Find Equation Of Plane Through Point And Parallel To Plane

Finding the Equation of a Plane Through a Point and Parallel to Another Plane

Determining the equation of a plane that passes through a specific point and runs parallel to another given plane is a fundamental concept in three-dimensional geometry. This process involves leveraging the properties of parallel planes and the standard form of a plane's equation. Understanding this concept is crucial for various applications in mathematics, physics, and computer graphics. This comprehensive guide will walk you through the steps, providing examples and explaining the underlying principles.

Understanding the Equation of a Plane

Before delving into the problem, let's review the equation of a plane. A plane in three-dimensional space can be represented by the equation:

Ax + By + Cz = D

Where:

• A, B, and C are constants representing the components of the normal vector to the plane (a vector perpendicular to the plane).
• x, y, and z are the coordinates of any point on the plane.
• D is a constant.

The normal vector n = <A, B, C> is a key element in defining the plane's orientation. All vectors lying within the plane are orthogonal (perpendicular) to the normal vector.

Parallel Planes: A Key Property

Two planes are parallel if and only if their normal vectors are parallel. This means that the normal vectors are scalar multiples of each other. In other words, if n1 is the normal vector of plane 1 and n2 is the normal vector of plane 2, then the planes are parallel if:

n1 = k * n2

where k is a non-zero scalar.

Finding the Equation: A Step-by-Step Approach

Let's assume we are given a point P(x₀, y₀, z₀) and a plane with equation Ax + By + Cz = D. Our goal is to find the equation of a plane that passes through P and is parallel to the given plane.

Step 1: Identify the Normal Vector

Since parallel planes have parallel normal vectors, the normal vector of our desired plane is the same as the normal vector of the given plane. Therefore, the normal vector of our new plane is n = <A, B, C>.

Step 2: Utilize the Point-Normal Form

The point-normal form of a plane's equation provides a convenient way to find the equation using a point on the plane and its normal vector. The point-normal form is:

A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

where (x₀, y₀, z₀) are the coordinates of the point P and <A, B, C> is the normal vector.

Step 3: Substitute and Simplify

Substitute the coordinates of point P(x₀, y₀, z₀) and the components of the normal vector <A, B, C> into the point-normal form:

A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

Simplify the equation to obtain the standard form:

Ax + By + Cz = Ax₀ + By₀ + Cz₀

The right-hand side of the equation represents the constant D for the new plane.

Worked Examples

Let's illustrate this process with a few examples.

Example 1:

Find the equation of the plane that passes through the point (2, -1, 3) and is parallel to the plane 2x - 5y + z = 7.

Solution:

1. Normal Vector: The normal vector of the given plane is <2, -5, 1>. Since our desired plane is parallel, its normal vector is also <2, -5, 1>.

2. Point-Normal Form: Using the point (2, -1, 3) and the normal vector <2, -5, 1>, the point-normal form is:

   2(x - 2) - 5(y + 1) + 1(z - 3) = 0

3. Standard Form: Simplifying the equation, we get:

   2x - 4 - 5y - 5 + z - 3 = 0 2x - 5y + z = 12

Therefore, the equation of the plane is 2x - 5y + z = 12.

Example 2:

Find the equation of the plane that passes through the point (1, 0, -2) and is parallel to the plane x + 3y - 2z = 5.

Solution:

1. Normal Vector: The normal vector of the given plane is <1, 3, -2>. This is also the normal vector for our new plane.

2. Point-Normal Form: Using the point (1, 0, -2) and the normal vector <1, 3, -2>, the point-normal form is:

   1(x - 1) + 3(y - 0) - 2(z + 2) = 0

3. Standard Form: Simplifying the equation, we get:

   x - 1 + 3y - 2z - 4 = 0 x + 3y - 2z = 5

Notice that in this case, the equation of the parallel plane is the same as the original plane except potentially for the constant term. This is because the given point (1, 0, -2) actually lies on the plane x + 3y - 2z = 5. If the point had been different and not on the original plane, we would have obtained a different constant term in the final equation.

Handling Cases with Different Normal Vector Representations

It's important to note that the normal vector can be expressed in different forms. For instance, if the equation of the given plane is provided in a non-standard form, you might need to manipulate it to get it into the standard form (Ax + By + Cz = D) to identify the normal vector correctly. Similarly, you might encounter parametric equations or other representations of the plane which would require conversion to Cartesian form before proceeding with this method.

Advanced Considerations and Applications

This method forms the foundation for solving more complex problems involving planes. For example:

• Finding the distance between parallel planes: Once you have the equations of two parallel planes, you can calculate the shortest distance between them using vector methods.
• Intersection of planes: Understanding plane equations is fundamental for determining the intersection lines or points of multiple planes.
• Computer Graphics: In computer graphics, plane equations are used extensively in tasks like clipping, collision detection, and rendering.

Conclusion

Finding the equation of a plane parallel to a given plane and passing through a specific point is a straightforward process, utilizing the properties of parallel vectors and the point-normal form of a plane’s equation. Mastering this technique is essential for a deeper understanding of three-dimensional geometry and its applications in various fields. By following the steps outlined in this guide and practicing with different examples, you can confidently tackle this type of problem and progress to more advanced concepts in three-dimensional geometry. Remember to always double-check your calculations and ensure you understand the underlying geometric principles.