Plane Passing Through Three Points (Vector Form)

In 3D geometry, a plane can be uniquely determined by three non-collinear points. Non-collinear means that the points do not all lie on a single straight line. The vector form is a useful way to represent this plane, as it allows us to use vector operations to solve various geometric problems.

Understanding the Vector Form of a Plane

The vector form of a plane is based on the concept of a normal vector. A normal vector to a plane is a vector that is perpendicular to every line lying on the plane. If we have a normal vector $\vec{n} = (a, b, c)$ and a point $P_0(x_0, y_0, z_0)$ on the plane, the equation of the plane in vector form is given by:

$$ \vec{n} \cdot (\vec{r} - \vec{r}_0) = 0 $$

where $\vec{r}$ is the position vector of any point $(x, y, z)$ on the plane, and $\vec{r}_0$ is the position vector of the point $P_0$.

Finding the Equation of a Plane Through Three Points

To find the equation of a plane that passes through three points $A$, $B$, and $C$, we can follow these steps:

1. Find the position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ corresponding to the points $A$, $B$, and $C$.
2. Compute two direction vectors $\vec{AB} = \vec{b} - \vec{a}$ and $\vec{AC} = \vec{c} - \vec{a}$.
3. Find the normal vector $\vec{n}$ to the plane by taking the cross product of $\vec{AB}$ and $\vec{AC}$: $\vec{n} = \vec{AB} \times \vec{AC}$.
4. Use the normal vector $\vec{n}$ and one of the points (say $A$) to write the equation of the plane.

Formula

Given points $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$, the equation of the plane in vector form is:

$$ (\vec{b} - \vec{a}) \times (\vec{c} - \vec{a}) \cdot (\vec{r} - \vec{a}) = 0 $$

Example

Let's find the equation of the plane passing through the points $A(1, 2, 3)$, $B(4, 5, 6)$, and $C(7, 8, 9)$.

1. Position vectors: $\vec{a} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}$, $\vec{b} = \begin{pmatrix} 4 \ 5 \ 6 \end{pmatrix}$, $\vec{c} = \begin{pmatrix} 7 \ 8 \ 9 \end{pmatrix}$
2. Direction vectors: $\vec{AB} = \vec{b} - \vec{a} = \begin{pmatrix} 3 \ 3 \ 3 \end{pmatrix}$, $\vec{AC} = \vec{c} - \vec{a} = \begin{pmatrix} 6 \ 6 \ 6 \end{pmatrix}$
3. Normal vector: $\vec{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$ (This is a zero vector, which means that points $A$, $B$, and $C$ are collinear, and there is no unique plane passing through them. For the sake of this example, let's choose different points.)

Let's choose different points $A(1, 0, 0)$, $B(0, 1, 0)$, and $C(0, 0, 1)$.

1. Position vectors: $\vec{a} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}$, $\vec{b} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}$, $\vec{c} = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$
2. Direction vectors: $\vec{AB} = \vec{b} - \vec{a} = \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix}$, $\vec{AC} = \vec{c} - \vec{a} = \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}$
3. Normal vector: $\vec{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix}$
4. Equation of the plane: $\vec{n} \cdot (\vec{r} - \vec{a}) = 0$, which simplifies to $x + y + z - 1 = 0$.

Differences and Important Points

Aspect	Description
Determination of a Plane	A plane is uniquely determined by three non-collinear points.
Normal Vector	A normal vector is perpendicular to the plane and is essential for defining the plane's equation.
Vector Form Equation	The vector form equation involves the dot product of the normal vector and the vector difference between any point on the plane and a known point on the plane.
Collinearity Issue	If the three points are collinear, they do not define a unique plane, and the cross product of the direction vectors will result in a zero vector.
Cross Product	The cross product of two direction vectors gives the normal vector to the plane.
Practical Applications	This method is widely used in computer graphics, engineering, and physics to define planes in 3D space for various calculations, such as reflections, projections, and determining the intersection of lines and other planes.

By understanding the vector form of a plane and how to derive it from three points, students can solve a wide range of geometric problems related to planes in three-dimensional space.