Suppose the equations of two planes P and P' are 3x+6y-5z=-3 and -2x+7y-z=24 respectively. Find the equation of intersection

To find the equation of the line of intersection between two planes \( P \) and \( P' \), we can follow these steps:

### Step 1: Write the equations of the planes
The equations of the planes are given as:
1. \( 3x + 6y - 5z = -3 \)  (Equation of \( P \))
2. \( -2x + 7y - z = 24 \)  (Equation of \( P' \))

### Step 2: Parametric form of the solution
The line of intersection of the two planes can be represented by solving the system of equations, which involves eliminating one variable and solving for the others. We use a method called **Gaussian elimination** or **substitution** to find parametric equations for the line.

We proceed by solving the system:

#### Multiply the second equation by 5 to align the \( z \)-coefficients:
\[
5(-2x + 7y - z) = 5(24)
\]
This gives:
\[
-10x + 35y - 5z = 120
\]

Now, subtract the first equation \( (3x + 6y - 5z = -3) \) from this new equation:
\[
(-10x + 35y - 5z) - (3x + 6y - 5z) = 120 - (-3)
\]
This simplifies to:
\[
-13x + 29y = 123
\]
Or:
\[
13x = 29y - 123
\]
\[
x = \frac{29y - 123}{13}
\]

### Step 3: Solve for \( z \)
Now, substitute this expression for \( x \) into one of the original plane equations (say the first plane \( 3x + 6y - 5z = -3 \)):

\[
3\left(\frac{29y - 123}{13}\right) + 6y - 5z = -3
\]
Simplify the expression:
\[
\frac{3(29y - 123)}{13} + 6y - 5z = -3
\]
\[
\frac{87y - 369}{13} + 6y - 5z = -3
\]
Multiply through by 13 to eliminate the denominator:
\[
87y - 369 + 78y - 65z = -39
\]
Simplify:
\[
165y - 65z = 330
\]
Divide by 5:
\[
33y - 13z = 66
\]
Solve for \( z \):
\[
13z = 33y - 66
\]
\[
z = \frac{33y - 66}{13}
\]
\[
z = \frac{33}{13}y - \frac{66}{13}
\]

### Step 4: Parametric form of the line
Thus, the parametric equations of the line of intersection are:
\[
x = \frac{29}{13}y - \frac{123}{13}
\]
\[
z = \frac{33}{13}y - \frac{66}{13}
\]

Therefore, the equation of the line of intersection in parametric form is:

\[
(x, y, z) = \left( \frac{29}{13}y - \frac{123}{13}, y, \frac{33}{13}y - \frac{66}{13} \right)
\]

This gives the parametric equation of the line of intersection between the two planes.

Would you like more details, or have any questions about this?

Here are 5 related questions for further exploration:
1. How can we represent the direction vector of the line of intersection?
2. What is the geometric meaning of the line of intersection?
3. How do we find the point where the line intersects the \( z \)-axis?
4. Can we find the angle between the two planes?
5. How do the normal vectors of the two planes relate to the line of intersection?

**Tip:** To solve systems of linear equations involving planes, using Gaussian elimination can often simplify the process significantly.

Suppose the equations of two planes P and P' are 3x+6y-5z=-3 and -2x+7y-z=24 respectively. Find the equation of intersection.

Learn how to find the equation of the line of intersection between two planes with equations 3x+6y-5z=-3 and -2x+7y-z=24. This solution involves Gaussian elimination and parametric equations. Suitable for Grades 10-12.

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