Plane Passing Through Two Points and Parallel to Two Vectors (Cartesian Form)

In 3D geometry, a plane can be uniquely determined by a point and a normal vector to the plane. However, there are situations where we need to find the equation of a plane that passes through two specific points and is parallel to two given vectors. This scenario is common in vector calculus and geometric modeling.

Understanding the Concept

To define a plane in 3D space, we need three non-collinear points or one point and two non-parallel vectors. The two vectors give us the direction of the plane, and the point fixes the plane in space.

Cartesian Form of a Plane

The general Cartesian form of a plane is given by:

$$ Ax + By + Cz + D = 0 $$

where (A), (B), and (C) are the components of the normal vector to the plane, and (D) is a constant.

Determining the Plane

Given two points (P_1(x_1, y_1, z_1)) and (P_2(x_2, y_2, z_2)), and two parallel vectors (\vec{v}) and (\vec{w}), we can find the equation of the plane as follows:

1. Find the vector (\vec{P_1P_2}) which is the displacement vector from (P_1) to (P_2).
2. Ensure that (\vec{v}) and (\vec{w}) are not parallel to (\vec{P_1P_2}) to avoid degeneracy.
3. Find the normal vector (\vec{n}) to the plane by taking the cross product of (\vec{v}) and (\vec{w}).
4. Use point (P_1) (or (P_2)) and the normal vector (\vec{n}) to find the equation of the plane.

The Cross Product

The cross product of two vectors (\vec{a} = \langle a_1, a_2, a_3 \rangle) and (\vec{b} = \langle b_1, b_2, b_3 \rangle) is given by:

$$ \vec{a} \times \vec{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle $$

This new vector is perpendicular to both (\vec{a}) and (\vec{b}).

Equation of the Plane

Once we have the normal vector (\vec{n} = \langle A, B, C \rangle), and a point (P_1(x_1, y_1, z_1)), the equation of the plane is:

$$ A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 $$

Differences and Important Points

Aspect	Point-Normal Form	Two Points and Two Vectors Form
Definition	A plane defined by a point and a normal vector.	A plane defined by two points and two vectors parallel to the plane.
Normal Vector	Given directly or computed from two non-parallel vectors.	Computed using the cross product of the two given vectors.
Equation	(Ax + By + Cz + D = 0) where (\vec{n} = \langle A, B, C \rangle) is the normal vector.	Derived from the normal vector and one of the given points.
Uniqueness	Unique if the normal vector is non-zero.	Unique if the vectors are not parallel and not parallel to the vector formed by the two points.
Geometric Interpretation	The normal vector is perpendicular to every vector lying on the plane.	The plane extends infinitely in the directions of the two given vectors.

Example

Let's find the equation of a plane that passes through the points (P_1(1, 2, 3)) and (P_2(4, 5, 6)), and is parallel to the vectors (\vec{v} = \langle 1, 0, 1 \rangle) and (\vec{w} = \langle 0, 1, 1 \rangle).

1. Compute the vector (\vec{P_1P_2} = \langle 4 - 1, 5 - 2, 6 - 3 \rangle = \langle 3, 3, 3 \rangle).
2. Compute the normal vector (\vec{n}) using the cross product of (\vec{v}) and (\vec{w}):

$$ \vec{n} = \vec{v} \times \vec{w} = \langle 0 \cdot 1 - 1 \cdot 1, 1 \cdot 0 - 1 \cdot 1, 1 \cdot 1 - 0 \cdot 0 \rangle = \langle -1, -1, 1 \rangle $$

1. Use point (P_1) and the normal vector (\vec{n}) to find the equation of the plane:

$$ -1(x - 1) - 1(y - 2) + 1(z - 3) = 0 \ -x + y + z - 1 - 2 + 3 = 0 \ -x + y + z = 0 $$

Therefore, the equation of the plane is (x - y - z = 0).

This example illustrates how to find the equation of a plane given two points and two vectors parallel to the plane. Remember to verify that the vectors are not parallel to each other and not parallel to the vector formed by the two points to ensure a unique plane is defined.