Plane in Normal Form (Cartesian Form)

In 3D geometry, a plane can be represented in various forms, one of which is the normal form (also known as the Cartesian form). This form is particularly useful because it directly relates the plane's orientation to a normal vector that is perpendicular to the plane.

Definition

A plane in normal form is defined by the equation:

$$ ax + by + cz = d $$

where ( (a, b, c) ) is the normal vector to the plane, and ( d ) is the scalar product of the normal vector with any point ( (x_0, y_0, z_0) ) on the plane. In other words, ( d = ax_0 + by_0 + cz_0 ).

Important Points

• The normal vector ( \vec{n} = (a, b, c) ) is perpendicular to the plane.
• The scalar ( d ) represents the perpendicular distance from the origin to the plane when the normal vector is a unit vector.
• If ( d = 0 ), the plane passes through the origin.

Differences and Important Points

Aspect	Description
Normal Vector	A vector perpendicular to the plane, given by ( (a, b, c) ).
Scalar ( d )	The scalar product of the normal vector with any point on the plane.
Distance from Origin	When the normal vector is a unit vector, ( d ) is the distance from the origin to the plane.
Plane through Origin	If ( d = 0 ), the plane passes through the origin.

Formulas

The distance ( D ) from a point ( P(x_1, y_1, z_1) ) to the plane ( ax + by + cz = d ) is given by:

$$ D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}} $$

Examples

Example 1: Plane Equation from a Normal Vector and a Point

Given a normal vector ( \vec{n} = (2, -3, 1) ) and a point ( P(1, -1, 2) ) on the plane, find the equation of the plane.

Solution:

1. Calculate ( d ) using the point ( P ):

$$ d = 2(1) - 3(-1) + 1(2) = 2 + 3 + 2 = 7 $$

1. Write the equation of the plane:

$$ 2x - 3y + z = 7 $$

Example 2: Distance from a Point to a Plane

Find the distance from the point ( Q(4, -2, 3) ) to the plane ( 2x - 3y + z = 7 ).

Solution:

1. Plug the coordinates of ( Q ) into the distance formula:

$$ D = \frac{|2(4) - 3(-2) + 1(3) - 7|}{\sqrt{2^2 + (-3)^2 + 1^2}} $$

1. Simplify the expression:

$$ D = \frac{|8 + 6 + 3 - 7|}{\sqrt{4 + 9 + 1}} = \frac{|10|}{\sqrt{14}} = \frac{10}{\sqrt{14}} = \frac{10\sqrt{14}}{14} = \frac{5\sqrt{14}}{7} $$

The distance from point ( Q ) to the plane is ( \frac{5\sqrt{14}}{7} ) units.

Understanding the plane in normal form is crucial for solving problems related to the orientation and position of planes in 3D space. It is a fundamental concept in fields such as physics, engineering, and computer graphics.