2) Write The Equation Of The Plane Containing The Points (3, 2, 1), (1, 2, -2), And (2, 0, 1) (Hint: You
Understanding how to find the equation of a plane given three points is an essential skill in coordinate geometry. This process involves using vector operations to determine the plane's normal vector and then applying the point-normal form of a plane equation. Whether you're a student working through algebra assignments or a professional dealing with 3D modeling, mastering this concept is vital. In this article, we'll walk through the step-by-step process of writing the equation of the plane that contains the points (3, 2, 1), (1, 2, -2), and (2, 0, 1).
Understanding the Fundamentals
Before diving into calculations, it's important to understand the key concepts involved in determining the plane's equation.
What is a Plane in 3D Space?
A plane in three-dimensional space is a flat, two-dimensional surface that extends infinitely in all directions within that space. It can be uniquely identified with a point through which it passes and a normal vector perpendicular to its surface.
Point-Normal Form of a Plane Equation
The standard form of a plane's equation is:
\[ a(x - x0) + b(y - y0) + c(z - z_0) = 0 \]
where:
- \( (x0, y0, z_0) \) is a point on the plane.
- \( \vec{n} = \langle a, b, c \rangle \) is the normal vector to the plane.
Alternatively, it can be written as:
\[ a x + b y + c z + d = 0 \]
where \( d \) can be found by substituting the point into the equation.
The main task in our problem is to find the normal vector \( \vec{n} \) using the given points, then write the plane's equation accordingly.
Step-by-Step Guide to Find the Plane Equation
Let's now proceed with the calculation steps to find the equation of the plane passing through the given points.
Step 1: Identify the Given Points
The points provided are:
- \( P_1 = (3, 2, 1) \)
- \( P_2 = (1, 2, -2) \)
- \( P_3 = (2, 0, 1) \)
These three points are guaranteed to define a unique plane if they are not collinear.
Step 2: Find Two Direction Vectors on the Plane
To determine the plane's normal vector, we need two vectors lying on the plane, which can be obtained by subtracting the position vectors of the given points:
- \( \vec{v}1 = P2 - P_1 \)
- \( \vec{v}2 = P3 - P_1 \)
Calculations:
\[
\begin{aligned}
\vec{v}_1 &= (1 - 3,\, 2 - 2,\, -2 - 1) = (-2, 0, -3) \\
\vec{v}_2 &= (2 - 3,\, 0 - 2,\, 1 - 1) = (-1, -2, 0)
\end{aligned}
\]
Step 3: Compute the Normal Vector Using Cross Product
The normal vector \( \vec{n} \) to the plane can be found by taking the cross product of \( \vec{v}1 \) and \( \vec{v}2 \):
\[
\vec{n} = \vec{v}1 \times \vec{v}2
\]
Recall the cross product formula for vectors in 3D:
\[
\vec{a} \times \vec{b} = (a2 b3 - a3 b2,\, a3 b1 - a1 b3,\, a1 b2 - a2 b1)
\]
Applying this:
\[
\begin{aligned}
a1 &= -2, \quad a2 = 0, \quad a_3 = -3 \\
b1 &= -1, \quad b2 = -2, \quad b_3 = 0
\end{aligned}
\]
Compute each component:
\[
\begin{aligned}
nx &= (a2 \times b3) - (a3 \times b_2) = (0 \times 0) - (-3 \times -2) = 0 - 6 = -6 \\
ny &= (a3 \times b1) - (a1 \times b_3) = (-3 \times -1) - (-2 \times 0) = 3 - 0 = 3 \\
nz &= (a1 \times b2) - (a2 \times b_1) = (-2 \times -2) - (0 \times -1) = 4 - 0 = 4
\end{aligned}
\]
Therefore, the normal vector is:
\[
\vec{n} = \langle -6, 3, 4 \rangle
\]
This vector is perpendicular to the plane.
Formulating the Equation of the Plane
With the normal vector \( \vec{n} = \langle -6, 3, 4 \rangle \) and a point on the plane (say, \( P_1 = (3, 2, 1) \)), we can write the plane's equation.
Step 4: Write the Equation Using Point-Normal Form
Plugging into the formula:
\[
a(x - x0) + b(y - y0) + c(z - z_0) = 0
\]
we get:
\[
-6(x - 3) + 3(y - 2) + 4(z - 1) = 0
\]
Expand:
\[
-6x + 18 + 3y - 6 + 4z - 4 = 0
\]
Combine like terms:
\[
-6x + 3y + 4z + (18 - 6 - 4) = 0
\]
\[
-6x + 3y + 4z + 8 = 0
\]
Multiplying through by -1 to make coefficients positive:
\[
6x - 3y - 4z - 8 = 0
\]
This is the equation of the plane containing the given points.
Final Equation of the Plane
\[
\boxed{
6x - 3y - 4z = 8
}
\]
This equation fully describes the plane passing through the points \( (3, 2, 1) \), \( (1, 2, -2) \), and \( (2, 0, 1) \).
Additional Tips for Writing Plane Equations
- Verify the points: Always confirm the points are not collinear; otherwise, they do not define a unique plane.
- Use vector operations: Cross products are powerful tools to find the perpendicular direction (normal vector) to the plane.
- Simplify equations: You can multiply the entire equation by a scalar to make coefficients cleaner or to eliminate fractions.
- Check your work: Substitute the original points into the final equation to ensure they satisfy it.
Applications of Plane Equations in Real Life
Understanding how to write the equation of a plane has practical applications across various fields:
- Computer Graphics & 3D Modeling: Defining surfaces, clipping, and rendering.
- Engineering: Analyzing structural components and surfaces.
- Geography & Cartography: Modeling terrains and land features.
- Physics: Describing planes of motion or interaction between objects.
Summary
In this article, we've demonstrated how to find the equation of a plane given three points in 3D space. The process involves:
- Calculating vectors between points.
- Computing the cross product to find the normal vector.
- Applying the point-normal form to derive the plane's equation.
For the specific points provided, the final plane equation is:
\[
6x - 3y - 4z = 8
\]
Mastering these steps enhances your understanding of spatial relationships and prepares you for tackling more complex geometry problems. Whether you’re solving homework problems or working on advanced engineering projects, knowing how to derive and interpret plane equations is a vital skill in the realm of three-dimensional geometry.
