Distance of a Point from a Plane (Cartesian Form)

Understanding the distance of a point from a plane in Cartesian form is a fundamental concept in 3D geometry. This concept is crucial in various fields such as physics, engineering, computer graphics, and more.

Cartesian Plane Equation

A plane in three-dimensional space can be defined by the Cartesian equation:

$$ ax + by + cz + d = 0 $$

where ( a ), ( b ), and ( c ) are the coefficients that represent the normal vector to the plane, and ( d ) is a constant term. The normal vector is perpendicular to the plane and is denoted as ( \vec{n} = (a, b, c) ).

Distance Formula

The distance ( D ) of a point ( P(x_1, y_1, z_1) ) from the plane ( ax + by + cz + d = 0 ) is given by the formula:

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

This formula is derived from the projection of the vector from the origin to the point ( P ) onto the normal vector of the plane.

Steps to Find the Distance

1. Identify the coefficients ( a ), ( b ), ( c ), and ( d ) from the plane equation.
2. Identify the coordinates ( x_1 ), ( y_1 ), and ( z_1 ) of the point ( P ).
3. Substitute these values into the distance formula.
4. Calculate the numerator and denominator separately.
5. Divide the absolute value of the numerator by the square root of the sum of the squares of ( a ), ( b ), and ( c ).
6. The result is the shortest distance from the point to the plane.

Table of Differences and Important Points

Aspect	Point-Plane Distance in Cartesian Form
Equation of Plane	( ax + by + cz + d = 0 )
Normal Vector	( \vec{n} = (a, b, c) )
Point Coordinates	( P(x_1, y_1, z_1) )
Distance Formula	( D = \frac{
Numerator	Absolute value of ( ax_1 + by_1 + cz_1 + d )
Denominator	( \sqrt{a^2 + b^2 + c^2} ), which is the magnitude of the normal vector
Interpretation	The shortest distance from the point to the plane

Examples

Example 1

Find the distance from the point ( P(1, 2, 3) ) to the plane given by the equation ( 2x - 3y + 6z - 18 = 0 ).

Solution:

1. Coefficients from the plane equation: ( a = 2 ), ( b = -3 ), ( c = 6 ), ( d = -18 ).
2. Coordinates of point ( P ): ( x_1 = 1 ), ( y_1 = 2 ), ( z_1 = 3 ).
3. Substitute into the distance formula:

$$ D = \frac{|2(1) - 3(2) + 6(3) - 18|}{\sqrt{2^2 + (-3)^2 + 6^2}} $$

1. Calculate the numerator and denominator:

$$ D = \frac{|2 - 6 + 18 - 18|}{\sqrt{4 + 9 + 36}} $$ $$ D = \frac{|-4|}{\sqrt{49}} $$ $$ D = \frac{4}{7} $$

The distance from the point ( P(1, 2, 3) ) to the plane is ( \frac{4}{7} ) units.

Example 2

Determine the distance of the point ( Q(-1, 0, 4) ) from the plane ( x + 2y - 2z + 5 = 0 ).

Solution:

1. Coefficients from the plane equation: ( a = 1 ), ( b = 2 ), ( c = -2 ), ( d = 5 ).
2. Coordinates of point ( Q ): ( x_1 = -1 ), ( y_1 = 0 ), ( z_1 = 4 ).
3. Substitute into the distance formula:

$$ D = \frac{|1(-1) + 2(0) - 2(4) + 5|}{\sqrt{1^2 + 2^2 + (-2)^2}} $$

1. Calculate the numerator and denominator:

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

The distance from the point ( Q(-1, 0, 4) ) to the plane is ( \frac{4}{3} ) units.

By understanding the formula and following these steps, one can calculate the distance of any point from a plane in Cartesian form, which is a valuable skill in various applications.