equation of a plane perpendicular to both of given two planes

To find the equation of a plane that is perpendicular to two given planes, we can follow a clear step-by-step approach using the normal vectors of the given planes.

### Steps:

1. **Understand the equation of a plane:**
   The general form of a plane equation is:
   \[
   ax + by + cz = d
   \]
   where \( (a, b, c) \) is the normal vector of the plane.

2. **Find the normal vectors of the given planes:**
   If two planes have equations:
   \[
   a_1x + b_1y + c_1z = d_1 \quad \text{and} \quad a_2x + b_2y + c_2z = d_2,
   \]
   their normal vectors are \( \vec{n_1} = (a_1, b_1, c_1) \) and \( \vec{n_2} = (a_2, b_2, c_2) \).

3. **Cross product of normal vectors:**
   To find the plane that is perpendicular to both planes, we need to find a normal vector to the new plane. The normal vector of the new plane will be the cross product of \( \vec{n_1} \) and \( \vec{n_2} \).

   The cross product of two vectors \( \vec{n_1} = (a_1, b_1, c_1) \) and \( \vec{n_2} = (a_2, b_2, c_2) \) is given by:
   \[
   \vec{n} = \vec{n_1} \times \vec{n_2} = 
   \begin{vmatrix}
   \hat{i} & \hat{j} & \hat{k} \\
   a_1 & b_1 & c_1 \\
   a_2 & b_2 & c_2
   \end{vmatrix}
   \]
   which results in:
   \[
   \vec{n} = \left( (b_1c_2 - b_2c_1), (c_1a_2 - c_2a_1), (a_1b_2 - a_2b_1) \right)
   \]

4. **Construct the equation of the new plane:**
   Using the components of the cross product, \( \vec{n} = (A, B, C) \), the equation of the plane will be:
   \[
   Ax + By + Cz = D
   \]
   where \( D \) is a constant that can be determined if a point on the plane is known. If no specific point is given, the equation can be left in terms of \( D \).

### Example:
Suppose the two planes are:
\[
2x + 3y + z = 4 \quad \text{and} \quad x - y + 2z = 5.
\]
The normal vectors are:
\[
\vec{n_1} = (2, 3, 1), \quad \vec{n_2} = (1, -1, 2).
\]
The cross product of \( \vec{n_1} \) and \( \vec{n_2} \) is:
\[
\vec{n} = \left( (3 \times 2 - (-1) \times 1), (1 \times 1 - 2 \times 2), (2 \times (-1) - 3 \times 1) \right) = (7, -3, -5).
\]
Thus, the equation of the plane perpendicular to both is:
\[
7x - 3y - 5z = D,
\]
where \( D \) is a constant depending on a specific point, or can be left undetermined.

Would you like further details or have any specific questions?

Here are 5 related questions for further exploration:

1. How do you find the intersection line of two planes?
2. What is the condition for two planes to be parallel?
3. How do you find the angle between two planes?
4. Can a plane be perpendicular to only one plane and not another? If so, how?
5. How can you determine the distance between a point and a plane?

**Tip:** The cross product of two vectors is always perpendicular to both original vectors!

Discover how to find the equation of a plane that is perpendicular to two given planes. This solution covers vector algebra concepts, including normal vectors and the cross product. Suitable for students in Grades 11-12 or early college level.

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