Finding The Length Of Projection Of A Line Segment On A Plane

In the realm of three-dimensional geometry, understanding the projection of line segments onto planes is a fundamental concept. This article delves into the process of determining the length of the projection of a line segment joining two given points onto a given plane. Specifically, we will explore the scenario where the line segment connects the points $(1, -1, 0)$ and $(-1, 0, 1)$, and the plane is defined by the equation $2x + y + 6z = 1$. Understanding this concept requires a blend of vector algebra, coordinate geometry, and a clear grasp of spatial visualization. The aim is to provide a comprehensive, step-by-step guide to solving this problem, ensuring clarity and ease of understanding for readers of all backgrounds.

Problem Statement

The problem at hand involves finding the length of the projection of the line segment that connects the points $A(1, -1, 0)$ and $B(-1, 0, 1)$ onto the plane defined by the equation $2x + y + 6z = 1$. This is a classic problem in 3D geometry that combines concepts from vector algebra and coordinate geometry. To solve this, we need to understand how to find the direction vector of the line segment, how to find the normal vector to the plane, and how to project a vector onto another vector. The final step involves finding the magnitude of this projection, which will give us the length we are looking for. Let's break down the steps involved in solving this problem systematically.

Step 1: Finding the Direction Vector of the Line Segment

The direction vector of the line segment joining two points is crucial for understanding its orientation in space. To find the direction vector $\vec{AB}$ of the line segment joining points $A(1, -1, 0)$ and $B(-1, 0, 1)$, we subtract the coordinates of point $A$ from the coordinates of point $B$. This gives us:

$$\vec{AB} = B - A = (-1 - 1, 0 - (-1), 1 - 0) = (-2, 1, 1)$$

This vector $\vec{AB} = (-2, 1, 1)$ represents the direction and magnitude of the displacement from point $A$ to point $B$. It's a fundamental element in our calculation, as it helps us understand the orientation of the line segment in 3D space. Now that we have the direction vector, the next step is to determine the normal vector to the plane, which will help us in understanding the plane's orientation.

Step 2: Determining the Normal Vector to the Plane

The normal vector to a plane is a vector that is perpendicular to the plane. This vector is crucial for determining the orientation of the plane in space and is directly derived from the equation of the plane. For a plane defined by the equation $ax + by + cz = d$, the normal vector $\vec{n}$ is given by the coefficients of $x$, $y$, and $z$, which is $\vec{n} = (a, b, c)$.

In our case, the equation of the plane is $2x + y + 6z = 1$. Therefore, the normal vector to the plane is:

$$\vec{n} = (2, 1, 6)$$

This vector $\vec{n} = (2, 1, 6)$ is perpendicular to the plane and provides us with the plane's orientation in 3D space. With both the direction vector of the line segment and the normal vector to the plane, we can now proceed to find the projection of the line segment onto the plane.

Step 3: Calculating the Projection of the Line Segment onto the Normal Vector

To find the projection of the line segment onto the plane, we first need to find the projection of the direction vector $\vec{AB}$ onto the normal vector $\vec{n}$. This projection tells us how much of the line segment's direction is aligned with the normal vector of the plane. The formula for the projection of vector $\vec{AB}$ onto vector $\vec{n}$ is given by:

$$\text{proj}_{\vec{n}} \vec{AB} = \frac{\vec{AB} \cdot \vec{n}}{\|\vec{n}\|^2} \vec{n}$$

First, let's calculate the dot product of $\vec{AB}$ and $\vec{n}$:

$$\vec{AB} \cdot \vec{n} = (-2)(2) + (1)(1) + (1)(6) = -4 + 1 + 6 = 3$$

Next, we calculate the magnitude squared of $\vec{n}$:

$$\|\vec{n}\|^2 = (2)^2 + (1)^2 + (6)^2 = 4 + 1 + 36 = 41$$

Now, we can find the projection:

$$\text{proj}_{\vec{n}} \vec{AB} = \frac{3}{41}(2, 1, 6) = \left(\frac{6}{41}, \frac{3}{41}, \frac{18}{41}\right)$$

This vector represents the component of $\vec{AB}$ that is parallel to the normal vector $\vec{n}$. The magnitude of this projection will give us the length of the component of the line segment that is perpendicular to the plane.

Step 4: Determining the Length of the Projection onto the Normal Vector

Having calculated the projection of the direction vector $\vec{AB}$ onto the normal vector $\vec{n}$, we now need to find the length of this projection. This length represents the component of the line segment that is perpendicular to the plane. The magnitude of the projection vector is given by:

$$\|\text{proj}_{\vec{n}} \vec{AB}\| = \left\| \left(\frac{6}{41}, \frac{3}{41}, \frac{18}{41}\right) \right\|$$

To find the magnitude, we use the formula:

$$\|(x, y, z)\| = \sqrt{x^2 + y^2 + z^2}$$

So,

$$\|\text{proj}_{\vec{n}} \vec{AB}\| = \sqrt{\left(\frac{6}{41}\right)^2 + \left(\frac{3}{41}\right)^2 + \left(\frac{18}{41}\right)^2} = \sqrt{\frac{36 + 9 + 324}{41^2}} = \sqrt{\frac{369}{41^2}} = \frac{\sqrt{369}}{41}$$

Simplifying further, we get:

$$\|\text{proj}_{\vec{n}} \vec{AB}\| = \frac{\sqrt{9 \cdot 41}}{41} = \frac{3\sqrt{41}}{41}$$

This value represents the length of the projection of the line segment onto the normal vector, which is the component perpendicular to the plane. To find the length of the projection of the line segment onto the plane itself, we will use the Pythagorean theorem.

Step 5: Calculating the Length of the Original Line Segment

Before we can determine the length of the projection of the line segment onto the plane, we need to know the length of the original line segment itself. This will serve as the hypotenuse in our right triangle, where the projection onto the normal vector is one leg, and the projection onto the plane is the other leg. The length of the line segment joining points $A(1, -1, 0)$ and $B(-1, 0, 1)$ is the magnitude of the direction vector $\vec{AB}$.

We already found $\vec{AB} = (-2, 1, 1)$. Now, we calculate its magnitude:

$$\|\vec{AB}\| = \sqrt{(-2)^2 + (1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$

So, the length of the original line segment is $\sqrt{6}$. This value is essential for our final calculation, as it represents the total length that we are projecting onto the plane.

Step 6: Finding the Length of the Projection onto the Plane Using the Pythagorean Theorem

Now that we have the length of the original line segment and the length of its projection onto the normal vector, we can use the Pythagorean theorem to find the length of the projection onto the plane. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the original line segment) is equal to the sum of the squares of the lengths of the other two sides (the projection onto the normal vector and the projection onto the plane).

Let $L$ be the length of the projection of the line segment onto the plane. Then:

$$\|\vec{AB}\|^2 = \|\text{proj}_{\vec{n}} \vec{AB}\|^2 + L^2$$

We know that $\|\vec{AB}\| = \sqrt{6}$ and $\|\text{proj}_{\vec{n}} \vec{AB}\| = \frac{3\sqrt{41}}{41}$. Plugging these values into the equation, we get:

$$(\sqrt{6})^2 = \left(\frac{3\sqrt{41}}{41}\right)^2 + L^2$$

$$6 = \frac{9 \cdot 41}{41^2} + L^2$$

$$6 = \frac{9}{41} + L^2$$

Now, we solve for $L^2$:

$$L^2 = 6 - \frac{9}{41} = \frac{6 \cdot 41 - 9}{41} = \frac{246 - 9}{41} = \frac{237}{41}$$

Finally, we find $L$ by taking the square root:

$$L = \sqrt{\frac{237}{41}} = \sqrt{\frac{3 \cdot 79}{41}}$$

Thus, the length of the projection of the line segment onto the plane is $\sqrt{\frac{237}{41}}$.

In conclusion, we have successfully determined the length of the projection of the line segment joining the points $(1, -1, 0)$ and $(-1, 0, 1)$ onto the plane $2x + y + 6z = 1$. This involved a series of steps, starting with finding the direction vector of the line segment and the normal vector to the plane. We then calculated the projection of the line segment onto the normal vector and found its magnitude. After determining the length of the original line segment, we used the Pythagorean theorem to find the length of the projection onto the plane. The final answer is $\sqrt{\frac{237}{41}}$, which represents the length of the shadow, or projection, of the line segment as it falls onto the plane. This problem showcases the power of combining vector algebra and coordinate geometry to solve complex spatial problems. Understanding these concepts is crucial for various applications in fields like computer graphics, engineering, and physics. By breaking down the problem into manageable steps, we have provided a clear and comprehensive solution that can be applied to similar problems in 3D geometry.