Plane Equation Through 3 Points: Detailed Solution

Hey guys! Let's dive into a common problem in geometry: finding the equation of a plane that passes through three given points. In this guide, we'll break down the process step-by-step, making sure you understand not just how to do it, but why it works. We'll use the specific example of points A₀(-1, 0, 0), B₀(0, 2, 0), and C₀(0, 0, 3) to illustrate the method, and we'll aim for the answer: 6x + 3y + 2z + 6 = 0. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into calculations, let's visualize what we're trying to achieve. Imagine three points floating in 3D space. These points define a unique plane, kind of like how three legs make a stable tripod. Our goal is to find the equation that describes this flat surface mathematically. Think of the equation as a secret code that tells you whether any given point in space lies on the plane or not. This equation of a plane is typically represented in the general form:

Ax + By + Cz + D = 0

Where A, B, C, and D are constants, and (x, y, z) are the coordinates of any point on the plane. Our mission is to find these constants using the information given by the three points.

Why Three Points?

You might wonder, why do we need three points? Well, two points only define a line, not a plane. A plane needs to be anchored in more than one direction, and three non-collinear points (points not lying on the same line) are the minimum requirement to uniquely define a plane in 3D space. Just picture trying to balance a plate – you need at least three fingers to hold it steady!

The Vector Approach: Our Roadmap

To find the equation, we'll use a clever trick involving vectors. Vectors are like arrows that have both magnitude (length) and direction. Here's the roadmap:

1. Create two vectors lying in the plane using the given points. These vectors will give us the 'direction' of the plane.
2. Calculate the normal vector to the plane. The normal vector is perpendicular to both vectors we created in step 1. Think of it as the plane's 'axis' – it points straight out from the surface. We'll find it using the cross product.
3. Use the normal vector and one of the points to find the equation of the plane. This is where we'll plug everything into the general form Ax + By + Cz + D = 0 and solve for the constants.

Sounds like a plan? Let's get started with step 1.

Step 1: Creating Vectors in the Plane

We have three points: A₀(-1, 0, 0), B₀(0, 2, 0), and C₀(0, 0, 3). To create vectors that lie in the plane, we can simply subtract the coordinates of one point from the others. This gives us two vectors that 'span' the plane. Let's create vectors A₀B₀ and A₀C₀.

Vector A₀B₀

To find the vector A₀B₀, we subtract the coordinates of A₀ from B₀:

A₀B₀ = B₀ - A₀ = (0 - (-1), 2 - 0, 0 - 0) = (1, 2, 0)

So, the vector A₀B₀ is (1, 2, 0). This vector points from point A₀ to point B₀ and lies entirely within our plane.

Vector A₀C₀

Similarly, to find the vector A₀C₀, we subtract the coordinates of A₀ from C₀:

A₀C₀ = C₀ - A₀ = (0 - (-1), 0 - 0, 3 - 0) = (1, 0, 3)

Thus, the vector A₀C₀ is (1, 0, 3). This vector points from point A₀ to point C₀ and also lies within the plane.

Now we have two vectors, A₀B₀ and A₀C₀, that lie in the plane. These vectors are crucial because they will help us find the direction perpendicular to the plane, which is our next step.

Step 2: Finding the Normal Vector

The normal vector is a vector that is perpendicular (at a 90-degree angle) to the plane. It's like the plane's unique identifier, telling us its orientation in space. To find the normal vector, we use a mathematical operation called the cross product.

The Cross Product: A Quick Primer

The cross product of two vectors results in a new vector that is perpendicular to both original vectors. If we have two vectors, u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃), their cross product, denoted as u × v, is calculated as follows:

u × v = (u₂v₃ - u₃v₂, u₃v₁ - u₁v₃, u₁v₂ - u₂v₁)

It might look a bit complicated, but it's just a matter of following the formula. A helpful way to remember it is to think of it as a determinant of a matrix, but we won't get into that here. For our purposes, the important thing is that the result is a vector perpendicular to both u and v.

Calculating the Normal Vector

We'll take the cross product of our two vectors, A₀B₀ and A₀C₀, to find the normal vector (n). Remember, A₀B₀ = (1, 2, 0) and A₀C₀ = (1, 0, 3).

n = A₀B₀ × A₀C₀ = (1, 2, 0) × (1, 0, 3)

Let's apply the cross-product formula:

n = (2 * 3 - 0 * 0, 0 * 1 - 1 * 3, 1 * 0 - 2 * 1) = (6, -3, -2)

So, our normal vector n is (6, -3, -2). This vector is perpendicular to the plane we're trying to define. We're getting closer to our goal!

Step 3: Finding the Equation of the Plane

Now we have the normal vector, n = (6, -3, -2), and we have a point on the plane, say A₀(-1, 0, 0). We can use this information to find the equation of the plane in the general form:

Ax + By + Cz + D = 0

The components of the normal vector, (6, -3, -2), directly correspond to the coefficients A, B, and C in the equation. So, we have:

A = 6, B = -3, C = -2

Our equation now looks like this:

6x - 3y - 2z + D = 0

We only need to find the value of D. To do this, we can plug in the coordinates of any point on the plane. We'll use point A₀(-1, 0, 0):

6*(-1) - 3*(0) - 2*(0) + D = 0 -6 + D = 0 D = 6

Now we have all the coefficients! The equation of the plane is:

6x - 3y - 2z + 6 = 0

Conclusion: Mission Accomplished!

We did it! We successfully found the equation of the plane passing through the points A₀(-1, 0, 0), B₀(0, 2, 0), and C₀(0, 0, 3). The equation is 6x - 3y - 2z + 6 = 0, just as we aimed for.

Let's recap the key steps:

1. We created two vectors in the plane using the given points.
2. We calculated the normal vector using the cross product.
3. We used the normal vector and a point to find the equation of the plane.

This method is a powerful tool for solving many geometry problems. Understanding the vector approach allows you to tackle similar challenges with confidence. So, the next time you need to find the equation of a plane, remember these steps, and you'll be well on your way to success! Keep practicing, and you'll become a geometry whiz in no time. You've got this!