Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Vectors | 3D Geometry
🔹 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.
🧭 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
🔹 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.
🔹 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.
🔹 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=(x1i+y1j+z1k)+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.
🔹 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).
🧮 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.
🔹 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.
🔹 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.
🔹 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
🔹 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=i12j−3−1k−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
🎯 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) |
🔹 Common Mistakes
- ❌ Confusing normal and direction vectors.
- ❌ Forgetting that plane’s vector form uses dot product.
- ❌ Missing parameter λ in line equation.
- ❌ Wrong sign of constant d in Cartesian plane equation.
🌟 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.
📘 Learn Beyond the Formula
At Math By Rishabh, we don’t just memorize equations — we build 3D reasoning skills.
In the Mathematics Elevate Mentorship Program, you’ll:
✅ Derive vector results geometrically,
✅ Visualize 3D objects through algebra,
✅ Solve real IB/A Level/STEP vector problems intuitively.
🚀 Think in 3D — not just calculate in 2D.
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