📘 Vector Equation of a Line and a Plane — The Language of 3D Geometry

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

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🔹 Introduction

In 3D geometry, lines and planes are the building blocks of space.
Their equations in vector form provide an elegant, coordinate-free way to represent direction, position, and geometry.

Every line is defined by a point and a direction,
and every plane by a point and a normal.

Once you understand their vector equations, all 3D concepts — intersection, distance, and angle — become simple logical extensions.

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🧭 1️⃣ Vector Equation of a Line

A line in 3D space can be defined if we know:

• A point A through which it passes (position vector a)
• A direction vector b that gives its orientation

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🔹 Vector Form

r=a+λb\boxed{\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}}r=a+λb​

where:

• r\mathbf{r}r → position vector of any general point P(x,y,z)P(x, y, z)P(x,y,z) on the line
• a\mathbf{a}a → position vector of a fixed point on the line
• b\mathbf{b}b → direction vector of the line
• λ∈R\lambda \in \mathbb{R}λ∈R → parameter that varies along the line

✅ As λ changes, r\mathbf{r}r traces the entire line.

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🔹 Geometric Meaning

(Illustrate a line passing through A with direction vector b and a variable point P on it.)

The line passes through A and moves along b.
When λ = 0 → point = A; when λ = 1 → point = A + b, etc.

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🔹 Cartesian Form

If a=(x1,y1,z1)\mathbf{a} = (x_1, y_1, z_1)a=(x1​,y1​,z1​) and b=(l,m,n)\mathbf{b} = (l, m, n)b=(l,m,n): r=(x1i+y1j+z1k)+λ(li+mj+nk)\mathbf{r} = (x_1\mathbf{i} + y_1\mathbf{j} + z_1\mathbf{k}) + \lambda(l\mathbf{i} + m\mathbf{j} + n\mathbf{k})r=(x1​i+y1​j+z1​k)+λ(li+mj+nk) ⇒x−x1l=y−y1m=z−z1n\Rightarrow \frac{x – x_1}{l} = \frac{y – y_1}{m} = \frac{z – z_1}{n}⇒lx−x1​​=my−y1​​=nz−z1​​

✅ This is the symmetric (Cartesian) form of a line.

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🔹 Example 1 — Equation of a Line

Find the vector and Cartesian equation of the line passing through A(2,−1,3)A(2, -1, 3)A(2,−1,3) and parallel to vector 3i−2j+6k3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k}3i−2j+6k.

Solution: a=2i−j+3k,b=3i−2j+6k\mathbf{a} = 2\mathbf{i} – \mathbf{j} + 3\mathbf{k}, \quad \mathbf{b} = 3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k}a=2i−j+3k,b=3i−2j+6k r=(2i−j+3k)+λ(3i−2j+6k)\boxed{\mathbf{r} = (2\mathbf{i} – \mathbf{j} + 3\mathbf{k}) + \lambda(3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k})}r=(2i−j+3k)+λ(3i−2j+6k)​

Cartesian form: x−23=y+1−2=z−36\frac{x – 2}{3} = \frac{y + 1}{-2} = \frac{z – 3}{6}3x−2​=−2y+1​=6z−3​

✅ Equation represents a line through (2, −1, 3) parallel to (3, −2, 6).

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🧮 2️⃣ Vector Equation of a Plane

A plane in space can be defined by:

• A point A with position vector a\mathbf{a}a, and
• A normal vector n=(a,b,c)\mathbf{n} = (a, b, c)n=(a,b,c), perpendicular to the plane.

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🔹 Vector Form

r⋅n=a⋅n\boxed{\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}}r⋅n=a⋅n​

where:

• r=xi+yj+zk\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}r=xi+yj+zk → position vector of a general point P(x,y,z)P(x, y, z)P(x,y,z) on the plane
• a\mathbf{a}a → position vector of known point A(x1,y1,z1)A(x_1, y_1, z_1)A(x1​,y1​,z1​)
• n\mathbf{n}n → normal vector to plane

✅ The dot product ensures that r−a\mathbf{r} – \mathbf{a}r−a is perpendicular to the normal vector.

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🔹 Cartesian Form

If n=(a,b,c)\mathbf{n} = (a, b, c)n=(a,b,c), and a=(x1,y1,z1)\mathbf{a} = (x_1, y_1, z_1)a=(x1​,y1​,z1​): a(x−x1)+b(y−y1)+c(z−z1)=0a(x – x_1) + b(y – y_1) + c(z – z_1) = 0a(x−x1​)+b(y−y1​)+c(z−z1​)=0

Simplify: ax+by+cz+d=0,where d=−(ax1+by1+cz1)\boxed{ax + by + cz + d = 0}, \quad \text{where } d = -(ax_1 + by_1 + cz_1)ax+by+cz+d=0​,where d=−(ax1​+by1​+cz1​)

✅ This is the Cartesian form of the plane equation.

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🔹 Example 2 — Equation of a Plane

Find the equation of a plane passing through (2,3,4)(2, 3, 4)(2,3,4) and perpendicular to 3i−2j+6k3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k}3i−2j+6k. a=2i+3j+4k,n=3i−2j+6k\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}, \quad \mathbf{n} = 3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k}a=2i+3j+4k,n=3i−2j+6k

Vector form: r⋅(3i−2j+6k)=(2i+3j+4k)⋅(3i−2j+6k)\mathbf{r} \cdot (3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k}) = (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \cdot (3\mathbf{i} – 2\mathbf{j} + 6\mathbf{k})r⋅(3i−2j+6k)=(2i+3j+4k)⋅(3i−2j+6k) ⇒r⋅(3,−2,6)=24\Rightarrow \mathbf{r} \cdot (3, -2, 6) = 24⇒r⋅(3,−2,6)=24

Cartesian form: 3x−2y+6z=243x – 2y + 6z = 243x−2y+6z=24

✅ Plane Equation: 3x−2y+6z−24=03x – 2y + 6z – 24 = 03x−2y+6z−24=0

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🔹 Example 3 — Plane Through 3 Points

Find equation of the plane passing through
A(1,2,3),B(2,−1,1),C(3,1,4)A(1, 2, 3), B(2, -1, 1), C(3, 1, 4)A(1,2,3),B(2,−1,1),C(3,1,4).

1️⃣ Find vectors AB and AC: AB=(1,−3,−2),AC=(2,−1,1)\mathbf{AB} = (1, -3, -2), \quad \mathbf{AC} = (2, -1, 1)AB=(1,−3,−2),AC=(2,−1,1)

2️⃣ Normal to plane = AB×AC\mathbf{AB} \times \mathbf{AC}AB×AC n=∣ijk1−3−22−11∣=(−5,−4,5)\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & -2 \\ 2 & -1 & 1 \end{vmatrix} = (-5, -4, 5)n=​i12​j−3−1​k−21​​=(−5,−4,5)

3️⃣ Equation: r⋅(−5i−4j+5k)=(1,2,3)⋅(−5,−4,5)\mathbf{r} \cdot (-5\mathbf{i} – 4\mathbf{j} + 5\mathbf{k}) = (1, 2, 3) \cdot (-5, -4, 5)r⋅(−5i−4j+5k)=(1,2,3)⋅(−5,−4,5) ⇒−5x−4y+5z=−5−8+15=2\Rightarrow -5x – 4y + 5z = -5 – 8 + 15 = 2⇒−5x−4y+5z=−5−8+15=2

✅ Simplified Cartesian form: 5x+4y−5z+2=05x + 4y – 5z + 2 = 05x+4y−5z+2=0

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🎯 3️⃣ Relation Between Line and Plane Equations

Concept	Line	Plane
Defining Info	A point + direction	A point + normal
Vector Equation	r=a+λb\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}r=a+λb	r⋅n=a⋅n\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}r⋅n=a⋅n
Cartesian Form	x−x1l=y−y1m=z−z1n\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}lx−x1​​=my−y1​​=nz−z1​​	ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0
Orientation Vector	Direction vector b	Normal vector n
Degrees of Freedom	1 (line)	2 (plane)

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🔹 Common Mistakes

1. ❌ Confusing normal and direction vectors.
2. ❌ Forgetting that plane’s vector form uses dot product.
3. ❌ Missing parameter λ in line equation.
4. ❌ Wrong sign of constant d in Cartesian plane equation.

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🌟 Why It Matters

Understanding vector equations lets you:

• Find intersections of lines and planes,
• Compute angles and distances,
• Model 3D geometry in mechanics and physics,
• Solve IB HL & A Level P4 vector geometry problems intuitively.

The power of vector equations lies in their simplicity —
one formula represents infinitely many points and directions.

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📘 Learn Beyond the Formula

At Math By Rishabh, we don’t just memorize equations — we build 3D reasoning skills.

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