Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: 3D Geometry | Vectors & Planes

🔹 Introduction

In 3D geometry, two planes can:
1️⃣ Be parallel,
2️⃣ Coincide, or
3️⃣ Intersect in a line.

That line of intersection is a crucial concept — it represents all points that lie on both planes simultaneously.

Two planes meet where their equations are simultaneously true.
The line of intersection = the common solution set of both planes.

⚡️ 1️⃣ Equation of a Plane

A plane in 3D space can be written as: a1x+b1y+c1z+d1=0a_1x + b_1y + c_1z + d_1 = 0a1​x+b1​y+c1​z+d1​=0

where n1=(a1,b1,c1)\mathbf{n_1} = (a_1, b_1, c_1)n1​=(a1​,b1​,c1​) is the normal vector to the plane.

Let’s consider two planes: π1:a1x+b1y+c1z+d1=0\pi_1: a_1x + b_1y + c_1z + d_1 = 0π1​:a1​x+b1​y+c1​z+d1​=0 π2:a2x+b2y+c2z+d2=0\pi_2: a_2x + b_2y + c_2z + d_2 = 0π2​:a2​x+b2​y+c2​z+d2​=0

🎯 2️⃣ Direction Vector of Line of Intersection

The direction vector of the line is perpendicular to both normals, n1\mathbf{n_1}n1​ and n2\mathbf{n_2}n2​.

Hence, d=n1×n2\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}d=n1​×n2​

✅ This gives the direction of the line (parallel to both planes but perpendicular to both normals).

🔹 Step-by-Step Example

Find the line of intersection of the planes: π1:x+2y+3z=4\pi_1: x + 2y + 3z = 4π1​:x+2y+3z=4 π2:2x−y+z=5\pi_2: 2x – y + z = 5π2​:2x−y+z=5

Step 1️⃣: Find Normal Vectors

n1=(1,2,3),n2=(2,−1,1)\mathbf{n_1} = (1, 2, 3), \quad \mathbf{n_2} = (2, -1, 1)n1​=(1,2,3),n2​=(2,−1,1)

Step 2️⃣: Find Direction Vector of Line

d=n1×n2\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}d=n1​×n2​ d=∣ijk1232−11∣=(2×1−3×(−1))i−(1×1−3×2)j+(1×(−1)−2×2)k\mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{vmatrix} = (2 \times 1 – 3 \times (-1))\mathbf{i} – (1 \times 1 – 3 \times 2)\mathbf{j} + (1 \times (-1) – 2 \times 2)\mathbf{k}d=​i12​j2−1​k31​​=(2×1−3×(−1))i−(1×1−3×2)j+(1×(−1)−2×2)k d=(5)i−(−5)j+(−5)k=(5,5,−5)\mathbf{d} = (5)\mathbf{i} – (-5)\mathbf{j} + (-5)\mathbf{k} = (5, 5, -5)d=(5)i−(−5)j+(−5)k=(5,5,−5)

✅ Direction vector d=(1,1,−1)\mathbf{d} = (1, 1, -1)d=(1,1,−1) (simplified).

Step 3️⃣: Find a Point on the Line

We can find one point that lies on both planes.
Let’s eliminate one variable.

From π1:x+2y+3z=4\pi_1: x + 2y + 3z = 4π1​:x+2y+3z=4
and π2:2x−y+z=5\pi_2: 2x – y + z = 5π2​:2x−y+z=5

Let’s set z=0z = 0z=0 to simplify.

Then: x+2y=4(1)x + 2y = 4 \quad (1)x+2y=4(1) 2x−y=5(2)2x – y = 5 \quad (2)2x−y=5(2)

Multiply (1) by 1 and (2) by 2 to eliminate xxx: x+2y=4x + 2y = 4x+2y=4 4x−2y=104x – 2y = 104x−2y=10

Add → 5x=14⇒x=2.85x = 14 \Rightarrow x = 2.85x=14⇒x=2.8
Substitute back: 2.8+2y=4⇒y=0.62.8 + 2y = 4 \Rightarrow y = 0.62.8+2y=4⇒y=0.6

✅ When z=0z = 0z=0, point P(2.8,0.6,0)P(2.8, 0.6, 0)P(2.8,0.6,0) lies on both planes.

Step 4️⃣: Write Line Equation

Using point P(2.8,0.6,0)P(2.8, 0.6, 0)P(2.8,0.6,0) and direction vector (1,1,−1)(1, 1, -1)(1,1,−1):

Vector Form: r=(2.8i+0.6j)+λ(i+j−k)\mathbf{r} = (2.8\mathbf{i} + 0.6\mathbf{j}) + \lambda(\mathbf{i} + \mathbf{j} – \mathbf{k})r=(2.8i+0.6j)+λ(i+j−k)

Parametric Form: x=2.8+λ,y=0.6+λ,z=−λx = 2.8 + \lambda, \quad y = 0.6 + \lambda, \quad z = -\lambdax=2.8+λ,y=0.6+λ,z=−λ

✅ Final answer: x=2.8+t,y=0.6+t,z=−t\boxed{x = 2.8 + t, \quad y = 0.6 + t, \quad z = -t}x=2.8+t,y=0.6+t,z=−t​

🔹 Alternative (Using Elimination Method)

To find the equation directly:
1️⃣ Multiply or add plane equations to eliminate one variable (say zzz).
2️⃣ Express the remaining two equations as xxx and yyy in terms of a parameter.
3️⃣ Back-substitute to find the third variable.

Both methods lead to the same result.

📘 3️⃣ Special Cases

🔹 Visual Interpretation

Picture two sheets of paper meeting at a straight edge — that edge is the line of intersection.
Mathematically, that line satisfies both plane equations simultaneously.

Each plane restricts space to a 2D surface; their intersection restricts it further to a 1D line.

(Insert visual: two semi-transparent planes intersecting in a line, with direction vector labeled as n1×n2\mathbf{n_1} \times \mathbf{n_2}n1​×n2​).

🔹 Common Mistakes

1. ❌ Forgetting to normalize or simplify the direction vector.
2. ❌ Assuming planes always intersect (check n1×n2\mathbf{n_1} \times \mathbf{n_2}n1​×n2​).
3. ❌ Picking invalid substitution (e.g., dividing by 0 when finding point).
4. ❌ Mixing up point vector and direction vector in vector form.

🌟 Why It Matters

Understanding the intersection of planes is essential in:

• 3D geometry and vector algebra,
• Engineering & computer graphics,
• Physics (force & motion planes),
• Advanced math competitions (STEP/MAT).

It’s also foundational for lines, distances, and planes intersection problems that follow.

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