COPLANAR POINTS: EQUATION, EXAMPLE AND SOLVED EXERCISES - MATHS - 2025

The coplanar points all belong to the same plane. Two points are always coplanar, since these points define a line through which infinite planes pass. Then, both points belong to each of the planes that pass through the line and therefore, they will always be coplanar.

On the other hand, three points define a single plane, from which it follows that three points will always be coplanar to the plane they determine.

Figure 1. A, B, C and D are coplanar to the (Ω) plane. E, F and G are not coplanar to (Ω) but they are coplanar to the plane that they define. Source: F. Zapata.

More than three points can be coplanar or not. For example in figure 1, points A, B, C and D are coplanar to the plane (Ω). But E, F and G are not coplanar to (Ω), although they are coplanar to the plane that they define.

Equation of a plane given three points

The equation of a plane determined by three known points A, B, C is a mathematical relation that guarantees that any point P with generic coordinates (x, y, z) that fulfills the equation belongs to said plane.

The previous statement is equivalent to saying that if P of coordinates (x, y, z) fulfills the equation of the plane, then said point will be coplanar with the three points A, B, C that determined the plane.

To find the equation of this plane, let's start by finding the vectors AB and AC:

AB =

AC =

The vector product AB X AC results in a vector perpendicular or normal to the plane determined by points A, B, C.

Any point P with coordinates (x, y, z) belongs to the plane if the vector AP is perpendicular to the vector AB X AC, which is guaranteed if:

AP • (AB X AC) = 0

This is equivalent to saying that the triple product of AP, AB, and AC is zero. The above equation can be written in matrix form:

Example

Let the points A (0, 1, 2); B (1, 2, 3); C (7, 2, 1) and D (a, 0, 1). What value must a have for the four points to be coplanar?

Solution

To find the value of a, point D must be part of the plane determined by A, B and C, which is guaranteed if it satisfies the equation of the plane.

Developing the determinant we have:

The previous equation tells us that a = -1 for the equality to be fulfilled. In other words, the only way that point D (a, 0,1) is coplanar with points A, B and C is for a to be -1. Otherwise it will not be coplanar.

Solved exercises

- Exercise 1

A plane intersects the Cartesian axes X, Y, Z at 1, 2, and 3 respectively. The intersection of this plane with the axes determines points A, B and C. Find the component Dz of a point D, whose Cartesian components are:

Provided that D is coplanar with points A, B and C.

Solution

When the intercepts of a plane with the Cartesian axes are known, the segmental form of the equation of the plane can be used:

x / 1 + y / 2 + z / 3 = 1

Since point D must belong to the previous plane, it has to:

-Dz / 1 + (Dz + 1) / 2 + Dz / 3 = 1

That is to say:

-Dz + Dz / 2 + ½ + Dz / 3 = 1

Dz (-1 + ½ + ⅓) = ½

Dz (-1 / 6⅙) = ½

Dz = -3

From the above it follows that point D (3, -2, -3) is coplanar with points A (1, 0, 0); B (0, 2, 0) and C (0, 0, 3).

- Exercise 2

Determine if the points A (0, 5, 3); B (0, 6, 4); C (2, 4, 2) and D (2, 3, 1) are coplanar.

Solution

We form the matrix whose rows are the coordinates of DA, BA, and CA. Then the determinant is calculated and it is verified whether or not it is zero.

After performing all the calculations, it is concluded that they are coplanar.

- Exercise 3

There are two lines in space. One of them is the line (R) whose parametric equation is:

And the other is the line (S) whose equation is:

Show that (R) and (S) are coplanar lines, that is, they lie in the same plane.

Solution

Let's start by arbitrarily taking two points on the line (R) and two on the line (S):

Line (R): λ = 0; A (1, 1, 1) and λ = 1; B (3, 0, 1)

Let x = 0 on the line (S) => y = ½; C (0, ½, -1). And on the other hand, if we make y = 0 => x = 1; D (1, 0, -1).

That is, we have taken the points A and B that belong to the line (R) and the points C and D that belong to the line (S). If those points are coplanar, then the two lines will be too.

Now we choose point A as the pivot and then we find the coordinates of the vectors AB, AC and AD. In this way you get:

B - A: (3-1, 0 -1, 1 - 1) => AB = (2, -1, 0)

C - A: (0-1, 1/2 -1, -1 - 1) => AC = (-1, -1/2, -2)

D - A: (1-1, 0 -1, -1 - 1) => AD = (0, -1, -2)

The next step is to construct and calculate the determinant whose first row are the coefficients of the vector AB, the second row are those of AC and the third row those of the vector AD:

Since the determinant turns out to be null, then we can conclude that the four points are coplanar. Additionally, it can be stated that the lines (R) and (S) are also coplanar.

- Exercise 4

The lines (R) and (S) are coplanar, as demonstrated in Exercise 3. Find the equation of the plane that contains them.

Solution

Points A, B, C completely define that plane, but we want to impose that any point X of coordinates (x, y, z) belongs to it.

For X to belong to the plane defined by A, B, C and in which the lines (R) and (S) are contained, it is necessary that the determinant formed in its first row by the components of AX, in the second row by those of AB and in the third by those of AC:

Following this result, we group in this way:

2 (x-1) + 4 (y-1) -2 (z-1) = 0

And immediately you see that it can be rewritten like this:

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

Therefore x + 2y - z = 2 is the equation of the plane that contains the lines (R) and (S).

References

1. Fleming, W. 1989. Precalculus Mathematics. Prentice Hall PTR.
2. Kolman, B. 2006. Linear Algebra. Pearson Education.
3. Leal, JM 2005. Flat Analytical Geometry. Mérida - Venezuela: Editorial Venezolana CA
4. Navarro, Rocio. Vectors. Recovered from: books.google.co.ve.
5. Pérez, CD 2006. Pre-calculation. Pearson Education.
6. Prenowitz, W. 2012. Basic Concepts of Geometry. Rowman & Littlefield.
7. Sullivan, M. 1997. Precalculus. Pearson Education.