To find the equation of a plane containing the points $A(1, -2, 3)$, $B(5, 1, 1)$, and $C(3, -3, 1)$, we can follow these steps:

Step 1: Find two vectors in the plane

We first need to find two vectors in the plane using the given points:

• Vector $\vec{AB}$ from $A$ to $B$: $\vec{AB} = B - A = (5 - 1, 1 - (-2), 1 - 3) = (4, 3, -2)$
• Vector $\vec{AC}$ from $A$ to $C$: $\vec{AC} = C - A = (3 - 1, -3 - (-2), 1 - 3) = (2, -1, -2)$

Step 2: Find the normal vector to the plane

The normal vector $\vec{n}$ to the plane is given by the cross product of vectors $\vec{AB}$ and $\vec{AC}$.

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & -2 \\ 2 & -1 & -2 \\ \end{vmatrix}$$ This is calculated as follows: $$\vec{n} = \mathbf{i} \left(3 \times -2 - (-1 \times -2)\right) - \mathbf{j} \left(4 \times -2 - 2 \times -2\right) + \mathbf{k} \left(4 \times -1 - 3 \times 2\right)$$ $$\vec{n} = \mathbf{i}(-6 - 2) - \mathbf{j}(-8 + 4) + \mathbf{k}(-4 - 6)$$ $$\vec{n} = \mathbf{i}(-8) - \mathbf{j}(-4) + \mathbf{k}(-10)$$ $$\vec{n} = (-8, 4, -10)$$ So the normal vector $$\vec{n}$$ to the plane is $$(-8, 4, -10)$$. ### Step 3: Write the equation of the plane The equation of a plane can be written in the form: $$Ax + By + Cz = D$$ Where $$(A, B, C)$$ is the normal vector and $$(x_0, y_0, z_0)$$ is a point on the plane. Using point $$A(1, -2, 3)$$ and the normal vector $$(-8, 4, -10)$$, we substitute into the equation: $$-8(x - 1) + 4(y + 2) - 10(z - 3) = 0$$ Simplifying: $$-8x + 8 + 4y + 8 - 10z + 30 = 0$$ $$-8x + 4y - 10z + 46 = 0$$ Thus, the equation of the plane is: $$8x - 4y + 10z = 46$$ Let me know if you need further clarification! ### Related Questions: 1. How do you compute the cross product of two vectors? 2. What is the geometric significance of the normal vector to a plane? 3. Can a plane in 3D space be represented using parametric equations? 4. How do you verify if a point lies on a given plane? 5. What are the different forms of representing the equation of a plane? ### Tip: When solving for the equation of a plane, the normal vector is key since it determines the plane's orientation in space.