Condition for a line to lie in a plane (cartesian form)
Condition for a Line to Lie in a Plane (Cartesian 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 if a line lies in a plane, we need to consider the equations that represent both the line and the plane in Cartesian form.
Equation of a Line in Space
A line in three-dimensional space can be represented in Cartesian form using parametric equations:
$$ \begin{align} x &= x_0 + at \ y &= y_0 + bt \ z &= z_0 + ct \end{align} $$
where $(x_0, y_0, z_0)$ is a point on the line, $(a, b, c)$ is the direction vector of the line, and $t$ is the parameter.
Equation of a Plane in Space
A plane in three-dimensional space can be represented in Cartesian form using the general equation:
$$ Ax + By + Cz + D = 0 $$
where $A$, $B$, $C$, and $D$ are constants, and $(x, y, z)$ is any point on 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 equation of the plane. This leads to the following conditions:
- The direction vector of the line must be parallel to the plane.
- A point on the line must satisfy the equation of the plane.
Condition 1: Direction Vector Parallel to Plane
The direction vector of the line $(a, b, c)$ must be perpendicular to the normal vector of the plane $(A, B, C)$. This is because if the direction vector is not perpendicular to the normal vector, the line will eventually leave the plane. The condition for the direction vector to be parallel to the plane is given by the cross product of the direction vector and the normal vector being zero:
$$ \begin{vmatrix} i & j & k \ a & b & c \ A & B & C \ \end{vmatrix} = \mathbf{0} $$
Condition 2: Point on Line Satisfying Plane Equation
A point $(x_0, y_0, z_0)$ on the line must satisfy the equation of the plane:
$$ Ax_0 + By_0 + Cz_0 + D = 0 $$
If both conditions are met, the line lies in the plane.
Table of Differences and Important Points
| Aspect | Line in Plane | Line Intersecting Plane | Line Parallel to Plane |
|---|---|---|---|
| Direction Vector | Parallel to plane | Not parallel to plane | Parallel to plane |
| Point on Line | Satisfies plane equation | Satisfies plane equation at intersection point | Does not satisfy plane equation |
| Cross Product | Zero | Non-zero | Zero |
| Condition | Both conditions must be met | Only intersection condition must be met | Line does not satisfy plane equation |
Examples
Example 1: Line Lying in a Plane
Consider a line with direction vector $(2, 3, 4)$ and passing through the point $(1, 2, 3)$. The plane has the equation $2x + 3y + 4z + 5 = 0$.
Condition 1: The direction vector is parallel to the plane since it is the same as the normal vector of the plane.
Condition 2: The point $(1, 2, 3)$ satisfies the plane equation:
$$ 2(1) + 3(2) + 4(3) + 5 = 2 + 6 + 12 + 5 = 25 \neq 0 $$
The point does not satisfy the plane equation, so the line does not lie in the plane.
Example 2: Line Lying in a Plane
Consider a line with direction vector $(1, -2, 1)$ and passing through the point $(3, -6, 3)$. The plane has the equation $x - 2y + z - 4 = 0$.
Condition 1: We need to check if the direction vector is parallel to the plane. Since there is no normal vector given, we cannot directly check this condition.
Condition 2: The point $(3, -6, 3)$ satisfies the plane equation:
$$ (3) - 2(-6) + (3) - 4 = 3 + 12 + 3 - 4 = 14 \neq 0 $$
The point does not satisfy the plane equation, so the line does not lie in the plane.
In conclusion, to determine if a line lies in a plane, we must check both conditions: the direction vector must be parallel to the plane, and a point on the line must satisfy the plane equation. If either condition is not met, the line does not lie in the plane.