Coplanarity of Two Lines: Definition, Conditions, Vector Form, Cartesian Form

Coplanarity of Two Lines: In \(3D\) geometry, coplanar lines are a prominent notion. Collinear and coplanar are two words in geometry that seem similar and might confuse the phrases used to describe them. In each of these terms, “co” means “together,” “linear” means “along a line,” and “planar” points “on a plane.” Thus, collinear indicates that they are parallel on a line, whereas coplanar implies that they are similar on a plane. The number of lines in the same plane determines the number of coplanar lines. The lines on the same plane are said to be coplanar, while lines that do not lay on the same plane are called non-coplanar.

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Equation of Line Passing Through a Point

Following are few different forms that explain equation of line passing through a point
Vector Form: Let the line pass through fixed point \(A\) whose position vector is \(\overrightarrow a \) and the line is parallel to the vector \(\overrightarrow b \) and \(P\left( {\overrightarrow r } \right)\) be a variable point on the line. Then, the equation of the line in vector form is
\(\vec r = \vec a + \lambda \vec b\)
Cartesian Form or Parametric Form: Let the line pass through a fixed point \(A\left( {{x_1},\,{y_1},\,{z_1}} \right)\) and be parallel to the vector \(\vec b\), whose Direction Cosines \(\left( {DC} \right)\) are \(l,\,m\), and \(n\).
If \(P\left( {x,\,y,\,z} \right)\) is a point on the line, then the equation of the line is given by
\(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n} = \lambda \)

Equation of a Line Passing Through Two Given Fixed Points

Following are few different forms that explain equation of line passing through two given fixed points.
Vector Form: Let the line passes through two fixed points \(A\) and \(B\) whose position vectors are \(\overrightarrow a \) and \(\overrightarrow b \), respectively. Let \(P\) be a variable point on the line whose position vector is \(\overrightarrow r \). Hence, the equation of the line in vector form is given by \(\vec r = \vec a + \lambda \left( {\vec b – \vec a} \right)\)

Cartesian Form or Parametric form: Let the line passes through a fixed point \(A\left( {{x_1},\,{y_1},\,{z_1}} \right)\) and \(B\left( {{x_2},\,{y_2},\,{z_2}} \right)\). If \(P\left( {x,\,\,y,\,\,z} \right)\) is a point on the line, then the equation of the line is given by \(\frac{{x – {x_1}}}{{{x_2} – {x_1}}} = \frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{z – {z_1}}}{{{z_2} – {z_1}}} = \lambda \)

Conditions to Prove Coplanarity

Coplanarity is defined as the condition of a particular number of lines lying on the same plane, as defined by mathematical ideas. In \(3\)-dimensional geometry, we can use the condition in cartesian and vector forms to prove that two lines are coplanar.

Condition for Coplanarity in Vector Form

Let’s write the equation in the vector form of two lines whose coplanarity must be determined.
\(\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \)
\(\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \)
The given line passes through the position vector \(a\) which lies in the \(3D\) plane and parallel to the position vector \(b\). So, the \(\overrightarrow {{r_1}} \) passes through the point \(A\) with the position \(\overrightarrow {{a_1}} \) and is parallel to the \(\overrightarrow {{b_1}} \), while the \(\overrightarrow {{r_2}} \) passes through the point \(B\) with position \(\overrightarrow {{a_2}} \) and is parallel to the \(\overrightarrow {{b_2}} \).
\(\therefore \overrightarrow {AB} = {\vec a_2} – {\vec a_1}\)
If \(\overrightarrow {AB} \) is perpendicular to the cross product of \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) then the given lines are coplanar, i.e.,
\(\overrightarrow {AB} \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\)
\( \Rightarrow \left( {{{\vec a}_2} – {{\vec a}_1}} \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\)
Here the cross product of \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) is a vector line that will be perpendicular to both \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) lines. \(\overrightarrow {AB} \) is a line vector joining the position \(\overrightarrow {{a_1}} \) and \(\overrightarrow {{a_2}} \) of two lines. Now, we can check whether two lines are coplanar or not by determining above dot product is zero or not.

Condition for Coplanarity in Cartesian Form

Let \(\left( {{x_1},\,{y_1},\,{z_1}} \right)\) and \(\left( {{x_2},\,{y_2},\,{z_2}} \right)\) be the coordinates of points \(A\) and \(B\), respectively. Let \({a_1},\,{b_1},\,{c_1}\) and \({a_2},\,{b_2},\,{c_2}\) be the direction ratios of vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \), respectively. Then
\(\overrightarrow {AB} = \left( {{x_2} – {x_1}} \right)\hat \imath + \left( {{y_2} – {y_1}} \right)\hat \jmath + \left( {{z_2} – {z_1}} \right)\hat k\)
\(\overrightarrow {{b_1}} = {a_1}\hat \imath + {b_1}\hat \jmath + {c_1}\hat k\) and \(\overrightarrow {{b_2}} = {a_2}\hat \imath + {b_2}\hat \jmath + {c_2}\hat k\)
The given lines are coplanar if \(\overrightarrow {AB} \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\).
In the Cartesian form, it can be expressed as
\(\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0\)
Therefore, both forms need position vectors \(\overrightarrow {{a_1}} \) and \(\overrightarrow {{a_2}} \) in input as \(\left( {{x_1},\,{y_1},\,{z_1}} \right)\) and \(\left( {{x_2},\,{y_2},\,{z_2}} \right)\) , respectively. The direction ratios of vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) as \(\left( {{a_1},\,\,{b_1},\,\,{c_1}} \right)\) and \(\left( {{a_2},\,\,{b_2},\,\,{c_2}} \right)\) respectively.

Steps to Solve the Problem Based on Coplanarity

Following are a few steps that help in solving the problems based on coplanarity.
Step 1: Initialize a \(3 \times 3\) matrix to store the elements of the determinant shown above.
Step 2: Calculate the cross product of \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) and the dot product of \(\left( {\overrightarrow {{a_2}} – \overrightarrow {{a_1}} } \right)\).
Step 3: If the value of the determinant is \(0\), the lines are coplanar. Otherwise, they are non-coplanar.

Points to Remember

• If two lines are present in the same plane, then the two lines are said to be coplanar.
• In mathematics, coplanarity is described as a condition in which the given number of lines lie on the same plane, thus, they are said to be coplanar.
• Coplanarity exists between any two locations.
• There are many real-world examples of coplanar lines such as grids on graph paper, lines of a notepad, etc.

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Real-World Examples of Coplanar Lines

Following are few examples that explain the coplanar lines

1. Graph paper is where you’ll find the grids. The grid’s vertical and horizontal lines are coplanar points since they are both on the same sheet of paper.
2. Coplanarity is also present in the hands-on watches and clocks. All hands for the second, minute, and hour are in the same circular space.
3. The lines of a notepad are parallel to one another. They are on the same plane because they are on the same page.

Solved Examples – Coplanarity of Two Lines

Below are a few solved examples that can help in getting a better idea.

Q.1. Prove that the lines \(\frac{{y + 3}}{5} = \frac{{z + 5}}{7}\) and \(\frac{{x – 2}}{1} = \frac{{y – 4}}{4} = \frac{{z – 6}}{7}\) are coplanar.
Ans: We know that the lines \(\frac{{x – {x_1}}}{{{a_1}}} = \frac{{y – {y_1}}}{{{b_1}}} = \frac{{z – {z_1}}}{{{c_1}}}\) and \(\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}\) are coplanar if
\(\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0\)
\(\therefore \left| {\begin{array}{*{20}{c}} 3&7&{11} \\ 3&5&7 \\ 1&4&7 \end{array}} \right| = 3\left( {35 – 28} \right) – 7\left( {21 – 7} \right) + 11\left( {12 – 5} \right)\)
\( = 21 – 98 + 77\)
\( = 0\)
So, the given lines are coplanar.

Q.2. Prove that the lines \(\frac{{x – 2}}{1} = \frac{{y – 5}}{2} = \frac{{z – 4}}{4}\) and \(\frac{{x – 6}}{{ – 1}} = \frac{{y – 1}}{3} = \frac{{z – 5}}{4}\) are coplanar or not.
Ans: We know that the lines \(\frac{{x – {x_1}}}{{{a_1}}} = \frac{{y – {y_1}}}{{{b_1}}} = \frac{{z – {z_1}}}{{{c_1}}}\) and \(\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}\) are coplanar if
\(\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0\)
\(\therefore \left| {\begin{array}{*{20}{c}} 4&{ – 4}&1 \\ 1&2&4 \\ { – 1}&3&4 \end{array}} \right| = 4\left( {8 – 12} \right) + 4\left( {4 + 4} \right) + 1\left( {3 + 2} \right)\)
\( = 21\)
\( \ne 0\)
Hence, the given lines are not coplanar.

Q.3. Show that the 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.
Ans: We know that the lines \(\frac{{x – {x_1}}}{{{a_1}}} = \frac{{y – {y_1}}}{{{b_1}}} = \frac{{z – {z_1}}}{{{c_1}}}\) and \(\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}\) are coplanar if
\(\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0\)
\(\therefore \left| {\begin{array}{*{20}{c}} 2&0&{ – 10} \\ { – 3}&4&5 \\ { – 3}&3&6 \end{array}} \right| = 2\left( {24 – 15} \right) – 0 – 10\left( { – 9 + 12} \right)\)
\( = – 12\)
\( \ne 0\)
So, the given lines are not coplanar.

Q.4. Show that the lines \(\frac{{x + 3}}{{ – 3}} = \frac{{y – 1}}{1} = \frac{{z – 5}}{5}\) and \(\frac{{x + 1}}{{ – 1}} = \frac{{y – 2}}{2} = \frac{{z – 5}}{5}\) are coplanar.
Ans: We know that the lines \(\frac{{x – {x_1}}}{{{a_1}}} = \frac{{y – {y_1}}}{{{b_1}}} = \frac{{z – {z_1}}}{{{c_1}}}\) and \(\frac{{x – {x_2}}}{{{a_2}}} = \frac{{y – {y_2}}}{{{b_2}}} = \frac{{z – {z_2}}}{{{c_2}}}\) are coplanar if
\(\left| {\begin{array}{*{20}{c}} {{x_2} – {x_1}}&{{y_2} – {y_1}}&{{z_2} – {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0\)
\(\therefore \left| {\begin{array}{*{20}{c}} 2&1&0 \\ { – 3}&1&5 \\ { – 1}&2&5 \end{array}} \right| = 2\left( {5 – 10} \right) – 1\left( { – 15 + 5} \right) + 0\)
\( – 10 + 10\)
\( = 0\)
So, the given lines are coplanar.

Q.5 Find the condition for the vector equations \(\overrightarrow {r_1} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \) and \(\overrightarrow {r_2} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) are coplanar.
Ans: The given lines are coplanar if the normal to the plane containing these lines are perpendicular to both of them. Since the given lines are parallel to the vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) the normal to the plane is parallel to \(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}}\ \) which is perpendicular to the line joining the points with position vectors \(\overrightarrow {a_1} \) and \(\overrightarrow {a_2} \).
\(\left( {\overrightarrow {a_2} – \overrightarrow {a_1}} \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\)

Summary

Lines lie on the same plane are said to be coplanar, while lines that do not lie on the same plane are called non-coplanar. Two and three points are always coplanar but four or more points will be coplanar only when they all lies on the same plane. Two lines \(\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \) and \(\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) are said to coplanar if \(\left( {{{\vec a}_2} – {{\vec a}_1}} \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\) Where \(\left( {{x_1},\,{y_1},\,\,{z_1}} \right)\) and \(\left( {{x_2},\,{y_2},\,\,{z_2}} \right)\) be the coordinates of points \(A\) and \(B\) respectively and \({a_1},\,\,{b_1},\,\,{c_1}\) and \({a_2},\,\,{b_2},\,\,{c_2}\) be the direction ratios of  \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) respectively. In cartesian form two lines are coplanar if

Frequently Asked Questions (FAQs)

Students might be having many questions regarding the Coplanarity of Two Lines. Here are a few commonly asked questions and answers.

Q.1. How do you prove coplanarity between two lines?
Ans: Lines that lie on the same plane are said to be coplanar, while lines that do not lay on the same plane are called non-coplanar. Coplanarity can be defined as the state of having a certain number of lines on the same plane. We may prove that two lines are coplanar in three dimensions by applying the condition in cartesian and vector forms.

Q.2. What is the condition of coplanarity?
Ans: Let’s write the equation in the vector form of two lines whose coplanarity needs to be determined.
\(\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \)
\(\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \)
Then the given lines are coplanar if, 
\(\left( {{{\vec a}_2} – {{\vec a}_1}} \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\)
Here the cross product of vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) will give a vector line that will be perpendicular to both \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) vector lines \(\overrightarrow {AB}\) is a line vector joining the position vectors \(\overrightarrow {{a_1}} \) and \(\overrightarrow {{a_2}} \) of two lines. Now, we have to check whether two lines are coplanar or not by determining above dot product is zero or not.

Q.3. What is an example of coplanar lines?
Ans: The coplanar lines are the lines which lies in the same plane. For example:
1. A notebook’s lines are parallel to each other and lying in the same plane so, they are coplanar.
2. On the surface of the planet, there are roads which are coplanar. They’re on the same plane because until they’re lifted or dropped to another plane, they’ll have to intersect.

Q.4. What is the formula of coplanar vectors?
Ans: Formula of coplanar vectors in cartesian form:
Let \(\left( {{x_1},\,\,{y_1},\,\,{z_1}} \right)\) and \(\left( {{x_2},\,\,{y_2},\,\,{z_2}} \right)\) be the coordinates of points \(A\) and \(B\) respectively. 
Let \({a_1},\,{b_1},\,{c_1}\) and \({a_2},\,{b_2},\,{c_2}\) be the direction ratios of vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) and respectively.
Then The given lines are coplanar if  

Formula of coplanar vectors in vector form:
Let’s write the equation in the vector form of two lines whose coplanarity needs to be determined.
\(\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \)
\(\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \)
Then the given lines are coplanar if 
\(\left( {{{\vec a}_2} – {{\vec a}_1}} \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = 0\)
Here the cross product of vectors \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) will give a vector line that will be perpendicular to both \(\overrightarrow {{b_1}} \) and \(\overrightarrow {{b_2}} \) vector lines. 

Q.5. Are any two points coplanar?
Ans: Any two or three points are always coplanar and four or more points are said to be coplanar if they all are present in the same plane. Similarly two or more lines are said to be coplanar if they all are present in one plane.

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