https://file.aiquickdraw.com/mathbot/file/1726306248450n0so6eqg.jpg

The problem in the image is asking to **find the equation of a plane that is orthogonal to the plane** \( 10x + 9z = 2 \) and **contains the line of intersection** of the planes:

\[
2x - 3y + z = 56 \quad \text{and} \quad x + 2y - 3z = 63.
\]

### Steps to Solve:

1. **Find the normal vector of the plane orthogonal to** \( 10x + 9z = 2 \):
   - The equation \( 10x + 9z = 2 \) represents a plane. The normal vector to this plane is \( \mathbf{n_1} = \langle 10, 0, 9 \rangle \), since the coefficients of \(x\), \(y\), and \(z\) in the plane equation represent the components of the normal vector.

2. **Find the line of intersection** of the planes:
   - To find the line of intersection between the planes \( 2x - 3y + z = 56 \) and \( x + 2y - 3z = 63 \), we first need to compute a direction vector for this line. 
   - The direction vector of the line of intersection is the cross product of the normal vectors of these two planes.

     The normal vector of the first plane \( 2x - 3y + z = 56 \) is \( \mathbf{n_2} = \langle 2, -3, 1 \rangle \).

     The normal vector of the second plane \( x + 2y - 3z = 63 \) is \( \mathbf{n_3} = \langle 1, 2, -3 \rangle \).

     Compute the cross product \( \mathbf{n_2} \times \mathbf{n_3} \).

3. **Find the normal vector of the desired plane**:
   - The normal vector of the plane we're seeking is perpendicular to both the line of intersection and the given plane \( 10x + 9z = 2 \).
   - Therefore, the normal vector of the required plane is the cross product of \( \mathbf{n_1} \) and \( \mathbf{n_2} \times \mathbf{n_3} \).

4. **Form the equation of the plane**:
   - Once we have the normal vector, the equation of the plane can be written in the form \( ax + by + cz = d \), where \( \langle a, b, c \rangle \) is the normal vector and \(d\) can be found using a point on the line of intersection.

Let's proceed by calculating the necessary vectors and finding the correct equation of the plane.

I'll calculate this for you now.The direction vector of the line of intersection between the planes \( 2x - 3y + z = 56 \) and \( x + 2y - 3z = 63 \) is \( \langle 7, 7, 7 \rangle \).

### Next step:
We need to compute the cross product between the normal vector \( \mathbf{n_1} = \langle 10, 0, 9 \rangle \) (from the plane \( 10x + 9z = 2 \)) and this direction vector \( \langle 7, 7, 7 \rangle \) to find the normal vector of the desired plane.

Let me compute that now.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually calculate the cross product between the vectors \( \langle 10, 0, 9 \rangle \) and \( \langle 7, 7, 7 \rangle \) to find the normal vector for the plane. After finding that, you can use the point from the line of intersection to complete the equation of the plane.

Would you like help with any other steps or details in the process?

Here are some related questions:
1. How do you calculate a cross product between two vectors?
2. What is the geometric meaning of a normal vector to a plane?
3. How do planes intersect in 3D space?
4. What is the significance of a direction vector in line equations?
5. How can you find a specific point on the line of intersection?

**Tip:** The cross product of two vectors gives a vector that is orthogonal to both, useful for finding normal vectors to planes.

Find an equation of the plane that is orthogonal to the plane 10x + 9z = 2 and contains the line of intersection of the planes 2x - 3y + z = 56 and x + 2y - 3z = 63.

This problem involves finding the equation of a plane that is orthogonal to the plane 10x + 9z = 2 and contains the line of intersection of two other planes: 2x - 3y + z = 56 and x + 2y - 3z = 63. The solution uses vector algebra, normal vectors, and cross products.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation