# Coplanarity of Two Lines In 3D Geometry

## Trigonometry # 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:

$$\vec{r}$$ = $$\vec{l_{1}}$$ + λ$$\vec{m_{1}}$$       ——————— (1)

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

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

This can be given as:

$$\vec{LM}$$ = $$\vec{l_{2}}$$$$\vec{l_{1}}$$

Thus condition of coplanarity is given by:

$$\vec{LM}$$.($$\vec{m_{1}}$$ x $$\vec{m_{2}}$$) = 0
($$\vec{l_{2}}$$$$\vec{l_{1}}$$).($$\vec{m_{1}}$$ x $$\vec{m_{2}}$$) = 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 $$\vec{m_{1}}$$ and $$\vec{m_{2}}$$ is given by a1, b1, c1 and a2, b2, c2 respectively.

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

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

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

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

$$\vec{LM}$$.($$\vec{m_{1}}$$ x $$\vec{m_{2}}$$) = 0

### Examples

Question: Show that lines $$\frac{x + 3}{-3}$$ = $$\frac{y – 2}{4}$$ = $$\frac{z – 5}{5}$$ and $$\frac{x + 1}{-3}$$ = $$\frac{y – 2}{3}$$ = $$\frac{z + 5}{6}$$ 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  