How to Find the Equation of a Plane Using Three Points
A plane is a smooth, two-dimensional surface, which stretches infinitely far. A plane is a two - dimensional representation of a point (zero dimensions), a line (one dimension), and a three-dimensional object. A plane in 3-dimensional space has the equation axe + by + cz + d = 0, where at least one of the coefficients a, b or c must be non-zero.
A vector is a physical quantity for which both direction and magnitude are defined. A position vector basically defines the position of a particular point in a three-dimensional cartesian plane system, with respect to an origin point. Consider a line on a plane. This line has a length and an arrow. Here, the length is the magnitude and the arrowhead shows the direction. Hence, in a plane, a line is a vector.
Perpendicular Planes to Vectors and Points
For one particular point on the vector, however, there is only one unique plane that passes through it and is also perpendicular to the vector. A vector can be thought of as a collection of points. So, for a particular vector, there are infinite planes that are perpendicular to it.
The vector equation for the following image is written as: (\[\overrightarrow{r}\] — \[\overrightarrow{r}_{0}\]). \[\overrightarrow{N}\] = 0, where \[\overrightarrow{r}\] and \[\overrightarrow{r}_{0}\] represent the position vectors. For this plane, the cartesian equation is written as:
\[A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0\], where A, B, and C are the direction ratios.
Equation of Plane Passing through 3 Non-Collinear Points
\[P(x_{1},y_{1},z_{1}),~Q(x_{2},y_{2},z_{2}),~and~R(x_{3},y_{3},z_{3})\] are three non-collinear points on a plane.
We know that: ax + by + cz + d = 0 —————(i)
By plugging in the values of the points P, Q, and R into equation (i), we get the following:
\[a(x_{1}) + b(y_{1}) + c(z_{1}) + d = 0\]
\[a(x_{2}) + b(y_{2}) + c(z_{2}) + d = 0\]
\[a(x_{3}) + b(y_{3}) + c(z_{3}) + d = 0\]
Suppose, P = (1,0,2), Q = (2,1,1), and R = (−1,2,1)
Then, by substituting the values in the above equations, we get the following:
a(1) + b(0) + c(2) + d = 0
a(2) + b(1) + c(1) + d = 0
a(-1) + b(2) + c(1) + d = 0
Solving these equations gives us b = 3a, c = 4a, and d = (-9)a. ———————(ii)
By plugging in the values from (ii) into (i), we end up with the following:
ax + by + cz + d = 0
ax + 3ay + 4az−9a
x + 3y + 4z−9
Therefore, the equation of the plane with the three non-collinear points P, Q, and R is x + 3y + 4z−9.
Solved Examples
Example 1: A (3,1,2), B (6,1,2), and C (0,2,0) are three non-collinear points on a plane. Find the equation of the plane.
Solution:
We know that: ax + by + cz + d = 0 —————(i)
By plugging in the values of the points A, B, and C into equation (i), we get the following:
a(3) + b(1) + c(2) + d = 0
a(6) + b(1) + c(2) + d = 0
a(0) + b(2) + c(0) + d = 0
Solving these equations gives us a = 0,
c =12
b, d = —2b ———————(ii)
By plugging in the values from (ii) into (i), we end up with the following:
ax + by + cz + d = 0
0x + (—by) +12bz — 2b = 0
x - y +12z —2 = 0
2x-2y + z-4 = 0
Therefore, the equation of the plane with the three non-collinear points A, B, and C is
2x-2y + z-4 = 0.
Example 2: S (0,0,2), U (1, 0, 1), and V (3, 1,1) are three non-collinear points on a plane. Find the equation of the plane.
Solution:
We know that: ax + by + cz + d = 0 —————(i)
By plugging in the values of the points S, U, and V into equation (i), we get the following:
a(0) + b(0) + c(2) + d = 0
a(1) + b(0) + c(1) + d = 0
a(3) + b(1) + c(1) + d = 0
Solving these equations gives us b = —2a, c = a, d = —2a ———————(ii)
By plugging in the values from (ii) into (i), we end up with the following:
ax + by + cz + d = 0
ax + —2ay + az — 2a = 0
x-2y + z-2 = 0
Therefore, the equation of the plane with the three non-collinear points A, B, and C is
x-2y + z-2 = 0.
FAQs on Equation of a Plane Passing Through 3 Non Collinear Points
1. What is the fundamental difference between collinear and non-collinear points in three-dimensional geometry?
In 3D space, points are collinear if they all lie on a single straight line. In contrast, points are non-collinear if they do not all lie on the same straight line. This distinction is crucial for defining planes: while infinitely many planes can pass through three collinear points (like the pages of a book rotating around its spine), only one unique plane can be defined by three non-collinear points.
2. Why is it necessary for three points to be non-collinear to define a unique plane?
Three non-collinear points are required to define a unique plane because they form a triangle, which is an inherently flat, two-dimensional shape. If the points were collinear (all on one line), you could pivot or rotate an infinite number of different planes around that line as an axis, meaning no single plane is uniquely defined. The third point being off the line 'locks' the plane into a fixed position. A classic example is a three-legged stool, which is always stable because its three feet (non-collinear points) define a unique plane on the floor.
3. What is the vector equation of a plane that passes through three non-collinear points?
To find the vector equation, let the three non-collinear points be A, B, and C with position vectors a, b, and c respectively. The equation is derived as follows:
- First, create two vectors that lie on the plane, for example, AB = (b - a) and AC = (c - a).
- The vector equation of the plane is then given by r = a + λ(b - a) + μ(c - a), where r is the position vector of any point on the plane, and λ and μ are scalar parameters.
4. How do you find the Cartesian equation of a plane passing through three non-collinear points?
If you have three non-collinear points P(x₁, y₁, z₁), Q(x₂, y₂, z₂), and R(x₃, y₃, z₃), the most direct method to find the Cartesian equation is by using a determinant. The equation of the plane is given by the condition that the vectors PQ, PR, and PX (where X is any point (x, y, z) on the plane) are coplanar:
| (x - x₁) (y - y₁) (z - z₁) |
| (x₂ - x₁) (y₂ - y₁) (z₂ - z₁) | = 0
| (x₃ - x₁) (y₃ - y₁) (z₃ - z₁) |
Solving this determinant gives a linear equation in the standard form Ax + By + Cz + D = 0, which is the required Cartesian equation.
5. How many unique planes can pass through three non-collinear points?
Exactly one unique plane can pass through any set of three non-collinear points. This is a fundamental axiom in three-dimensional geometry. The three points act as fixed anchors that determine the precise position and orientation of the plane in space, preventing it from being tilted or shifted.
6. How does the scalar triple product help in finding the equation of a plane through three points?
The scalar triple product provides an elegant way to derive the plane's equation. Given three non-collinear points A, B, and C, we can define two vectors in the plane, AB and AC. Their cross product, n = AB × AC, yields a normal vector perpendicular to the plane.
For any general point R on the plane, the vector AR must also lie in the plane. Consequently, AR must be perpendicular to the normal vector n. This means their dot product is zero: AR ⋅ (AB × AC) = 0. This expression is the scalar triple product, and setting it to zero directly provides the Cartesian equation of the plane.
7. What is a simple real-world example that shows a plane defined by three non-collinear points?
A very common real-world example is a camera tripod. The tips of its three legs form three non-collinear points. Regardless of the ground's evenness, the three feet will always rest stably because they define a single, unique plane. This stability is the practical application of the geometric principle that three non-collinear points define a plane, ensuring the camera remains steady.
