Find an equation of the plane that is orthogonal to the plane 5x + 4z = 4 and contains the line of intersection of the planes 2x - 3y + z = 35 and x + 2y -3z = 42

To find the equation of the plane that is orthogonal to the plane \( 5x + 4z = 4 \) and contains the line of intersection of the planes \( 2x - 3y + z = 35 \) and \( x + 2y - 3z = 42 \), we can proceed step by step.

### Step 1: Equation of the plane orthogonal to \( 5x + 4z = 4 \)
The normal vector to the plane \( 5x + 4z = 4 \) is given by \( \mathbf{n_1} = \langle 5, 0, 4 \rangle \). 

Thus, the required plane should have a normal vector that is parallel to this one, i.e., the form of the plane will be:
\[
5x + 0y + 4z + d = 0
\]
or simply:
\[
5x + 4z + d = 0
\]
We need to determine \( d \) by using the next steps.

### Step 2: Find the line of intersection of the two planes
The line of intersection of the planes \( 2x - 3y + z = 35 \) and \( x + 2y - 3z = 42 \) can be found by solving these two plane equations simultaneously.

#### Step 2.1: Eliminate one variable
Let's solve these equations step by step to eliminate one variable and find a parametric form of the line.

1. Multiply the second equation by 2 to align the coefficients of \( x \):
   \[
   2(x + 2y - 3z) = 2 \times 42
   \]
   This gives:
   \[
   2x + 4y - 6z = 84
   \]

2. Subtract the first equation \( 2x - 3y + z = 35 \) from this new equation:
   \[
   (2x + 4y - 6z) - (2x - 3y + z) = 84 - 35
   \]
   Simplifying:
   \[
   7y - 7z = 49
   \]
   Dividing by 7:
   \[
   y - z = 7
   \]
   Hence, \( y = z + 7 \).

#### Step 2.2: Substitute \( y = z + 7 \) into one of the original equations
Substitute \( y = z + 7 \) into \( 2x - 3y + z = 35 \):
\[
2x - 3(z + 7) + z = 35
\]
Simplifying:
\[
2x - 3z - 21 + z = 35
\]
\[
2x - 2z = 56
\]
\[
x - z = 28
\]
Thus, \( x = z + 28 \).

#### Step 2.3: Parametric equations of the line of intersection
Now, we have:
\[
x = z + 28, \quad y = z + 7
\]
Let \( z = t \) be the parameter. Then the parametric equations of the line are:
\[
x = t + 28, \quad y = t + 7, \quad z = t
\]
These represent the line of intersection.

### Step 3: Find \( d \) using a point on the line
We now substitute a point on the line into the equation \( 5x + 4z + d = 0 \) to find \( d \).

Using \( t = 0 \) in the parametric equations, we get the point \( (28, 7, 0) \).

Substitute \( x = 28 \) and \( z = 0 \) into \( 5x + 4z + d = 0 \):
\[
5(28) + 4(0) + d = 0
\]
\[
140 + d = 0
\]
\[
d = -140
\]

### Step 4: Final equation of the plane
The equation of the required plane is:
\[
5x + 4z - 140 = 0
\]
or equivalently:
\[
5x + 4z = 140
\]

This is the equation of the plane that is orthogonal to \( 5x + 4z = 4 \) and contains the line of intersection of the given planes.

---

Would you like more details or have any questions on this solution?

### 5 Related Questions:
1. How do you find the normal vector to a plane given its equation?
2. What is the geometric interpretation of the line of intersection of two planes?
3. How can parametric equations of a line be derived from simultaneous plane equations?
4. How do you solve systems of linear equations involving three variables?
5. How does the orientation of a plane change if its normal vector changes?

### Tip:
When finding the line of intersection between two planes, using elimination to solve for one variable simplifies the process significantly.

Find an equation of the plane that is orthogonal to the plane 5x + 4z = 4 and contains the line of intersection of the planes 2x - 3y + z = 35 and x + 2y - 3z = 42

This problem explores finding the equation of a plane orthogonal to 5x + 4z = 4 that contains the line of intersection of the planes 2x - 3y + z = 35 and x + 2y - 3z = 42. Using parametric equations and elimination, the solution guides you step-by-step through solving for the required plane.

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