Plane Passing Through a Point and Perpendicular to a Vector (Cartesian Form)

In three-dimensional geometry, a plane can be defined in various ways. One common definition is by specifying a point through which the plane passes and a vector that is perpendicular to the plane. This is known as the normal form of the equation of a plane.

Understanding the Concept

A plane in 3D space can be uniquely determined if we know:

1. A point ( P(x_1, y_1, z_1) ) through which the plane passes.
2. A normal vector ( \vec{N} = (a, b, c) ) which is perpendicular to the plane.

The normal vector provides the direction that is orthogonal to every line lying on the plane.

Cartesian Form of the Plane Equation

The equation of a plane in Cartesian form that passes through a point ( P(x_1, y_1, z_1) ) and is perpendicular to a vector ( \vec{N} = (a, b, c) ) is given by:

[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 ]

Here, ( (x, y, z) ) represents any point on the plane.

Derivation of the Equation

To derive this equation, consider a point ( Q(x, y, z) ) on the plane. The vector ( \overrightarrow{PQ} ) drawn from ( P ) to ( Q ) lies on the plane and hence is perpendicular to the normal vector ( \vec{N} ). The dot product of ( \overrightarrow{PQ} ) and ( \vec{N} ) must be zero:

[ \vec{N} \cdot \overrightarrow{PQ} = 0 ]

Expanding this, we get:

[ (a, b, c) \cdot ((x - x_1), (y - y_1), (z - z_1)) = 0 ]

Which simplifies to the Cartesian form of the plane equation.

Important Points and Differences

Aspect	Description
Point on the plane	The plane must pass through a specific point ( P(x_1, y_1, z_1) ).
Normal vector	A vector ( \vec{N} = (a, b, c) ) that is perpendicular to the plane.
Cartesian equation	The general form is ( ax + by + cz + d = 0 ), where ( d ) is the constant term.
Equation through a point	The form ( a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 ) includes the coordinates of point ( P ).
Perpendicularity condition	The dot product of the normal vector and any vector on the plane is zero.

Examples

Example 1: Finding the Equation of a Plane

Given: A plane passes through the point ( P(2, -1, 3) ) and is perpendicular to the vector ( \vec{N} = (4, 5, 6) ).

Find: The equation of the plane.

Solution:

Using the Cartesian form of the plane equation:

[ 4(x - 2) + 5(y + 1) + 6(z - 3) = 0 ]

Expanding and simplifying:

[ 4x - 8 + 5y + 5 + 6z - 18 = 0 ]

[ 4x + 5y + 6z - 21 = 0 ]

This is the equation of the plane in Cartesian form.

Example 2: Verifying a Point Lies on the Plane

Given: A plane with the equation ( 2x - 3y + z - 6 = 0 ) and a point ( Q(3, 0, -6) ).

Verify: Whether point ( Q ) lies on the plane.

Solution:

Substitute the coordinates of ( Q ) into the plane equation:

[ 2(3) - 3(0) + (-6) - 6 = 0 ]

Simplify:

[ 6 - 6 = 0 ]

Since the equation holds true, point ( Q ) lies on the plane.

Understanding the concept of a plane passing through a point and perpendicular to a vector is crucial for solving problems in 3D geometry. By using the Cartesian form of the plane equation, one can easily determine the equation of such a plane and verify points and directions related to the plane.