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 coplanarity of two lines.

#### 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 (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) be the coordinates of the points respectively. The direction cosines of two vectors \(\vec{m_{1}}\) and \(\vec{m_{2}}\) is given by a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} respectively.

\(\vec{LM}\) = (x_{2} − x_{1})\(\hat{i}\) + (y_{2} − y_{1})\(\hat{j}\) + (z_{2} − z_{1})\(\hat{k}\)

\(\vec{m_{1}}\) = a_{1}\(\hat{i}\) +b_{1}\(\hat{j}\) + c_{1}\(\hat{k}\)

\(\vec{m_{2}}\) = a_{2}\(\hat{i}\) +b_{2}\(\hat{j}\) + c_{2}\(\hat{k}\)

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

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

This can be represented in cartesian form as,

#### Problems related to coplanarity of two lines:

__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:

x_{1} = -3, y_{1} = 2, z_{1} = 5, x_{2} = -1 , y_{2} = 2, z_{2} = -5, a_{1} = -3 , b_{1} = 4, c_{1} = 5, a_{2} = -3 , b_{2} = 3, c_{2} = 6

Thus, the given lines are not coplanar. To learn more about coplanarity of two lines download Byju’s- The Learning App.