Condition for two lines to be intersecting, coplanar and the plane containing them (vector form)
Condition for Two Lines to be Intersecting, Coplanar and the Plane Containing Them (Vector Form)
Understanding the relationship between two lines in three-dimensional space is a fundamental aspect of 3D geometry. Specifically, we are interested in the conditions that determine whether two lines intersect, are coplanar, or the plane that contains them. We will explore these concepts using vector notation.
Vector Equation of a Line
Before we delve into the conditions for intersection and coplanarity, let's recall the vector equation of a line in 3D space:
$$ \vec{r} = \vec{a} + \lambda \vec{b} $$
where:
- $\vec{r}$ is the position vector of any point on the line,
- $\vec{a}$ is the position vector of a fixed point on the line,
- $\vec{b}$ is the direction vector of the line (non-zero),
- $\lambda$ is a scalar parameter.
Intersecting Lines
Two lines are said to intersect if there is a common point that lies on both lines. For two lines with equations:
$$ \vec{r}_1 = \vec{a}_1 + \lambda_1 \vec{b}_1 $$ $$ \vec{r}_2 = \vec{a}_2 + \lambda_2 \vec{b}_2 $$
the condition for them to intersect is that there exist values of $\lambda_1$ and $\lambda_2$ such that:
$$ \vec{a}_1 + \lambda_1 \vec{b}_1 = \vec{a}_2 + \lambda_2 \vec{b}_2 $$
This equation implies that the two lines share a common point.
Coplanar Lines
Two lines are coplanar if they lie in the same plane. For two lines to be coplanar, the direction vectors $\vec{b}_1$ and $\vec{b}_2$, and the vector connecting a point on one line to a point on the other line, say $\vec{a}_2 - \vec{a}_1$, must be coplanar. This is equivalent to saying that the scalar triple product of these vectors must be zero:
$$ (\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = 0 $$
If this condition is satisfied, the lines are either parallel or intersecting, but certainly coplanar.
Plane Containing Two Lines
If two lines are coplanar, there exists a plane that contains both lines. The equation of the plane containing the two lines can be written using the normal vector of the plane, which is the cross product of the direction vectors of the lines:
$$ \vec{n} = \vec{b}_1 \times \vec{b}_2 $$
The equation of the plane is then given by:
$$ \vec{n} \cdot (\vec{r} - \vec{a}_1) = 0 $$
where $\vec{r}$ is the position vector of any point on the plane, and $\vec{a}_1$ is a point on one of the lines.
Table of Differences and Important Points
| Property | Intersecting Lines | Coplanar Lines | Plane Containing Two Lines |
|---|---|---|---|
| Condition | $\exists \lambda_1, \lambda_2: \vec{a}_1 + \lambda_1 \vec{b}_1 = \vec{a}_2 + \lambda_2 \vec{b}_2$ | $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = 0$ | $\vec{n} = \vec{b}_1 \times \vec{b}_2$ |
| Geometric Meaning | Common point exists | Lines lie in the same plane | Plane defined by a normal vector |
| Normal Vector | Not applicable | Not directly used | $\vec{n} = \vec{b}_1 \times \vec{b}_2$ |
| Equation Form | Parametric equations equal | Scalar triple product is zero | $\vec{n} \cdot (\vec{r} - \vec{a}_1) = 0$ |
Examples
Example 1: Intersecting Lines
Consider two lines with equations:
$$ \vec{r}_1 = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} + \lambda_1 \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} $$ $$ \vec{r}_2 = \begin{pmatrix} 3 \ 4 \ 5 \end{pmatrix} + \lambda_2 \begin{pmatrix} -1 \ 0 \ -1 \end{pmatrix} $$
To find if they intersect, we set $\vec{r}_1 = \vec{r}_2$ and solve for $\lambda_1$ and $\lambda_2$. If a solution exists, the lines intersect.
Example 2: Coplanar Lines
For the same lines as in Example 1, we calculate the scalar triple product:
$$ (\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \times \begin{pmatrix} -1 \ 0 \ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 2 \ 2 \end{pmatrix} $$
If this product equals zero, the lines are coplanar.
Example 3: Plane Containing Two Lines
Assuming the lines from Example 1 are coplanar, we find the normal vector $\vec{n}$:
$$ \vec{n} = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \times \begin{pmatrix} -1 \ 0 \ -1 \end{pmatrix} $$
Then, the equation of the plane containing the two lines is:
$$ \vec{n} \cdot (\vec{r} - \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}) = 0 $$
By understanding these conditions and applying the vector form equations, one can determine the spatial relationship between any two lines in 3D geometry.