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.

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