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# Coplanarity of Two Lines In 3D Geometry

Two lines are said to be coplanar when they both lie on the same plane in a three-dimensional space. We have learnt how to represent the equation of a line in three-dimensional space using vector notations. In this article, we will learn about the coplanarity of two lines in 3D geometry.

## Condition for coplanarity of two lines in vector form

Using vector notations equation of line is given by:

$$\begin{array}{l}\vec{r}\end{array}$$
=
$$\begin{array}{l}\vec{l_{1}}\end{array}$$
+ λ
$$\begin{array}{l}\vec{m_{1}}\end{array}$$
——————— (1)

$$\begin{array}{l}\vec{r}\end{array}$$
=
$$\begin{array}{l}\vec{l_{2}}\end{array}$$
+ μ
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
——————– (2)

Here, the line (1) passes through a point L having position vector

$$\begin{array}{l}\vec{l_{1}}\end{array}$$
and is parallel to
$$\begin{array}{l}\vec{m_{1}}\end{array}$$
and the line (2) passes through a point M having position vector
$$\begin{array}{l}\vec{l_{2}}\end{array}$$
and is parallel to
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
. These two lines are coplanar if and only if
$$\begin{array}{l}\vec{LM}\end{array}$$
is perpendicular to
$$\begin{array}{l}\vec{m_{1}}\end{array}$$
x
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
.

This can be given as:

$$\begin{array}{l}\vec{LM}\end{array}$$
=
$$\begin{array}{l}\vec{l_{2}}\end{array}$$
$$\begin{array}{l}\vec{l_{1}}\end{array}$$

Thus condition of coplanarity is given by:

$$\begin{array}{l}\vec{LM}\end{array}$$
.(
$$\begin{array}{l}\vec{m_{1}}\end{array}$$
x
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
) = 0
(
$$\begin{array}{l}\vec{l_{2}}\end{array}$$
$$\begin{array}{l}\vec{l_{1}}\end{array}$$
).(
$$\begin{array}{l}\vec{m_{1}}\end{array}$$
x
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
) = 0

## Condition for coplanarity of two lines in cartesian form

Let us take two points L and M such that (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points respectively. The direction cosines of two vectors

$$\begin{array}{l}\vec{m_{1}}\end{array}$$
and
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
is given by a1, b1, c1 and a2, b2, c2 respectively.

$$\begin{array}{l}\vec{LM}\end{array}$$
= (x2 − x1)
$$\begin{array}{l}\hat{i}\end{array}$$
+ (y2 − y1)
$$\begin{array}{l}\hat{j}\end{array}$$
+ (z2 − z1)
$$\begin{array}{l}\hat{k}\end{array}$$

$$\begin{array}{l}\vec{m_{1}}\end{array}$$
= a1
$$\begin{array}{l}\hat{i}\end{array}$$
+b1
$$\begin{array}{l}\hat{j}\end{array}$$
+ c1
$$\begin{array}{l}\hat{k}\end{array}$$

$$\begin{array}{l}\vec{m_{2}}\end{array}$$
= a2
$$\begin{array}{l}\hat{i}\end{array}$$
+b2
$$\begin{array}{l}\hat{j}\end{array}$$
+ c2
$$\begin{array}{l}\hat{k}\end{array}$$

By the above condition two lines can be coplanar if and only if,

$$\begin{array}{l}\vec{LM}\end{array}$$
.(
$$\begin{array}{l}\vec{m_{1}}\end{array}$$
x
$$\begin{array}{l}\vec{m_{2}}\end{array}$$
) = 0

### Examples

Question: Show that lines

$$\begin{array}{l}\frac{x + 3}{-3}\end{array}$$
=
$$\begin{array}{l}\frac{y – 2}{4}\end{array}$$
=
$$\begin{array}{l}\frac{z – 5}{5}\end{array}$$
and
$$\begin{array}{l}\frac{x + 1}{-3}\end{array}$$
=
$$\begin{array}{l}\frac{y – 2}{3}\end{array}$$
=
$$\begin{array}{l}\frac{z + 5}{6}\end{array}$$
are coplanar.

Solution:

According to the question:

x1 = -3, y1 = 2, z1 = 5, x2 = -1 , y2 = 2, z2 = -5, a1 = -3 , b1 = 4, c1 = 5, a2 = -3 , b2 = 3, c2 = 6