Calculating The Length Of Projection Of A Line Segment Onto A Plane

In the realm of three-dimensional geometry, understanding the projection of line segments onto planes is a fundamental concept with various applications in fields like computer graphics, engineering, and physics. This article delves into the process of determining the length of the projection of a line segment onto a plane. Specifically, we will explore the case of the line segment joining the points (1, -1, 0) and (-1, 0, 1) projected onto the plane 2x + y + 6z = 1. This problem requires a combination of vector algebra, plane geometry, and a clear understanding of projections. This detailed exploration will offer a comprehensive guide, ensuring a solid grasp of the methods and principles involved. The length of projection is a critical concept, playing a pivotal role in understanding spatial relationships and geometric transformations. We'll navigate through each step, clarifying the underlying logic and calculations, to illuminate the path to the solution. Mastering this concept enhances problem-solving skills in spatial geometry and related disciplines. Our approach will break down the complexities, offering insights that make tackling similar problems more manageable. This journey through the geometry of projections will empower you with the knowledge and confidence to address related challenges with greater ease and precision.

Understanding the Problem

Before diving into the solution, it's crucial to dissect the problem statement. We are given two points, A(1, -1, 0) and B(-1, 0, 1), which define a line segment. Additionally, we have a plane defined by the equation 2x + y + 6z = 1. The core task is to find the length of the projection of the line segment AB onto this plane. This involves several key steps: first, we need to determine the vector representing the line segment AB. Next, we need to find the normal vector to the plane, which will help us understand the orientation of the plane. Then, we'll calculate the projection of the line segment onto the normal vector. Finally, using the Pythagorean theorem or a similar approach, we can find the length of the projection of the line segment onto the plane. This process necessitates a firm understanding of vector operations, including dot products and vector projections, as well as the ability to interpret the geometric significance of these operations. The projection of a line segment onto a plane is not just a theoretical exercise; it has practical implications in various fields. For instance, in computer graphics, projections are used to render 3D objects onto a 2D screen. In engineering, understanding projections is crucial for designing structures and analyzing forces. Thus, mastering this concept is not only essential for academic purposes but also for real-world applications. By carefully examining the problem and breaking it down into smaller, manageable steps, we can approach the solution with clarity and precision. This methodical approach ensures that we do not overlook any critical details and that our final answer is accurate and well-supported.

Step-by-Step Solution

To find the length of the projection, let's break down the solution into a step-by-step process:

1. Finding the Vector Representing the Line Segment

First, determine the vector AB that represents the line segment joining points A(1, -1, 0) and B(-1, 0, 1). The vector AB can be found by subtracting the coordinates of point A from the coordinates of point B:

AB = B - A = (-1 - 1, 0 - (-1), 1 - 0) = (-2, 1, 1)

This vector AB now represents the direction and magnitude of the line segment in three-dimensional space. Understanding how to compute this vector is fundamental, as it forms the basis for further calculations involving projections and distances. The vector representation allows us to apply algebraic operations to geometric entities, making it possible to solve problems involving lines, planes, and other shapes. This step is not just a mechanical calculation; it's a crucial transformation that allows us to work with the line segment mathematically. The accuracy of this step is paramount, as any error here will propagate through the rest of the solution. Therefore, double-checking the subtraction and ensuring the correct order of operations is essential. With the vector AB correctly determined, we can proceed to the next step, which involves understanding the plane onto which we will project this line segment.

2. Determining the Normal Vector to the Plane

The equation of the plane is given as 2x + y + 6z = 1. The coefficients of x, y, and z in the equation of the plane directly give us the normal vector n to the plane. The normal vector is a vector perpendicular to the plane, and it plays a crucial role in determining the projection of a line segment onto the plane. In this case, the normal vector n is:

n = (2, 1, 6)

The normal vector's direction is critical because it dictates the orientation of the plane in space. This vector is used to calculate the projection of the line segment onto a line perpendicular to the plane. Understanding the significance of the normal vector is key to solving problems involving planes in three-dimensional space. It provides a reference direction that allows us to determine angles and distances relative to the plane. This step is relatively straightforward, but it is crucial to correctly identify the coefficients in the plane equation to avoid errors. The normal vector, once determined, provides a crucial piece of information that will be used in subsequent steps to calculate the projection. The normal vector is not just a mathematical construct; it has a clear geometric interpretation, representing the direction in which the plane is "facing." This intuitive understanding helps in visualizing the problem and ensuring that the calculations make sense geometrically.

3. Calculating the Projection of the Line Segment onto the Normal Vector

The projection of the vector AB onto the normal vector n gives us the component of AB that is perpendicular to the plane. This projection can be calculated using the formula:

proj_n AB = (AB · n) / ||n||² * n

First, calculate the dot product of AB and n:

AB · n = (-2 * 2) + (1 * 1) + (1 * 6) = -4 + 1 + 6 = 3

Next, calculate the magnitude squared of n:

||n||² = 2² + 1² + 6² = 4 + 1 + 36 = 41

Now, calculate the projection:

proj_n AB = (3 / 41) * (2, 1, 6) = (6/41, 3/41, 18/41)

The projection of a vector onto another vector is a fundamental concept in linear algebra and vector calculus. It allows us to decompose a vector into components that are parallel and perpendicular to a given direction. In this case, the projection of AB onto n gives us the component of AB that is perpendicular to the plane. This calculation involves the dot product, which measures the alignment of two vectors, and the magnitude squared, which provides a scaling factor. Understanding the formula for vector projection is essential for solving a wide range of problems in geometry and physics. This step requires careful attention to detail, as errors in the dot product or magnitude calculation will lead to an incorrect projection. The result of this calculation, the vector (6/41, 3/41, 18/41), represents the component of the line segment AB that is perpendicular to the plane. This information is crucial for determining the length of the projection of AB onto the plane.

4. Finding the Length of the Projection of AB onto n

The length of the projection of AB onto n is the magnitude of the vector proj_n AB:

||proj_n AB|| = √((6/41)² + (3/41)² + (18/41)²) = √(36/1681 + 9/1681 + 324/1681) = √(369/1681) = √(9 * 41) / 41 = 3√41 / 41 = 3 / √41

This value represents the length of the component of the line segment that is perpendicular to the plane. The magnitude of a vector is its length, and it is calculated using the Pythagorean theorem in three dimensions. This calculation is a straightforward application of the formula for the magnitude of a vector, but it is important to ensure that the arithmetic is accurate. This value, 3/√41, represents the length of the projection of the line segment onto the normal vector, which is a crucial piece of information for finding the length of the projection onto the plane. Understanding how to calculate the magnitude of a vector is a fundamental skill in vector algebra and is used extensively in various applications. The result of this calculation provides a scalar value that represents the length of the perpendicular component of the line segment relative to the plane.

5. Calculating the Length of the Line Segment AB

The length of the line segment AB is the magnitude of the vector AB:

||AB|| = √((-2)² + 1² + 1²) = √(4 + 1 + 1) = √6

This is the actual length of the line segment in 3D space, irrespective of the plane. Calculating the length of a line segment in three dimensions involves applying the Pythagorean theorem. This step is relatively straightforward, but it provides a crucial piece of information for the final calculation. The length of the line segment, √6, represents the hypotenuse of a right triangle formed by the projection of the line segment onto the plane and the projection onto the normal vector. This value serves as a reference for determining the length of the projection onto the plane. Understanding how to calculate the length of a line segment is a fundamental skill in geometry and is used extensively in various applications. The result of this calculation provides a scalar value that represents the total length of the line segment in three-dimensional space.

6. Finding the Length of the Projection onto the Plane

Let the length of the projection of the line segment onto the plane be L. We can use the Pythagorean theorem in 3D to relate the length of the line segment, the length of its projection onto the normal vector, and the length of its projection onto the plane:

||AB||² = ||proj_n AB||² + L²

L² = ||AB||² - ||proj_n AB||²

L² = (√6)² - (3 / √41)² = 6 - 9/41 = (246 - 9) / 41 = 237 / 41

L = √(237 / 41) = √(237) / √41 = (√237 * √41) / 41 = √(9717) / 41 = √(9 * 1079) / 41 = 3√1079 / 41

However, we can simplify this further:

L = √(6 - 9/41) = √(246/41 - 9/41) = √(237/41)

L = √237 / √41

We can rationalize the denominator by multiplying the numerator and denominator by √41:

L = (√237 * √41) / 41

Since 237 = 3 * 79, there are no perfect square factors to simplify further. So,

L = √(237 * 41) / 41 = √9717 / 41

Now, let's approximate the square root of 9717:

√9717 ≈ 98.57

So, L ≈ 98.57 / 41 ≈ 2.404

We have L = √(237/41). Let's try simplifying it:

L = √237 / √41 = √(237/41) ≈ √5.78 ≈ 2.404

Thus, the length of the projection of the line segment onto the plane is approximately 2.404 units.

This final step combines all the previous calculations to arrive at the answer. The Pythagorean theorem provides a powerful tool for relating the lengths of the sides of a right triangle. In this case, the right triangle is formed by the line segment AB, its projection onto the plane, and its projection onto the normal vector. By applying the Pythagorean theorem, we can solve for the length of the projection onto the plane. This calculation requires careful attention to detail, as there are several steps involved in simplifying the expression. The final result, approximately 2.404 units, represents the length of the shadow that the line segment AB would cast onto the plane if a light source were positioned directly above the plane. This answer provides a concrete measure of the length of the projection, which is the ultimate goal of the problem.

Conclusion

The length of the projection of the line segment joining the points (1, -1, 0) and (-1, 0, 1) onto the plane 2x + y + 6z = 1 is found to be approximately 2.404 units. This problem demonstrates the application of vector algebra and geometry in solving 3D spatial problems. Understanding vector projections, normal vectors, and the Pythagorean theorem is crucial in tackling such problems. The step-by-step approach outlined in this article provides a clear and methodical way to solve similar problems in the future. This exercise reinforces the importance of spatial reasoning and the ability to translate geometric concepts into algebraic expressions. The solution not only provides a numerical answer but also enhances our understanding of the underlying principles of 3D geometry. Mastering these concepts opens doors to more advanced topics in mathematics, physics, and engineering. The ability to visualize and manipulate objects in three dimensions is a valuable skill that has numerous practical applications. This problem serves as a valuable learning experience, highlighting the power of mathematical tools in solving real-world problems. The journey through this solution reinforces the interconnectedness of various mathematical concepts and the importance of a systematic approach to problem-solving. The final answer, approximately 2.404 units, represents a precise measure of the length of the projection, showcasing the effectiveness of the methods and principles employed in this analysis.