Plane Passing Through Two Points and Parallel to Two Vectors (Vector 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 given points and is parallel to two given vectors. This scenario is common in vector calculus and 3D geometry problems.

Understanding the Concept

To define a plane in 3D space, we need three non-collinear points or one point and two non-parallel vectors lying on the plane. If we have two points, we can find a vector that connects them, and if we have two vectors, we can find their cross product to get a normal vector to the plane.

Vector Form of a Plane

The vector equation of a plane is given by:

$$ \vec{r} \cdot \vec{n} = d $$

where:

• $\vec{r}$ is the position vector of any point on the plane,
• $\vec{n}$ is the normal vector to the plane,
• $d$ is the scalar product of the normal vector with the position vector of a known point on the plane.

Plane Through Two Points and Parallel to Two Vectors

Given two points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$, and two vectors $\vec{p}$ and $\vec{q}$ that are parallel to the plane, we can determine the equation of the plane by following these steps:

1. Find the vector connecting points $A$ and $B$, which is $\vec{AB} = \vec{b} - \vec{a}$.
2. Since $\vec{p}$ and $\vec{q}$ are parallel to the plane, their cross product $\vec{n} = \vec{p} \times \vec{q}$ will give us a normal vector to the plane.
3. Use point $A$ (or $B$) and the normal vector $\vec{n}$ to write the plane's equation.

Formulas

Given two points $A(\vec{a})$ and $B(\vec{b})$, and two vectors $\vec{p}$ and $\vec{q}$, the equation of the plane can be written as:

$$ (\vec{r} - \vec{a}) \cdot (\vec{p} \times \vec{q}) = 0 $$

where $\vec{r}$ is the position vector of any point on the plane.

Examples

Let's consider an example to illustrate the concept.

Example 1

Find the equation of the plane passing through the points $A(1, 2, 3)$ and $B(4, 5, 6)$, and parallel to the vectors $\vec{p} = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}$ and $\vec{q} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix}$.

Solution:

1. Calculate $\vec{AB}$:

$$ \vec{AB} = \vec{b} - \vec{a} = \begin{bmatrix} 4 - 1 \ 5 - 2 \ 6 - 3 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \ 3 \end{bmatrix} $$

1. Find the normal vector $\vec{n}$ using the cross product of $\vec{p}$ and $\vec{q}$:

$$ \vec{n} = \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 1 \ 0 & 1 & 1 \ \end{vmatrix} = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} $$

1. Write the equation of the plane using point $A$ and normal vector $\vec{n}$:

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

$$ (\vec{r} - \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}) \cdot \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} = 0 $$

$$ \vec{r} \cdot \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} - \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \cdot \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} = 0 $$

$$ \vec{r} \cdot \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} - (1 - 2 + 3) = 0 $$

$$ \vec{r} \cdot \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} - 2 = 0 $$

Therefore, the vector equation of the plane is:

$$ \vec{r} \cdot \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix} = 2 $$

Table of Differences and Important Points

Feature	Point-Normal Form	Two Points and Two Vectors Form
Known Quantities	One point and a normal vector	Two points and two vectors parallel to the plane
Normal Vector	Given directly	Obtained by the cross product of the two vectors
Equation Form	$\vec{r} \cdot \vec{n} = d$	$(\vec{r} - \vec{a}) \cdot (\vec{p} \times \vec{q}) = 0$
Geometric Interpretation	Plane is perpendicular to the normal vector	Plane contains the two vectors and the line segment joining the two points

Conclusion

Understanding how to find the equation of a plane passing through two points and parallel to two vectors is crucial for solving problems in 3D geometry. By following the steps outlined above and using the cross product to find a normal vector, one can easily derive the plane's equation in vector form. Remember that the cross product of the two vectors gives a direction perpendicular to both, which is essential for defining the plane's orientation in space.