Condition for a Line to Lie in a Plane (Vector Form)

In three-dimensional geometry, a line can either lie in a plane, intersect a plane at a single point, or be parallel to a plane without intersecting it. To determine the condition for a line to lie in a plane using vector form, we need to understand the equations that represent lines and planes in three-dimensional space.

Vector Equation of a Line

A line in three-dimensional space can be represented in vector form as:

$$ \vec{r} = \vec{a} + \lambda \vec{b} $$

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.

Vector Equation of a Plane

A plane in three-dimensional space can be represented in vector form as:

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

where:

• $\vec{r}$ is the position vector of any point on the plane,
• $\vec{n}$ is the normal vector to the plane (non-zero),
• $d$ is the perpendicular distance from the origin to the plane.

Condition for a Line to Lie in a Plane

For a line to lie entirely within a plane, every point on the line must satisfy the plane's equation. This leads to two main conditions:

1. The direction vector of the line must be perpendicular to the normal vector of the plane.
2. Any point on the line must satisfy the plane's equation.

Condition 1: Perpendicular Direction Vector

The direction vector $\vec{b}$ of the line must be perpendicular to the normal vector $\vec{n}$ of the plane. This is expressed as:

$$ \vec{b} \cdot \vec{n} = 0 $$

This condition ensures that the line is not crossing the plane at any point.

Condition 2: Point on the Plane

A point $\vec{a}$ on the line must satisfy the plane's equation:

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

This condition ensures that the line is not parallel and outside the plane.

Summary Table

Condition	Description	Mathematical Expression
1	Direction vector of the line is perpendicular to the normal vector of the plane	$\vec{b} \cdot \vec{n} = 0$
2	A point on the line satisfies the plane's equation	$\vec{a} \cdot \vec{n} = d$

Examples

Example 1: Verifying a Line Lies in a Plane

Consider a line with direction vector $\vec{b} = \begin{pmatrix} 2 \ 3 \ -1 \end{pmatrix}$ and a point on the line with position vector $\vec{a} = \begin{pmatrix} 1 \ -2 \ 3 \end{pmatrix}$. Let's check if this line lies in the plane with normal vector $\vec{n} = \begin{pmatrix} 6 \ 9 \ -3 \end{pmatrix}$ and $d = 6$.

Condition 1:

$$ \vec{b} \cdot \vec{n} = \begin{pmatrix} 2 \ 3 \ -1 \end{pmatrix} \cdot \begin{pmatrix} 6 \ 9 \ -3 \end{pmatrix} = 12 + 27 + 3 = 42 \neq 0 $$

Since $\vec{b} \cdot \vec{n} \neq 0$, the direction vector is not perpendicular to the normal vector, and the line does not lie in the plane.

Example 2: Line Lying in a Plane

Consider a line with direction vector $\vec{b} = \begin{pmatrix} 1 \ -2 \ 1 \end{pmatrix}$ and a point on the line with position vector $\vec{a} = \begin{pmatrix} 3 \ 2 \ -1 \end{pmatrix}$. Let's check if this line lies in the plane with normal vector $\vec{n} = \begin{pmatrix} 2 \ -4 \ 2 \end{pmatrix}$ and $d = 6$.

Condition 1:

$$ \vec{b} \cdot \vec{n} = \begin{pmatrix} 1 \ -2 \ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ -4 \ 2 \end{pmatrix} = 2 - 8 + 2 = 0 $$

The direction vector is perpendicular to the normal vector, satisfying the first condition.

Condition 2:

$$ \vec{a} \cdot \vec{n} = \begin{pmatrix} 3 \ 2 \ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ -4 \ 2 \end{pmatrix} = 6 - 8 - 2 = -4 \neq 6 $$

The point on the line does not satisfy the plane's equation, so the second condition is not met, and the line does not lie in the plane.

In conclusion, for a line to lie in a plane, it must satisfy both conditions: the direction vector must be perpendicular to the normal vector of the plane, and any point on the line must satisfy the plane's equation. If either condition is not met, the line does not lie in the plane.