Find an equation of the plane.
the plane that passes through the line of intersection of the planes 
x − z = 2 and y + 4z = 2
 and is perpendicular to the plane 
x + y − 4z = 4

To find the equation of a plane that passes through the line of intersection of two given planes and is also perpendicular to a third plane, we follow these steps:

### 1. General form of a plane through the intersection of two planes
Given the planes:

\[
P_1: x - z = 2
\]
\[
P_2: y + 4z = 2
\]

The equation of the plane passing through their line of intersection can be written as a linear combination of these two planes:

\[
\text{Plane equation: } (x - z - 2) + \lambda(y + 4z - 2) = 0
\]

Here, \(\lambda\) is a scalar parameter.

### 2. Perpendicular condition to the plane \( x + y - 4z = 4 \)
The normal vector of the plane we want to be perpendicular to is the normal vector of the plane \( x + y - 4z = 4 \), which is \( \mathbf{n}_3 = (1, 1, -4) \).

The normal vector of the plane we're trying to find will be the combination of the normal vectors of the two planes:

- Normal vector of \( P_1: x - z = 2 \) is \( \mathbf{n}_1 = (1, 0, -1) \)
- Normal vector of \( P_2: y + 4z = 2 \) is \( \mathbf{n}_2 = (0, 1, 4) \)

Thus, the normal vector of the desired plane is:

\[
\mathbf{n} = \mathbf{n}_1 + \lambda \mathbf{n}_2 = (1, 0, -1) + \lambda(0, 1, 4) = (1, \lambda, -1 + 4\lambda)
\]

For the plane to be perpendicular to the plane \( x + y - 4z = 4 \), the dot product of the normal vectors \( \mathbf{n} \) and \( \mathbf{n}_3 = (1, 1, -4) \) must be zero:

\[
\mathbf{n} \cdot \mathbf{n}_3 = (1, \lambda, -1 + 4\lambda) \cdot (1, 1, -4) = 1 \cdot 1 + \lambda \cdot 1 + (-1 + 4\lambda) \cdot (-4) = 0
\]

### 3. Solve for \(\lambda\)
Expanding the dot product:

\[
1 + \lambda + (-1 + 4\lambda)(-4) = 0
\]
\[
1 + \lambda + 4 - 16\lambda = 0
\]
\[
5 - 15\lambda = 0
\]
\[
\lambda = \frac{5}{15} = \frac{1}{3}
\]

### 4. Equation of the plane
Now substitute \(\lambda = \frac{1}{3}\) into the equation of the plane:

\[
(x - z - 2) + \frac{1}{3}(y + 4z - 2) = 0
\]

Multiplying everything by 3 to eliminate the fraction:

\[
3(x - z - 2) + (y + 4z - 2) = 0
\]
\[
3x - 3z - 6 + y + 4z - 2 = 0
\]
\[
3x + y + z - 8 = 0
\]

Thus, the equation of the required plane is:

\[
\boxed{3x + y + z = 8}
\]

---

Would you like further clarification on any step or additional details?

Here are 5 related questions to explore:

1. How do you find the intersection line of two planes?
2. What is the geometric meaning of the dot product between two vectors?
3. How can you check if a line lies on a plane?
4. Can two planes be perpendicular to each other? How would you prove it?
5. What are the conditions for a plane to be parallel to another plane?

**Tip:** The normal vector of a plane is a key element in understanding the plane's orientation in space, and it helps to solve problems related to perpendicularity and parallelism.

Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 2 and y + 4z = 2 and is perpendicular to the plane x + y − 4z = 4.

Learn how to find the equation of a plane that passes through the line of intersection of two planes and is perpendicular to a third plane. This solution includes key concepts like vector algebra, dot product, and plane geometry. Ideal for students in Grades 11-12 and college.

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