https://file.aiquickdraw.com/mathbot/file/17254016882931376styb.jpg

To solve the problem in the image, we need to determine the equation of the plane that passes through the given points \( P(1, -1, 3) \), \( Q(2, -1, 2) \), and \( R(3, 0, 3) \).

### Step 1: Find the vectors
We need two vectors that lie on the plane. These can be obtained from the given points:
- Vector \( \mathbf{PQ} = Q - P = (2 - 1, -1 - (-1), 2 - 3) = (1, 0, -1) \)
- Vector \( \mathbf{PR} = R - P = (3 - 1, 0 - (-1), 3 - 3) = (2, 1, 0) \)

### Step 2: Find the normal vector
The normal vector to the plane can be found by taking the cross product of \( \mathbf{PQ} \) and \( \mathbf{PR} \).

\[
\mathbf{n} = \mathbf{PQ} \times \mathbf{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{vmatrix}
\]

\[
\mathbf{n} = \mathbf{i} \left( 0(0) - (-1)(1) \right) - \mathbf{j} \left( 1(0) - (-1)(2) \right) + \mathbf{k} \left( 1(1) - 0(2) \right)
\]

\[
\mathbf{n} = \mathbf{i}(1) - \mathbf{j}(2) + \mathbf{k}(1)
\]

So the normal vector is \( \mathbf{n} = (1, -2, 1) \).

### Step 3: Use the normal vector to find the plane equation
The general form of the plane equation is:
\[
a(x - x_1) + b(y - y_1) + c(z - z_1) = 0
\]
where \( (x_1, y_1, z_1) \) is a point on the plane (e.g., point \( P(1, -1, 3) \)) and \( \mathbf{n} = (a, b, c) = (1, -2, 1) \).

Substitute the values:
\[
1(x - 1) - 2(y + 1) + 1(z - 3) = 0
\]
Simplifying:
\[
x - 1 - 2y - 2 + z - 3 = 0
\]
\[
x - 2y + z - 6 = 0
\]

### Step 4: Conclusion
The equation of the plane is \( x - 2y + z = 6 \), which corresponds to the second option in the provided choices:
\[
x - 2y + z - 6 = 0
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you calculate the cross product of two vectors?
2. What is the geometric meaning of the normal vector to a plane?
3. How do you find the equation of a plane using three points in space?
4. How do you verify if a point lies on a plane?
5. What is the significance of the coefficients in the equation of a plane?

### Tip:
When solving plane equations, always check the normal vector, as it determines the orientation of the plane in space.

Find an equation in x, y, z for the plane that passes through the points: P(1, -1, 3), Q(2, -1, 2), R(3, 0, 3).

Discover how to find the equation of a plane passing through three points P(1, -1, 3), Q(2, -1, 2), and R(3, 0, 3). This solution uses key linear algebra concepts, including vector subtraction, the cross product, and the plane equation. Ideal for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation