Real-World Hook: Architecture and Design
When architects design a flat roof, all the support beams must be coplanar (lie in the same plane). If even one beam is out of plane, the structure becomes unstable! Similarly, in circuit board design, certain traces must be coplanar for proper electrical contact. Understanding coplanarity is crucial in engineering and design.
What is Coplanarity?
Definition
Points, lines, or vectors are coplanar if they all lie in the same plane.
Types of Coplanarity Problems
- Four points coplanar
- Two lines coplanar
- Three vectors coplanar
- Line and point coplanar with another line
Condition for Four Points to be Coplanar
Method 1: Using Determinant
Four points A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃), D(x₄, y₄, z₄) are coplanar if:
$$\boxed{\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{vmatrix} = 0}$$Interactive Demo: Visualize Coplanar Points
Explore how four points can lie in the same plane.
Method 2: Using Vectors (Scalar Triple Product)
If A, B, C, D are four points, then they are coplanar if:
$$\boxed{[\vec{AB} \quad \vec{AC} \quad \vec{AD}] = 0}$$where [u v w] denotes scalar triple product = u · (v × w)
Equivalently:
$$\boxed{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \\ x_4-x_1 & y_4-y_1 & z_4-z_1 \end{vmatrix} = 0}$$Memory Trick 🧠
“Determinant of Differences Must Vanish”
- Form vectors from one point (say A) to the other three points
- Put these vectors as rows in a determinant
- If determinant = 0, points are coplanar
- Geometric meaning: The parallelepiped has zero volume!
Why This Works?
The scalar triple product AB · (AC × AD) represents the volume of parallelepiped formed by vectors AB, AC, AD.
If volume = 0, the parallelepiped is flat (degenerate) → all four points lie in the same plane!
Condition for Two Lines to be Coplanar
Definition
Two lines are coplanar if they:
- Intersect, OR
- Are parallel
If they’re neither (skew lines), they’re NOT coplanar.
Condition (Vector Form)
Lines r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar if:
$$\boxed{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0}$$Or equivalently:
$$\boxed{[\vec{a_2} - \vec{a_1} \quad \vec{b_1} \quad \vec{b_2}] = 0}$$Condition (Cartesian Form)
Lines through (x₁, y₁, z₁) with DR (a₁, b₁, c₁) and through (x₂, y₂, z₂) with DR (a₂, b₂, c₂) are coplanar if:
$$\boxed{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0}$$Memory Trick 🧠
“Same as Skew Lines Formula Numerator!”
- This is exactly the numerator of the shortest distance formula
- For coplanar lines: Shortest distance = 0
- Therefore, numerator must be 0!
Geometric Interpretation
Four Points Coplanar
Three vectors AB, AC, AD starting from point A are coplanar if they form a degenerate parallelepiped (zero volume).
This happens when one vector is a linear combination of the other two:
$$\vec{AD} = \alpha\vec{AB} + \beta\vec{AC}$$Two Lines Coplanar
The vector joining a point on line 1 to a point on line 2 (a₂ - a₁) must be perpendicular to the cross product of direction vectors (b₁ × b₂).
Why? Because b₁ × b₂ is perpendicular to the plane containing the lines. If a₂ - a₁ is also perpendicular to this, it lies in the plane!
Special Cases
Case 1: Parallel Lines
If lines are parallel: b₁ × b₂ = 0
The coplanarity condition becomes 0 = 0 (always satisfied).
Conclusion: Parallel lines are always coplanar.
Case 2: Intersecting Lines
If lines intersect: There exists a common point, so (a₂ - a₁) can be written as a linear combination of b₁ and b₂.
Therefore, (a₂ - a₁) · (b₁ × b₂) = 0.
Conclusion: Intersecting lines are always coplanar.
Case 3: Skew Lines
If the determinant ≠ 0, lines are skew (not coplanar).
Equation of Plane Containing Coplanar Lines
Method
If two lines are coplanar and we need to find the plane containing them:
Step 1: Verify coplanarity (determinant = 0)
Step 2: Find normal to plane: n = b₁ × b₂
Step 3: Use point-normal form with any point from either line:
$$\vec{n} \cdot (\vec{r} - \vec{a_1}) = 0$$Or in Cartesian form:
$$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$where (a, b, c) are components of n.
Worked Examples
Example 1: Four Points Coplanar (JEE Main Level)
Problem: Show that points A(1, 2, 3), B(2, 3, 1), C(3, 1, 2), D(0, 4, 4) are coplanar.
Solution: Using vector method:
AB = (2-1, 3-2, 1-3) = (1, 1, -2) AC = (3-1, 1-2, 2-3) = (2, -1, -1) AD = (0-1, 4-2, 4-3) = (-1, 2, 1)
Scalar triple product:
$$[\vec{AB} \quad \vec{AC} \quad \vec{AD}] = \begin{vmatrix} 1 & 1 & -2 \\ 2 & -1 & -1 \\ -1 & 2 & 1 \end{vmatrix}$$Expanding along R1:
$$= 1(-1-(-2)) - 1(2-1) + (-2)(4-1)$$ $$= 1(1) - 1(1) + (-2)(3)$$ $$= 1 - 1 - 6 = -6$$Wait, this is ≠ 0, so points are NOT coplanar!
Let me recalculate:
$$= 1(-1+2) - 1(2+1) + (-2)(4+1)$$ $$= 1(1) - 1(3) - 2(5)$$ $$= 1 - 3 - 10 = -12 \neq 0$$Conclusion: Points are NOT coplanar (the problem statement was incorrect).
Example 2: Two Lines Coplanar (JEE Main Level)
Problem: Show that lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5 are coplanar.
Solution: Line 1: Point (1, 2, 3), DR (2, 3, 4) Line 2: Point (2, 3, 4), DR (3, 4, 5)
Check coplanarity:
$$\begin{vmatrix} 2-1 & 3-2 & 4-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$Expanding along R1:
$$= 1(15-16) - 1(10-12) + 1(8-9)$$ $$= 1(-1) - 1(-2) + 1(-1)$$ $$= -1 + 2 - 1 = 0$$Since determinant = 0, lines are coplanar.
Example 3: Finding Value of k (JEE Main Level)
Problem: Find k if the lines (x-1)/1 = (y-2)/2 = (z-3)/3 and (x-2)/1 = (y-k)/2 = (z-4)/3 are coplanar.
Solution: Line 1: (1, 2, 3), DR (1, 2, 3) Line 2: (2, k, 4), DR (1, 2, 3)
Direction ratios are proportional (1/1 = 2/2 = 3/3), so lines are parallel.
Parallel lines are always coplanar, so k can be any value.
Wait - but the question asks for a specific k. Let me reconsider.
Actually, if they’re asking for coplanarity, they want the lines to be in the same plane. For parallel lines to be in the same plane, the vector joining points must be perpendicular to the common direction.
Let me check: If lines have the same direction ratios, they’re parallel and always coplanar regardless of k.
But let me verify differently. The question might have different DR for line 2.
Actually, looking again, DR for both are (1, 2, 3) - identical! So they’re parallel.
For any k, the lines are coplanar (parallel lines are always coplanar).
Answer: k can be any real value (or the problem might need correction).
Example 4: Equation of Plane (JEE Advanced Level)
Problem: Find equation of plane containing lines:
$$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3} \quad \text{and} \quad \frac{x-2}{2} = \frac{y-3}{3} = \frac{z-1}{1}$$Solution: Step 1: Verify coplanarity
Line 1: (1, 2, 3), DR (1, 2, 3) Line 2: (2, 3, 1), DR (2, 3, 1)
$$\begin{vmatrix} 2-1 & 3-2 & 1-3 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 1 & -2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$$ $$= 1(2-9) - 1(1-6) + (-2)(3-4)$$ $$= 1(-7) - 1(-5) + (-2)(-1)$$ $$= -7 + 5 + 2 = 0$$Lines are coplanar ✓
Step 2: Find normal to plane
$$\vec{n} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}$$ $$= \hat{i}(2-9) - \hat{j}(1-6) + \hat{k}(3-4)$$ $$= -7\hat{i} + 5\hat{j} - \hat{k}$$Normal: (-7, 5, -1) or (7, -5, 1)
Step 3: Plane equation using point (1, 2, 3):
$$7(x-1) - 5(y-2) + 1(z-3) = 0$$ $$7x - 7 - 5y + 10 + z - 3 = 0$$ $$\boxed{7x - 5y + z = 0}$$Verification: Substitute both points:
- Point (1,2,3): 7(1) - 5(2) + 3 = 7 - 10 + 3 = 0 ✓
- Point (2,3,1): 7(2) - 5(3) + 1 = 14 - 15 + 1 = 0 ✓
Example 5: Using Coplanarity for Unknown (JEE Advanced Level)
Problem: Find the value of k for which the lines:
$$\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{k} \quad \text{and} \quad \frac{x-2}{k} = \frac{y-3}{1} = \frac{z-4}{2}$$are coplanar.
Solution: Line 1: (3, 4, 5), DR (1, 2, k) Line 2: (2, 3, 4), DR (k, 1, 2)
Coplanarity condition:
$$\begin{vmatrix} 2-3 & 3-4 & 4-5 \\ 1 & 2 & k \\ k & 1 & 2 \end{vmatrix} = 0$$ $$\begin{vmatrix} -1 & -1 & -1 \\ 1 & 2 & k \\ k & 1 & 2 \end{vmatrix} = 0$$Expanding along R1:
$$-1(4-k) - (-1)(2-k^2) + (-1)(1-2k) = 0$$ $$-4 + k + 2 - k^2 - 1 + 2k = 0$$ $$-k^2 + 3k - 3 = 0$$ $$k^2 - 3k + 3 = 0$$Using quadratic formula:
$$k = \frac{3 \pm \sqrt{9-12}}{2} = \frac{3 \pm \sqrt{-3}}{2}$$This gives complex values, which means no real value of k makes the lines coplanar!
Wait, let me recalculate the determinant:
$$-1(4-k) + 1(2-k^2) - 1(1-2k) = 0$$ $$-4 + k + 2 - k^2 - 1 + 2k = 0$$ $$-k^2 + 3k - 3 = 0$$ $$k^2 - 3k + 3 = 0$$Discriminant = 9 - 12 = -3 < 0
Conclusion: No real value of k exists for which lines are coplanar.
Practice Problems
Level 1: JEE Main Basics
Show that points (0, 0, 0), (1, 2, 3), (2, 4, 6), (3, 6, 9) are coplanar.
Check if lines (x-1)/1 = (y-1)/1 = (z-1)/1 and (x-2)/2 = (y-2)/2 = (z-2)/2 are coplanar.
Find if points A(1, 2, 3), B(2, 3, 4), C(3, 4, 5), D(4, 5, 6) lie in the same plane.
Level 2: JEE Main Standard
Prove that the lines r = (2i + 3j + 4k) + λ(i + 2j + 3k) and r = (3i + 5j + 7k) + μ(2i + 4j + 6k) are coplanar.
Find the equation of plane containing lines:
- x/1 = y/2 = z/3
- (x-1)/2 = (y-2)/3 = (z-3)/4
Find k if points (1, 2, 3), (2, k, 5), (3, 4, 6), (4, 5, k) are coplanar.
Level 3: JEE Advanced
Show that the four points with position vectors a, b, c, d are coplanar if and only if:
$$[\vec{b} - \vec{a} \quad \vec{c} - \vec{a} \quad \vec{d} - \vec{a}] = 0$$Find the condition that the lines:
$$\frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1} \quad \text{and} \quad \frac{x-x_2}{l_2} = \frac{y-y_2}{m_2} = \frac{z-z_2}{n_2}$$should be coplanar.
If the lines r = a₁ + λb and r = a₂ + μc are coplanar, prove that (a₂ - a₁) · (b × c) = 0.
Find the equation of plane through the line (x-1)/1 = (y-2)/2 = (z-3)/3 and parallel to line (x-2)/2 = (y-3)/3 = (z-4)/4.
Quick Reference Formulas
| Concept | Condition |
|---|---|
| 4 points coplanar | [AB AC AD] = 0 |
| 2 lines coplanar | [a₂-a₁ b₁ b₂] = 0 |
| 3 vectors coplanar | [u v w] = 0 |
| Normal to plane (2 lines) | n = b₁ × b₂ |
Connection to Other Concepts
Coplanarity ↔ Shortest Distance
- Lines are coplanar ⟺ Shortest distance = 0
- Lines are skew ⟺ Shortest distance > 0
Coplanarity ↔ Volume
- Four points coplanar ⟺ Tetrahedron volume = 0
- Three vectors coplanar ⟺ Parallelepiped volume = 0
Algorithm: Checking Coplanarity
For Four Points:
- Form three vectors from one point to the other three
- Calculate scalar triple product (determinant)
- If = 0, coplanar; otherwise, not coplanar
For Two Lines:
- Extract points (x₁,y₁,z₁) and (x₂,y₂,z₂)
- Extract DRs (a₁,b₁,c₁) and (a₂,b₂,c₂)
- Form determinant with (x₂-x₁, y₂-y₁, z₂-z₁) in R1
- If = 0, coplanar; otherwise, skew
Special Check:
- If DRs proportional → Parallel → Always coplanar
Cross-Links
- Previous Topic: Skew Lines - Non-coplanar lines
- Related: Plane Equations - Equation of plane containing lines
- Foundation: Scalar Triple Product - Key to coplanarity
- Foundation: Determinants - Calculation method
- Related: 3D Coordinates - Basic setup
Common Exam Patterns
JEE Main typically asks:
- Verifying coplanarity of four points (2-3 marks)
- Checking if two lines are coplanar (3 marks)
- Finding unknown parameter using coplanarity (3-4 marks)
JEE Advanced patterns:
- Finding equation of plane containing coplanar lines
- Proving general coplanarity theorems
- Combined with vector algebra
- Locus problems involving coplanarity
Pro Tip:
- Always check for parallel lines first - automatic coplanarity!
- Coplanarity determinant is same as SD numerator
- For four points, any three can be chosen as base vectors
Last updated: November 2025