We know that, the points and lines can be drawn on a plane surface. Plane surfaces are those which are two dimensional surfaces. Points are said to be collinear if they lie on the same line. Set of lines which pass through the same point are called the concurrent lines. In this section we shall discuss about coplanarity. Also when we study three dimensional geometry we learn about the direction ratios, direction cosines and the equation of the planes and the lines. We shall also discuss few examples on coplanar points, lines etc.

## Definition of Coplanar

Set of points, or lines or line segments or rays or geometrical shapes which lie on the same plane are said to be co-planar.

Coplanar points

In the above diagram, we can see that two planes intersect ( meet ) along a line. The points A and B are on one plane and the points M and N are another plane.

Therefore, Points A and B are coplanar.

similarly the points M and N are co-planar.

Coplanar lines
The lines which lie on the same plane are said to be coplanar lines.

In the above figure, we can see that the lines, l, m and n lie on the same plane, hence they are coplanar.

Coplanar line segments

In the above diagram, the line segments, $\overline{AB}$, $\overline{BC}$, $\overline{CA}$ and $\overline{PQ}$ are on the same plane and hence they are called as coplanar line segments.

Coplanar rays

In the above figure, the rays $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{PQ}$ lie on the same plane and hence they are called as coplanar rays.

Coplanar geometrical shapes

In the above figure, the shapes rectangle and rhombus lie on plane 1 and the shapes, square and triangle lie on plane 2.
The shapes on the same plane are coplanar.Therefore, rectangle and the rhombus are coplanar.

similarly the square and the triangle are coplanar.

## Coplanar Lines

Lines which lie on the same plane are said to be coplanar lines.

#### Similarly the lines which lie on different planes are said to be non-co-planar lines.

In the following figure, we can see that the lines l and m are on the same plane and the lines p and q are on another plane.

Therefore, the lines l and m are co-planar and the lines p and q are co-planar.
where as the lines, l and p or l and q, m and p or m and q are non-coplanar lines.

## Coplanar Planes

When two planes coincide, we say that the planes are coplanar.

For example,

We can see that the Plane coincide with the Plane 1, therefore, the two planes are co-planar.
similarly we can have more that two planes coincide with the same plane, hence we can say that the planes are coplanar.

## Coplanar Example

Let us verify the coplanarity using vector form.

### Solved Examples

Question 1: Prove that the points with position vectors
( - $\hat{j}$ - $\hat{k}$ ),
( 4 $\hat{i}$ + 5 $\hat{j}$ + $\hat{k}$ ),
( 3 $\hat{i}$ + 9 $\hat{j}$ + 4 $\hat{k}$ )
and ( -4 $\hat{i}$ + 4 $\hat{j}$ + 4 $\hat{k}$ ) are coplanar.
Solution:

Let the given points be A, B, C and D respectively.
The given points will be coplanar, if the vectors, $\overrightarrow{DA}$, $\overrightarrow{DB}$, $\overrightarrow{DC}$ are coplanar.
$\overrightarrow{DA}$ = position vector of A - position vector of D
= ( - $\hat{j}$ - $\hat{k}$ ) - ( -4 $\hat{i}$ + 4 $\hat{j}$ + 4 $\hat{k}$ ) = ( 4 $\hat{i}$ - 5 $\hat{j}$ - 5 $\hat{k}$ )
$\overrightarrow{DB}$ = position vector of B - position vector of D
= ( 4 $\hat{i}$ + 5 $\hat{j}$ + $\hat{k}$ ) - ( -4 $\hat{i}$ + 4 $\hat{j}$ + 4 $\hat{k}$ ) = (8 $\hat{i}$ +  $\hat{j}$ - 3 $\hat{k}$ )
$\overrightarrow{DC}$ = position vector of C - position vector of D
= ( 3 $\hat{i}$ + 9 $\hat{j}$ + 4 $\hat{k}$ ) - ( -4 $\hat{i}$ + 4 $\hat{j}$ + 4 $\hat{k}$ ) = 7 $\hat{i}$ + 5 $\hat{j}$
To verify that the vectors $\overrightarrow{DA}$, $\overrightarrow{DB}$, $\overrightarrow{DC}$ are coplanar,
let us prove that the determinant formed by these vectors = 0.
$\begin{vmatrix} 4&-5 & -5\\ 8&1 & -3\\ 7& 5 & 0 \end{vmatrix}$                    = 4 ( 0 -  ( -15 ) ) + 5 ( 0 - ( - 21 ) ) - 5 ( 40 - 7 )
=4 ( 15 ) + 5 ( 21 ) - 5 ( 33 )
= 60 +- 105 - 165
= 165 - 165 = 0
As the determinant is 0, we can say that the four points are coplanar.
Hence the four points are coplanar.

Question 2: Find the Equation of the plane containing the lines,
$\frac{x-1}{2}$ = $\frac{y-2}{-4}$ = $\frac{z-5}{5}$
$\frac{x-3}{2}$ = $\frac{y-4}{4}$ = $\frac{z+1}{5}$
Solution:

We are given the lines,
$\frac{x-1}{2}$ = $\frac{y-2}{-4}$ = $\frac{z-5}{5}$       ------------------------ ( 1 )
$\frac{x-3}{2}$ = $\frac{y-4}{4}$ = $\frac{z+1}{5}$      ------------------------- ( 2 )
The direction ratio of the line ( 1 ) is 2, - 4, 5
and the direction ratio of the line ( 2 ) is 2, 4, 5
Line ( 1 ) passes through the point ( 1, 2, 5 )
Therefore, the equation of the plane passing through the point ( 1, 2, 5 ) and containing direction ratios 2, - 4 , 5 and 2, 4, 5 is given by the following determinant.
$\begin{vmatrix} x-1 &x-2 &x-5 \\ 2 &-4 &5 \\ 2 &4 &5 \end{vmatrix}$                  =  0
=> ( x - 1 ) ( - 20 - 20 ) - ( y - 2 ) ( 10 - 10 ) + ( z - 5 ) ( 8 + 8 ) = 0
=>                        ( x - 1 ) ( - 40 ) - ( y - 2 ) ( 0 ) + ( z - 5 ) 16 = 0
=>                                         - 40 x + 40 + 16 z - 80           = 0
=>                                          - 40 x + 16 z - 40                  = 0
=>                                               5 x - 2 z + 5                   = 0 [ dividing both sides of the equation by - 8 ]

therefore, the equation of the plane containing the two given lines is, 5 x - 2 z + 5 = 0

## Non Coplanar

### Set of points or lines are said to non-coplanar if all of them do not like on the same plane.

In the above diagram, the lines $\vec{p}$, $\vec{q}$ and $\vec{r}$ are on different planes, and hence are called as non-coplanar lines.

### Solved Example

Question: Prove that the vectors ( $\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$ ),
( 2 $\hat{i}$ + $\hat{j}$ + 3 $\hat{k}$ )
and ( $\hat{i}$    + $\hat{j}$  +    $\hat{k}$ )           are non coplanar.
Solution:

To prove that the vectors are co-planar, let us find the determinant of the vectors.
$\begin{vmatrix} 1 &2 &3 \\ 2 &1 &3 \\ 1 &1 &1 \end{vmatrix}$              = 1 ( 1 - 3 ) - 2 ( 2 - 3 ) + 3 ( 2 - 1 )
= 1 ( - 2 ) - 2 ( - 1 ) + 3 ( 1 )
= - 2 + 2 + 3
= 3 $\neq$  0

Therefore, the above vectors are non coplanar.