To find the equation of the plane passing through the points A(2, 1, -3), B(-3, -2, 1), and C(2, 4, -1), we can follow these steps:
### Step 1: Identify the Points
We have the points:
- A(2, 1, -3)
- B(-3, -2, 1)
- C(2, 4, -1)
### Step 2: Find Two Vectors in the Plane
We can find two vectors that lie in the plane by subtracting the coordinates of these points:
- Vector AB = B - A = (-3 - 2, -2 - 1, 1 - (-3)) = (-5, -3, 4)
- Vector AC = C - A = (2 - 2, 4 - 1, -1 - (-3)) = (0, 3, 2)
### Step 3: Find the Normal Vector
The normal vector to the plane can be found using the cross product of vectors AB and AC:
Let \( \mathbf{AB} = (-5, -3, 4) \) and \( \mathbf{AC} = (0, 3, 2) \).
The cross product \( \mathbf{N} = \mathbf{AB} \times \mathbf{AC} \) is calculated as follows:
\[
\mathbf{N} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-5 & -3 & 4 \\
0 & 3 & 2
\end{vmatrix}
\]
Calculating the determinant:
\[
\mathbf{N} = \mathbf{i}((-3)(2) - (4)(3)) - \mathbf{j}((-5)(2) - (4)(0)) + \mathbf{k}((-5)(3) - (-3)(0))
\]
\[
= \mathbf{i}(-6 - 12) - \mathbf{j}(-10 - 0) + \mathbf{k}(-15 - 0)
\]
\[
= \mathbf{i}(-18) + \mathbf{j}(10) + \mathbf{k}(-15)
\]
Thus, the normal vector \( \mathbf{N} = (-18, 10, -15) \).
### Step 4: Write the Equation of the Plane
The equation of a plane can be written as:
\[
A(x - x_1) + B(y - y_1) + C(z - z_1) = 0
\]
where \( (x_1, y_1, z_1) \) is a point on the plane (we can use point A) and \( (A, B, C) \) are the components of the normal vector.
Using point A(2, 1, -3) and normal vector \( (-18, 10, -15) \):
\[
-18(x - 2) + 10(y - 1) - 15(z + 3) = 0
\]
### Step 5: Simplify the Equation
Expanding this equation:
\[
-18x + 36 + 10y - 10 - 15z - 45 = 0
\]
Combining like terms:
\[
-18x + 10y - 15z + (36 - 10 - 45) = 0
\]
\[
-18x + 10y - 15z - 19 = 0
\]
### Final Equation
Thus, the equation of the plane is:
\[
18x - 10y + 15z + 19 = 0
\]
