Find The Distance From The Point To The Given Plane

Finding the Distance from a Point to a Plane: A Comprehensive Guide

Finding the distance from a point to a plane is a fundamental concept in three-dimensional geometry with applications spanning various fields, including computer graphics, physics, and engineering. This comprehensive guide will delve into the mathematical principles behind this calculation, explore different methods of solving the problem, and offer practical examples to solidify your understanding.

Understanding the Problem

Before we delve into the solutions, let's clearly define the problem. We are given a point P(x₀, y₀, z₀) in three-dimensional space and a plane defined by the equation Ax + By + Cz + D = 0, where A, B, C are the components of the normal vector to the plane, and D is a constant. Our goal is to determine the shortest distance between the point P and the plane. This shortest distance will always be along a line perpendicular to the plane, passing through the point P.

Method 1: Using the Normal Vector

This is the most common and straightforward method. The key lies in understanding that the shortest distance from a point to a plane is the length of the projection of the vector connecting the point to any point on the plane onto the plane's normal vector.

Steps:

1. Find a point on the plane: Choose any point Q(x₁, y₁, z₁) that satisfies the plane's equation Ax + By + Cz + D = 0. Often, setting two of the coordinates to zero and solving for the third is easiest.

2. Create a vector from the point to the plane: Construct the vector PQ = ⟨x₁ - x₀, y₁ - y₀, z₁ - z₀⟩. This vector connects the given point P to the point Q on the plane.

3. Find the unit normal vector: Normalize the normal vector n = ⟨A, B, C⟩ to obtain the unit normal vector û = n/||**n||, where ||n|| represents the magnitude of the normal vector (√(A² + B² + C²)).

4. Project PQ onto the normal vector: The projection of PQ onto û is given by the dot product: proj_û(PQ) = PQ • û. This projection represents the vector component of PQ parallel to the normal vector. Its magnitude is the distance we seek.

5. Calculate the distance: The distance 'd' from the point P to the plane is the absolute value of the projection's magnitude: d = |PQ • û|. The absolute value ensures a positive distance.

Example:

Let's find the distance from the point P(1, 2, 3) to the plane 2x - y + 2z - 5 = 0.

1. Point on the plane: Let's set x = 0 and y = 0. This gives 2(0) - 0 + 2z - 5 = 0, which solves to z = 5/2. So, Q(0, 0, 5/2) is a point on the plane.

2. Vector PQ: PQ = ⟨0 - 1, 0 - 2, 5/2 - 3⟩ = ⟨-1, -2, -1/2⟩

3. Unit normal vector: n = ⟨2, -1, 2⟩. ||n|| = √(2² + (-1)² + 2²) = 3. Therefore, û = ⟨2/3, -1/3, 2/3⟩.

4. Projection: PQ • û = (-1)(2/3) + (-2)(-1/3) + (-1/2)(2/3) = -2/3 + 2/3 - 1/3 = -1/3

5. Distance: d = |-1/3| = 1/3

Therefore, the distance from the point (1, 2, 3) to the plane 2x - y + 2z - 5 = 0 is 1/3.

Method 2: Using the Point-Plane Distance Formula

This method provides a more concise formula for directly calculating the distance. It's derived from the principles outlined in Method 1 but offers a more streamlined approach.

The formula is:

d = |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²)

Where:

• (x₀, y₀, z₀) are the coordinates of the point P
• A, B, C, and D are the coefficients of the plane equation Ax + By + Cz + D = 0

This formula directly calculates the distance without needing to find a point on the plane.

Example:

Using the same example as before, P(1, 2, 3) and the plane 2x - y + 2z - 5 = 0:

d = |2(1) - 2 + 2(3) - 5| / √(2² + (-1)² + 2²) = |2 - 2 + 6 - 5| / 3 = 1/3

This method confirms the distance calculated using Method 1.

Method 3: Using Vector Projection (Alternative Approach)

This method uses the concept of vector projection but approaches it slightly differently. We find a vector from a point on the plane to our given point and project this vector onto the normal vector of the plane.

1. Find a point on the plane: As before, find a convenient point Q on the plane.

2. Form the vector QP: Create the vector QP which points from a point on the plane (Q) to the given point (P). This is the opposite of PQ in Method 1.

3. Find the unit normal vector: Calculate the unit normal vector û as before.

4. Calculate the scalar projection: The scalar projection of QP onto û is given by QP • û. This is the signed distance from the plane to the point. The absolute value gives the distance.

5. Find the distance: The distance d = | QP • û |

Example:

Let’s use the same example again. We have Q(0, 0, 5/2) and P(1, 2, 3).

1. Vector QP: QP = ⟨1-0, 2-0, 3-5/2⟩ = ⟨1, 2, 1/2⟩

2. Unit normal vector: As before, û = ⟨2/3, -1/3, 2/3⟩

3. Scalar projection: QP • û = (1)(2/3) + (2)(-1/3) + (1/2)(2/3) = 2/3 - 2/3 + 1/3 = 1/3

4. Distance: d = |1/3| = 1/3

This method yields the same result, demonstrating the equivalence of different approaches.

Applications and Further Considerations

The ability to calculate the distance from a point to a plane has numerous applications:

• Collision detection in computer graphics: Determining whether an object has collided with a surface.
• Robotics: Calculating the distance of a robot arm from an obstacle.
• Physics: Determining the distance of a charged particle from a charged plane.
• Computer-aided design (CAD): Measuring distances in three-dimensional models.
• Geographic Information Systems (GIS): Calculating distances between points and surfaces.

Understanding the underlying principles and applying different methods allows for flexibility and robustness in solving various problems involving points and planes in three-dimensional space. Remember that the choice of method depends on the specific context and the information readily available. The key is to understand the fundamental geometric relationships between the point, the plane, and the normal vector. This understanding empowers you to solve diverse problems efficiently and accurately. Furthermore, consider exploring more advanced concepts such as the distance between two parallel planes or the distance from a line to a plane, which build upon the fundamental principles discussed here. The more you delve into these concepts, the more proficient you'll become in manipulating and understanding three-dimensional geometry.