When we have three or more points that lie in the same plane in three-dimensional space, then these points would be called coplanar. We need to remember here that a plane is a flat surface that extends infinitely in all directions. A plane is normally denoted by italics upper case alphabet. The coplanar points in the plane are usually denoted by normal upper case alphabets. **This is shown in the picture below:**

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We know that there can be infinitely many lines that can pass through one point in a plane. But if we have two points in a plane then there can be exactly one line only that passes through both the points. Such points are called collinear points. Similarly, in three-dimensional space, there can be infinitely many planes that pass through a line joining two points. However, if we have three non-collinear points, then there can be exactly only one plane that passes through all the three points. Such points are called coplanar points. It is important to note here that the points have to be non-collinear. If the points are collinear then there is one single line that passes through all the points, in that case, there can be infinitely many planes that can pass through that line. However, the points lying on the same line are still coplanar. In other words, collinear points are always coplanar, but coplanar points may or may not be collinear. Collinearity is a concept of two-dimensional plane geometry and coplanarity is a concept of three-dimensional spatial geometry.

In the above figure, note that the points B, D, and A are collinear as well as coplanar. All these three points lie in both the planes $\wp$ and $\Re$. These points are coplanar with C and E in plane $\wp$ and with G and F in plane $\Re$. However, C, E, F, and G are non-coplanar points. Again the points H and I do not lie in any of the two planes shown in the figure.

The condition for defining a plane is that there should be a minimum of three non-collinear points. Then we can use these points to describe the plane that they lie in. When we have four or more points, obviously we can define more than one planes in three-dimensional space taking three points at a time. This concept is similar to a two-dimensional plane. If we have 2 points in a plane, we can define a line that joins the two points. If we have 3 points in a plane, we can define 3 lines that join the points taking two at a time. If we have 4 points in a plane, we can define 6 lines joining them taking again two at a time. Similarly in three-dimensional space, if we have 4 points we can define 4 planes. We use a bit of combinatorics here. We need 3 points to define one plane. So if we have 4 points, then the number of planes that can be defined can be calculated as:

$nC_r$ = $\frac{n!}{r!(n\ -\ r)!}$

$4C_3$ = $\frac{4!}{3!(4\ -\ 3)!}$ = $4$

If we have exactly three points, they are obviously going to be coplanar. However, if we have 4 points then we may need to determine whether they are coplanar or not. Suppose our four points are:

($x_1$, $y_1$, $z_1$),($x_2$, $y_2$, $z_2$),($x_3$, $y_3$, $z_3$) and ($x_4$, $y_4$, $z_4$)

Using the first three points given to us, we derive the equation of a plane that is defined by those three points. Then we plug in the fourth point into that equation. If it satisfies the equation, it means that the fourth point also lies in the same plane as the other three points. Thus making the four points coplanar. A simpler way to do this is using the triple product of vectors. This method may not be too useful if we have more than 4 points.

In this method, we write a matrix of the relative vectors of the given four points. If the rank of this matrix is two or less, then the four points are coplanar.

Let:

A(-1,0,1)

B(3,2,0)

C(1,1,2)

D(3,2,4)

Therefore the vectors,

$AB$ = (4,2,-1)

$AC$ =(2,1,1)

$AD$ = (4,2,3)

Now we find the triple scalar product:

$AB\ •\ (AC\ \times\ AD)$ = $\begin{bmatrix} 4 & 2 & -1\\ 2 & 1 & 1\\ 4 & 2 & 3 \end{bmatrix}$

= $4\ \begin{bmatrix} 1 & 1\\ 2 & 3 \end{bmatrix}$ $-2\ \begin{bmatrix} 2 & 1\\ 4 & 3 \end{bmatrix}$ $-1\ \begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$

= $4(3\ -\ 2)\ -\ 2(6\ -\ 4)\ -\ 1(4\ -\ 4)$

= $4(1)\ -\ 2\ (2)\ -\ 1\ (0)$

= $4\ -\ 4\ -\ 0$ = $0$

Since the triple cross product is zero, the four points are coplanar!

This time let us use the matrix rank method. Just so that we know how to use both the methods. The first two steps of finding the relative vectors and writing the matrix are same for both the methods. So here goes:

Let,

A(0, 1, 5)

B(4, 6, 1)

C(- 2, 2, 1)

D(2, -1, 0)

The vectors AB, AC, and AD would be:

$AB$ = (4, 5, -4)

$AC$ = (- 2, 1, - 4)

$AD$ = (2, - 2, - 5)

The matrix of these vectors would look like this:

$\begin{bmatrix} 4 & 5 & -4\\ -2 & 1 & -4\\ 2 & -2 & -5 \end{bmatrix}$

Let we now try to find the rank of this matrix.

Row 2 $\rightarrow$ Row 2 + $(\frac{1}{2})$ Row 1

Row 3 $\rightarrow$ Row 3 - $(\frac{1}{2})$ Row 1

= $\begin{bmatrix} 4 & 5 & -4\\ 0 & \frac{7}{2} & -6\\ 0 & - \frac{9}{2} & -3 \end{bmatrix}$

Next we do a similar operation to eliminate the $- \frac{9}{2}$. We have:

Row 3 $\rightarrow$ Row 3 + $(\frac{9}{7})$ Row 2

= $\begin{bmatrix} 4 & 5 & -4\\ 0 & \frac{7}{2} & -6\\ 0 & 0 & - \frac{75}{7} \end{bmatrix}$

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