of \(x\), \(y\), and \(z\) from the equation of the planes.
- \(D_1\) and \(D_2\) are the constants from the equations of the two planes.
- \(d\) is the distance between the planes.

Given the two parallel planes:

\[
9x + 5y + 3z = 69 \quad \text{(Plane 1)}
\]
\[
9x + 5y + 3z = 19 \quad \text{(Plane 2)}
\]

The coefficients of \(x\), \(y\), and \(z\) are \(A = 9\), \(B = 5\), and \(C = 3\). The constants are \(D_1 = 69\) and \(D_2 = 19\).

Now, applying the formula:

\[
d = \frac{|69 - 19|}{\sqrt{9^2 + 5^2 + 3^2}} = \frac{50}{\sqrt{81 + 25 + 9}} = \frac{50}{\sqrt{115}}
\]

Simplifying further:

\[
d = \frac{50}{\sqrt{115}} \approx \frac{50}{10.723} \approx 4.66
\]

Thus, the distance between the two parallel planes is approximately **4.66 units**.

Would you like more details or have any questions?

Here are some related questions:
1. How do we find the equation of a plane given a point and a normal vector?
2. What is the geometric interpretation of the distance between parallel planes?
3. How do we calculate the distance between a point and a plane?
4. Can the distance between non-parallel planes be defined?
5. How do the coefficients of the plane equation affect the normal vector?

**Tip:** Always make sure to simplify square roots when calculating distances between planes for an exact value.

Learn how to calculate the distance between two parallel planes 9x + 5y + 3z = 69 and 9x + 5y + 3z = 19. This solution includes a detailed step-by-step process using the distance formula for parallel planes and involves geometric concepts. Suitable for students in Grades 10-12.

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