Transformation of General Form to Normal Form

In 3D geometry, the equation of a plane can be represented in various forms. Two of the most common representations are the General Form and the Normal Form. Understanding how to transform an equation of a plane from the General Form to the Normal Form is essential for solving many geometric problems. In this content, we will explore the transformation process in detail.

General Form of a Plane

The General Form of a plane's equation is given by:

$$ Ax + By + Cz + D = 0 $$

where ( A ), ( B ), ( C ), and ( D ) are real numbers, and ( A^2 + B^2 + C^2 > 0 ).

Normal Form of a Plane

The Normal Form of a plane's equation is given by:

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

where ( \vec{r} ) is the position vector of any point on the plane, ( \vec{n} ) is a unit normal vector to the plane, and ( p ) is the perpendicular distance from the origin to the plane.

Transformation Process

To transform the General Form to the Normal Form, we need to follow these steps:

1. Identify the coefficients: Extract the coefficients ( A ), ( B ), ( C ), and ( D ) from the General Form.
2. Find the normal vector: The coefficients ( A ), ( B ), and ( C ) in the General Form represent the components of a normal vector to the plane, ( \vec{N} = A\vec{i} + B\vec{j} + C\vec{k} ).
3. Normalize the normal vector: Calculate the magnitude of ( \vec{N} ) and divide each component by this magnitude to obtain the unit normal vector ( \vec{n} ).
4. Calculate the perpendicular distance: The perpendicular distance ( p ) from the origin to the plane can be found using the formula ( p = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} ).
5. Determine the sign of ( p ): The sign of ( p ) should be the same as the sign of ( D ) in the General Form.
6. Write the Normal Form: Use the unit normal vector ( \vec{n} ) and the perpendicular distance ( p ) to write the equation in Normal Form.

Differences and Important Points

Aspect	General Form	Normal Form
Equation	( Ax + By + Cz + D = 0 )	( \vec{r} \cdot \vec{n} = p )
Normal Vector	( \vec{N} = A\vec{i} + B\vec{j} + C\vec{k} )	( \vec{n} ) (unit vector)
Perpendicular Distance	Not directly given	( p ) (explicitly given)
Ease of Use	Less intuitive for certain calculations	More intuitive for distance and angle calculations

Examples

Example 1: Transformation

Given the General Form of a plane:

$$ 2x - 3y + 6z - 12 = 0 $$

Step 1: Identify the coefficients: ( A = 2 ), ( B = -3 ), ( C = 6 ), ( D = -12 ).

Step 2: Find the normal vector: ( \vec{N} = 2\vec{i} - 3\vec{j} + 6\vec{k} ).

Step 3: Normalize the normal vector:

$$ |\vec{N}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 $$

$$ \vec{n} = \frac{1}{7}(2\vec{i} - 3\vec{j} + 6\vec{k}) = \frac{2}{7}\vec{i} - \frac{3}{7}\vec{j} + \frac{6}{7}\vec{k} $$

Step 4: Calculate the perpendicular distance:

$$ p = \frac{|-12|}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{12}{7} $$

Step 5: Determine the sign of ( p ): Since ( D = -12 ), ( p ) is positive.

Step 6: Write the Normal Form:

$$ \vec{r} \cdot \left(\frac{2}{7}\vec{i} - \frac{3}{7}\vec{j} + \frac{6}{7}\vec{k}\right) = \frac{12}{7} $$

Example 2: Using the Normal Form

Suppose we have the Normal Form of a plane from Example 1:

$$ \vec{r} \cdot \left(\frac{2}{7}\vec{i} - \frac{3}{7}\vec{j} + \frac{6}{7}\vec{k}\right) = \frac{12}{7} $$

To find the distance from a point ( P(1, 2, -3) ) to the plane, we can use the Normal Form directly:

$$ d = |\vec{r}_P \cdot \vec{n} - p| $$

where ( \vec{r}_P ) is the position vector of point ( P ).

$$ d = \left|\left(1\vec{i} + 2\vec{j} - 3\vec{k}\right) \cdot \left(\frac{2}{7}\vec{i} - \frac{3}{7}\vec{j} + \frac{6}{7}\vec{k}\right) - \frac{12}{7}\right| $$

$$ d = \left|\frac{2}{7} - \frac{6}{7} - \frac{18}{7} - \frac{12}{7}\right| $$

$$ d = \left|-\frac{34}{7}\right| = \frac{34}{7} $$

The distance from point ( P ) to the plane is ( \frac{34}{7} ) units.

By transforming the General Form to the Normal Form, we can easily solve problems related to distances and angles involving planes in 3D geometry.