Equation Of A Line Perpendicular To A Plane Through A Point

In mathematics, particularly in three-dimensional geometry, determining the equation of a line that is perpendicular to a given plane and passes through a specific point is a fundamental problem. This article provides a comprehensive explanation of the process involved, focusing on the underlying principles and step-by-step calculations. We will explore how to extract the normal vector from the plane equation, utilize it to define the direction vector of the line, and then construct the parametric equations of the line. The example we will use is finding the equation of the line perpendicular to the plane 2x + 2y - 3z = 6 that passes through the point A(-2, 3, 4).

Understanding the Fundamentals

Before diving into the calculations, it's crucial to grasp the fundamental concepts that underpin this problem. A plane in three-dimensional space can be defined by a linear equation of the form Ax + By + Cz = D, where A, B, and C are the coefficients of the variables x, y, and z, respectively, and D is a constant. The vector (A, B, C) is known as the normal vector to the plane. This vector is perpendicular to every line lying in the plane. In our given equation, 2x + 2y - 3z = 6, the normal vector is (2, 2, -3). Understanding the normal vector is paramount, as it dictates the direction of any line perpendicular to the plane.

A line in three-dimensional space can be defined in several ways, but for this problem, the parametric form is the most convenient. The parametric equations of a line are expressed as:

• x = x₀ + at
• y = y₀ + bt
• z = z₀ + ct

where (x₀, y₀, z₀) is a point on the line, (a, b, c) is the direction vector of the line, and t is a parameter that varies over all real numbers. The direction vector determines the orientation of the line in space. For a line to be perpendicular to a plane, its direction vector must be parallel to the normal vector of the plane. This is a critical insight that allows us to connect the plane's equation to the line's equation.

Step-by-Step Solution

1. Identify the Normal Vector of the Plane

The first step in finding the equation of the line is to identify the normal vector of the given plane. The equation of the plane is 2x + 2y - 3z = 6. By comparing this equation to the general form Ax + By + Cz = D, we can identify the coefficients A, B, and C as the components of the normal vector. In this case, A = 2, B = 2, and C = -3. Therefore, the normal vector n to the plane is (2, 2, -3). The normal vector is crucial because it provides the direction perpendicular to the plane, which is the same direction the line we're trying to find will have.

2. Determine the Direction Vector of the Line

Since the line we are looking for is perpendicular to the plane, its direction vector will be parallel to the normal vector of the plane. This means that the direction vector of the line can be taken as the normal vector of the plane. Thus, the direction vector v of the line is also (2, 2, -3). The direction vector is essential for defining the line's orientation in space. It dictates how the x, y, and z coordinates change as we move along the line.

3. Use the Given Point to Form the Parametric Equations

We are given that the line passes through the point A(-2, 3, 4). This point will serve as our (x₀, y₀, z₀) in the parametric equations of the line. Now that we have the direction vector v = (2, 2, -3) and a point A(-2, 3, 4) on the line, we can construct the parametric equations. The parametric equations of a line are given by:

• x = x₀ + at
• y = y₀ + bt
• z = z₀ + ct

where (a, b, c) are the components of the direction vector and t is a parameter. Substituting the values we have, we get:

• x = -2 + 2t
• y = 3 + 2t
• z = 4 - 3t

These are the parametric equations of the line perpendicular to the plane 2x + 2y - 3z = 6 and passing through the point A(-2, 3, 4). The parametric equations provide a complete description of the line in three-dimensional space. By varying the parameter t, we can generate any point on the line.

4. Express the Equation in Symmetric Form (Optional)

While the parametric equations fully define the line, it is sometimes useful to express the equation in symmetric form. The symmetric form is derived from the parametric equations by solving each equation for the parameter t and then equating them. From the parametric equations we found:

• x = -2 + 2t => t = (x + 2) / 2
• y = 3 + 2t => t = (y - 3) / 2
• z = 4 - 3t => t = (4 - z) / 3

Equating these expressions for t, we get the symmetric equation of the line:

(x + 2) / 2 = (y - 3) / 2 = (4 - z) / 3

The symmetric equation offers an alternative representation of the line, which can be useful in certain contexts. It highlights the ratios between the differences in coordinates and the components of the direction vector.

Common Mistakes to Avoid

When solving problems of this nature, several common mistakes can lead to incorrect answers. Understanding these pitfalls is essential for achieving accuracy. One frequent error is incorrectly identifying the normal vector from the plane equation. Ensure you correctly extract the coefficients of x, y, and z to form the normal vector. Another mistake is confusing the direction vector with a point on the line. The direction vector provides the line's orientation, while a point on the line anchors it in space. Failing to correctly substitute the point and direction vector into the parametric equations is also a common source of error. Always double-check your substitutions to avoid mistakes.

Practical Applications

Finding the equation of a line perpendicular to a plane has numerous practical applications in various fields. In computer graphics, it is used for lighting calculations, determining surface normals, and rendering 3D objects. In physics, it is applied in mechanics for calculating forces and moments, as well as in electromagnetism for analyzing electric and magnetic fields. In engineering, it is crucial for structural analysis, determining the stability of structures, and designing efficient support systems. Understanding these applications underscores the significance of this mathematical concept in real-world scenarios.

Conclusion

The equation of the line perpendicular to the plane 2x + 2y - 3z = 6 that passes through the point A(-2, 3, 4) is given by the parametric equations:

• x = -2 + 2t
• y = 3 + 2t
• z = 4 - 3t

This article has provided a detailed step-by-step solution, emphasizing the underlying principles and potential pitfalls. By understanding the concepts of normal vectors, direction vectors, and parametric equations, you can confidently tackle similar problems in three-dimensional geometry. The process involves identifying the normal vector from the plane equation, using it as the direction vector for the line, and then constructing the parametric equations using the given point. This fundamental technique is essential for various applications in mathematics, physics, engineering, and computer graphics. Mastering these concepts will empower you to solve a wide range of geometric problems and apply them to real-world scenarios.

By following this methodology, one can determine the equation of a line perpendicular to a plane passing through a specific point, a critical skill in various mathematical and applied contexts. The ability to accurately compute such equations is invaluable in fields ranging from computer graphics to structural engineering, showcasing the far-reaching applications of this mathematical principle. The key to success lies in a solid grasp of vector geometry and a meticulous approach to calculations. With practice and a thorough understanding of the concepts, one can confidently tackle these problems and appreciate their significance in various scientific and engineering disciplines.