find the line of the intersection between the plane that has normal vector (2,-4,-3) and passes through the point (2,0,4) and the plane 6(x-2)-33(y-1)-9(z-1)=0

We are tasked with finding the line of intersection between two planes. 

### Step 1: Define the planes.
1. The first plane has a **normal vector** \( \mathbf{n_1} = (2, -4, -3) \) and passes through the point \( P_1(2, 0, 4) \).
   
   The equation of a plane is given by:
   \[
   2(x - 2) - 4(y - 0) - 3(z - 4) = 0
   \]
   Expanding this:
   \[
   2(x - 2) - 4y - 3(z - 4) = 0
   \]
   \[
   2x - 4 - 4y - 3z + 12 = 0
   \]
   \[
   2x - 4y - 3z + 8 = 0
   \]
   So, the equation of the first plane is:
   \[
   2x - 4y - 3z + 8 = 0
   \]

2. The second plane is already provided in expanded form:
   \[
   6(x - 2) - 33(y - 1) - 9(z - 1) = 0
   \]
   Expanding this:
   \[
   6x - 12 - 33y + 33 - 9z + 9 = 0
   \]
   \[
   6x - 33y - 9z + 30 = 0
   \]

### Step 2: Find the line of intersection.

To find the line of intersection, we first find the **direction vector** of the line, which is the **cross product** of the normal vectors of the two planes.

- The normal vector of the first plane is \( \mathbf{n_1} = (2, -4, -3) \).
- The normal vector of the second plane is \( \mathbf{n_2} = (6, -33, -9) \).

The cross product \( \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \) will give the direction vector of the line of intersection.

### Step 3: Solve the system of equations for a specific point.

We now find a specific point on the line by solving the system of plane equations simultaneously. Once we find this point and have the direction vector, we can write the parametric equations for the line.

Let me calculate these details.The direction vector of the line of intersection is \( \mathbf{d} = (-63, 0, -42) \).

### Step 4: Solve for a specific point on the line

Now, to find a specific point on the line, we solve the system of the two plane equations. The two planes are:

1. \( 2x - 4y - 3z + 8 = 0 \)
2. \( 6x - 33y - 9z + 30 = 0 \)

We can set one of the variables (e.g., \( y = 0 \)) and solve for \( x \) and \( z \).

Let me calculate the coordinates of a point on the line.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, I can guide you on how to find the point manually:

1. **Set \( y = 0 \)** and substitute into both plane equations:
   \[
   2x - 3z + 8 = 0 \quad \text{(Plane 1)}
   \]
   \[
   6x - 9z + 30 = 0 \quad \text{(Plane 2)}
   \]

2. **Solve the system of two equations** to find \( x \) and \( z \). This will give you a specific point on the line of intersection.

If you need further clarification or have any questions, feel free to ask!

Here are 5 related questions to explore:
1. How do you find the parametric equations of a line in 3D?
2. What is the geometric meaning of the cross product of two vectors?
3. How can you determine if two planes are parallel?
4. How do you interpret the direction vector of a line?
5. How can you find the angle between two planes?

**Tip:** To solve systems of equations, substitution and elimination methods are commonly used in algebra.

Find the line of the intersection between the plane that has normal vector (2,-4,-3) and passes through the point (2,0,4) and the plane 6(x-2)-33(y-1)-9(z-1)=0.

Explore how to find the line of intersection between two planes, including the use of cross product to find the direction vector. This example uses planes with normal vectors (2, -4, -3) and (6, -33, -9) to solve for the parametric equations of the line. Suitable for advanced high school and college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation