# Equation of a Plane Passing Through 3 Non Collinear 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 ax + 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 show 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 which 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 which 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−x1) + B (y−y1) + C (z−z1) = 0, where A, B, and C are the direction ratios.

### Equation of Plane Passing Through 3 Non - Collinear Points

P(x1, y1, z1), Q(x2, y2, z2), and R (x3, y3, z3) 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(x1) + b(y1) + c(z1) + d = 0

a(x2) + b(y2) + c(z2) + d = 0

a(x3) + b(y3) + c(z3) + 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 = $\frac{1}{2}$ 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) + $\frac{1}{2}$ bz — 2b = 0

x - y + $\frac{1}{2}$ z —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.