Finding The Plane Equation Containing A Straight Line

Finding the Equation of a Plane Containing a Straight Line: A Step-by-Step Guide

Hey everyone! Today, we're diving into the cool world of vector spaces and analytic geometry, and we're going to figure out how to find the equation of a plane that includes a straight line. This is super useful for all sorts of problems, and it's actually not as hard as it might sound! Let's break it down, step by step.

First off, let's talk about the problem we're tackling. Imagine you have a straight line floating in 3D space, and you want to find the equation of a plane that perfectly contains this line. The line passes through two points, let's say, (1, 2, 3) and (-3, -2, -1). The question is: how do you find the equation of that plane? Keep in mind that a plane can be defined by a point on the plane and a vector that is normal (perpendicular) to the plane.

Finding the Equation of a Plane: The Basics

To find the equation of a plane, we need a few key ingredients. First, we need a point that lies on the plane. We know our line is on the plane, so any point on the line will do. Then, we need a normal vector to the plane. This vector is perpendicular to the plane, and it's crucial for defining the plane's orientation in space. Once we have these two things, we can use the following equation of a plane:

A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

Where (x₀, y₀, z₀) is a point on the plane, and is the normal vector to the plane with components (A, B, C). The equation of the plane is a linear equation in terms of x, y, and z, where the coefficients of x, y, and z are the components of the normal vector. The constant term is determined by the point on the plane and the normal vector. Let's use the initial point (1, 2, 3) on the line, and let's use the directional vector of the line, which we will compute later. The directional vector can be taken from the parametric equation of the line to get the normal vector.

Setting Up the Problem

Let's start with our given information. We have two points on the line: (1, 2, 3) and (-3, -2, -1). We can calculate the direction vector of the line using these points. This vector will give us the direction the line is traveling in space, and it's a super important part of finding the plane's equation. To find the direction vector, subtract the coordinates of the first point from the coordinates of the second point, or vice versa. For example, we can subtract the coordinates of the first point from the coordinates of the second point:

(-3 - 1, -2 - 2, -1 - 3) = (-4, -4, -4)

So, our direction vector is (-4, -4, -4). Notice that the components of the vector are all the same; we can scale this to something simple. We can use the direction vector (-4, -4, -4), but to keep the numbers a little simpler, we can use a simplified version of this direction vector. For example, we can scale it down by dividing by -4. So, a simplified version of the direction vector is (1, 1, 1). This means that the x, y, and z components of the line increase (or decrease) at the same rate. Next, we need to find a second vector that is in the plane. We can get it by subtracting the coordinates of a point on the line from another point not on the line. If we're given two points on the line, we can take the difference between them to find a vector that lies in the plane. Once we have the direction vector and the other vector, we can compute the normal vector to the plane by taking the cross product of these two vectors. It might be that a variety of answers are possible based on different point choices. This is because an infinite number of planes can contain a single line.

Finding the Parametric Equation of a Line

We're told that the system of equations of this line is:

x = 1 - t, y = 2 - t, z = 3 - t

This is called the parametric form of the line. Here, t is a parameter that can take any real value. Each value of t gives you a different point on the line. The coefficients of t in each equation represent the direction vector of the line. From this parametric form, you can quickly find the direction vector by looking at the coefficients of t, which is (-1, -1, -1). Note that we can also choose (-1, -1, -1) to be the direction vector, but for simplicity, we can scale it to (1, 1, 1). The point (1, 2, 3) is also a point on the line, as the parametric equations will show when t = 0. Another point on the line can be found by choosing a value for t. For example, when t = 1, (0, 1, 2) is also a point on the line. Let's use (1, 2, 3) as our known point, and (0, 1, 2) to calculate another vector.

Determining the Normal Vector

To define the plane, we need a normal vector. Since we know the line lies in the plane, the direction vector of the line will be parallel to the plane. However, the normal vector is perpendicular to the plane. To get the normal vector, we can take the cross product of two non-parallel vectors in the plane. In our case, we can use the direction vector of the line, (1, 1, 1), and another vector from a point on the line to some other point in the plane. Since we only have two points on the line, we can't readily find a second vector in the plane. If we pick a point not on the line, such as (0, 0, 0), which is known to be not on the line, we can construct the vector between (1, 2, 3) and (0, 0, 0) to get another vector in the plane. Then, taking the cross product of the direction vector (1, 1, 1) and the vector between (1, 2, 3) and (0, 0, 0) = (-1, -2, -3), we get the normal vector, by computing the cross product:

(1, 1, 1) x (-1, -2, -3) = (-1, -2, 1)

The vector (-1, -2, 1) is normal to the plane that contains the line.

Writing the Plane Equation

Now that we have a point on the plane (1, 2, 3) and the normal vector (-1, -2, 1), we can write the equation of the plane using the formula we mentioned earlier:

A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

Plug in the values:

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

Simplify it:

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

Which simplifies to:

-x - 2y + z + 2 = 0

Or, we can multiply by -1 to make the equation more readable:

x + 2y - z - 2 = 0

And there you have it! The equation of the plane that contains the line is x + 2y - z - 2 = 0.

Key Takeaways

• To find the equation of a plane containing a line, you need a point on the plane (which can be a point on the line) and a normal vector to the plane. Keep in mind that infinite planes can contain a line. The answer we got is only one of the solutions.
• You can find the direction vector of the line from its parametric equations.
• The normal vector can be found by taking the cross product of two vectors that lie in the plane.
• Always double-check your calculations to make sure everything is correct.

That's the basic idea! Let me know if you have any questions or want to explore other examples. Happy calculating, guys!