To find the equation of a plane passing through three given points, we need to determine a normal vector to the plane using the cross product of two vectors lying on the plane. Let's go through the step-by-step process.

Given Points

• $A = [1, -1, -1]$
• $B = [-2, 2, -1]$
• $C = [-3, 0, 2]$

Step 1: Calculate Vectors on the Plane

Let's find two vectors that lie on the plane by taking the difference between the given points:

Vector $\mathbf{AB}$:

[ \mathbf{AB} = B - A = [-2, 2, -1] - [1, -1, -1] = [-3, 3, 0] ]

Vector $\mathbf{AC}$:

[ \mathbf{AC} = C - A = [-3, 0, 2] - [1, -1, -1] = [-4, 1, 3] ]

Step 2: Find the Normal Vector $\mathbf{n}$

The normal vector $\mathbf{n}$ to the plane can be found using the cross product of $\mathbf{AB}$ and $\mathbf{AC}$:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 3 & 0 \\ -4 & 1 & 3 \end{vmatrix}$$ #### Calculating the determinant: $$\mathbf{n} = \mathbf{i}(3 \cdot 3 - 0 \cdot 1) - \mathbf{j}(-3 \cdot 3 - 0 \cdot -4) + \mathbf{k}(-3 \cdot 1 - 3 \cdot -4)$$ $$= \mathbf{i}(9) - \mathbf{j}(-9) + \mathbf{k}(-3 + 12)$$ $$= \mathbf{i}(9) + \mathbf{j}(9) + \mathbf{k}(9)$$ $$= [9, 9, 9]$$ **Simplifying**, $$\mathbf{n}$$ becomes $$[1, 1, 1]$$. ### Step 3: Equation of the Plane The general equation of a plane with normal vector $$\mathbf{n} = [a, b, c]$$ passing through a point $$A(x_0, y_0, z_0)$$ is: $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$ Using point $$A = [1, -1, -1]$$ and $$\mathbf{n} = [1, 1, 1]$$: $$1(x - 1) + 1(y + 1) + 1(z + 1) = 0$$ Simplify this equation: $$(x - 1) + (y + 1) + (z + 1) = 0$$ $$x + y + z + 1 = 0$$ $$x + y + z = -1$$ ### Final Equation of the Plane The equation of the plane passing through the points $$[1, -1, -1]$$, $$[-2, 2, -1]$$, and $$[-3, 0, 2]$$ is: $$\boxed{x + y + z = -1}$$ ### Verification Let's verify that the plane equation satisfies all three given points. #### Point $$A = [1, -1, -1]$$: $$1 + (-1) + (-1) = -1 \quad \text{(True)}$$ #### Point $$B = [-2, 2, -1]$$: $$-2 + 2 + (-1) = -1 \quad \text{(True)}$$ #### Point $$C = [-3, 0, 2]$$: $$-3 + 0 + 2 = -1 \quad \text{(True)}$$ Thus, the plane equation $$x + y + z = -1$$ satisfies all three points, confirming it is correct. --- If you have any further questions or need more clarification, feel free to ask! Here are some suggested questions you might consider exploring: 1. **How does the cross product help in determining the normal vector?** 2. **Can the equation of a plane be simplified further?** 3. **What if two of the given points are the same?** 4. **How to find the distance from a point to this plane?** 5. **What happens if the points are collinear?** ### Tip: When determining the equation of a plane, ensure that your normal vector is non-zero; otherwise, the points might be collinear.