Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: 3D Geometry | Vectors & Distances
🔹 Introduction
In 3D geometry, measuring distances isn’t as straightforward as in 2D.
We often need to find:
1️⃣ The shortest distance from a point to a line, or
2️⃣ The perpendicular distance from a point to a plane.
Both are perpendicular measurements — the shortest possible path from the point to the geometric object.
In essence, we’re finding how far a point lies “off” the line or plane.
🧭 1️⃣ Distance from a Point to a Line
Given:
A line passing through point A(x1,y1,z1)A(x_1, y_1, z_1)A(x1,y1,z1)
with direction ratios (l,m,n)(l, m, n)(l,m,n).
Equation of line (vector form): r=a+λb\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}r=a+λb
where 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).
A point P(x2,y2,z2)P(x_2, y_2, z_2)P(x2,y2,z2) in space.
Formula
The shortest distance (D) from point PPP to the line is: D=∣ (b)×(AP) ∣∣b∣\boxed{D = \frac{|\ (\mathbf{b}) \times (\mathbf{AP})\ |}{|\mathbf{b}|}}D=∣b∣∣ (b)×(AP) ∣
Or equivalently (in coordinate form): D=∣(rP−rA)×b∣∣b∣D = \frac{|(\mathbf{r_P – r_A}) \times \mathbf{b}|}{|\mathbf{b}|}D=∣b∣∣(rP−rA)×b∣
🔹 Derivation (Conceptual View)
Let AP=rP−rA\mathbf{AP} = \mathbf{r_P} – \mathbf{r_A}AP=rP−rA.
We can split AP\mathbf{AP}AP into two components:
- One along the line (parallel to b\mathbf{b}b),
- One perpendicular to the line (the distance we seek).
That perpendicular component has magnitude: ∣AP∣sinθ=∣AP×b∣∣b∣|\mathbf{AP}| \sin\theta = \frac{|\mathbf{AP} \times \mathbf{b}|}{|\mathbf{b}|}∣AP∣sinθ=∣b∣∣AP×b∣
✅ Hence: D=∣AP×b∣∣b∣\boxed{D = \frac{|\mathbf{AP} \times \mathbf{b}|}{|\mathbf{b}|}}D=∣b∣∣AP×b∣
🔹 Example 1 — Point to Line
Find the perpendicular distance from P(1,2,3)P(1, 2, 3)P(1,2,3) to the line 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
Step 1️⃣: Identify knowns a=(2,−1,3),b=(3,−2,6)\mathbf{a} = (2, -1, 3), \quad \mathbf{b} = (3, -2, 6)a=(2,−1,3),b=(3,−2,6) rP=(1,2,3)\mathbf{r_P} = (1, 2, 3)rP=(1,2,3)
Step 2️⃣: Find AP=rP−a=(−1,3,0)\mathbf{AP} = \mathbf{r_P} – \mathbf{a} = (-1, 3, 0)AP=rP−a=(−1,3,0)
Step 3️⃣: Cross product AP×b=∣ijk−1303−26∣=(18)i+(6)j+(−7)k\mathbf{AP} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 0 \\ 3 & -2 & 6 \end{vmatrix} = (18)\mathbf{i} + (6)\mathbf{j} + (-7)\mathbf{k}AP×b=i−13j3−2k06=(18)i+(6)j+(−7)k ∣AP×b∣=182+62+(−7)2=409|\mathbf{AP} \times \mathbf{b}| = \sqrt{18^2 + 6^2 + (-7)^2} = \sqrt{409}∣AP×b∣=182+62+(−7)2=409 ∣b∣=32+(−2)2+62=7|\mathbf{b}| = \sqrt{3^2 + (-2)^2 + 6^2} = 7∣b∣=32+(−2)2+62=7
✅ Distance: D=4097≈2.89D = \frac{\sqrt{409}}{7} \approx 2.89D=7409≈2.89
🧮 2️⃣ Distance from a Point to a Plane
Given:
Equation of plane: ax+by+cz+d=0ax + by + cz + d = 0ax+by+cz+d=0
Point P(x1,y1,z1)P(x_1, y_1, z_1)P(x1,y1,z1).
Formula
D=∣ax1+by1+cz1+d∣a2+b2+c2\boxed{D = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}}D=a2+b2+c2∣ax1+by1+cz1+d∣
🔹 Derivation (Vector Approach)
Let the plane have normal vector n=(a,b,c)\mathbf{n} = (a, b, c)n=(a,b,c),
and let r0\mathbf{r_0}r0 be a point on the plane.
The shortest distance from PPP to plane = component of rP−r0\mathbf{r_P – r_0}rP−r0 along n\mathbf{n}n: D=∣(rP−r0)⋅n∣∣n∣D = \frac{|(\mathbf{r_P – r_0})\cdot \mathbf{n}|}{|\mathbf{n}|}D=∣n∣∣(rP−r0)⋅n∣
Using plane equation ax0+by0+cz0+d=0a x_0 + b y_0 + c z_0 + d = 0ax0+by0+cz0+d=0,
we simplify to: D=∣ax1+by1+cz1+d∣a2+b2+c2\boxed{D = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}}D=a2+b2+c2∣ax1+by1+cz1+d∣
🔹 Example 2 — Point to Plane
Find the perpendicular distance from P(3,−2,1)P(3, -2, 1)P(3,−2,1) to the plane 2x−3y+6z−4=02x – 3y + 6z – 4 = 02x−3y+6z−4=0 a=2, b=−3, c=6, d=−4a = 2, \ b = -3, \ c = 6, \ d = -4a=2, b=−3, c=6, d=−4 D=∣2(3)−3(−2)+6(1)−4∣22+(−3)2+62=∣6+6+6−4∣49=147=2D = \frac{|2(3) – 3(-2) + 6(1) – 4|}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{|6 + 6 + 6 – 4|}{\sqrt{49}} = \frac{14}{7} = 2D=22+(−3)2+62∣2(3)−3(−2)+6(1)−4∣=49∣6+6+6−4∣=714=2
✅ Distance = 2 units
🔹 Example 3 — Special Case
For plane x+y+z=6x + y + z = 6x+y+z=6 and point P(1,1,1)P(1, 1, 1)P(1,1,1): D=∣1+1+1−6∣12+12+12=43=433D = \frac{|1 + 1 + 1 – 6|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}D=12+12+12∣1+1+1−6∣=34=343
✅ Distance = 433\frac{4\sqrt{3}}{3}343
🧩 3️⃣ Summary of Distance Formulas
| Type | Formula | Vector Form | Key Idea |
|---|---|---|---|
| Point → Line | ( D = \frac{ | (\mathbf{b}) \times (\mathbf{AP}) | }{ |
| Point → Plane | ( D = \frac{ | ax_1 + by_1 + cz_1 + d | }{\sqrt{a^2 + b^2 + c^2}} ) |
🔹 Common Mistakes
- ❌ Forgetting the absolute value — distance is always positive.
- ❌ Using wrong direction vector (for line) or coefficients (for plane).
- ❌ Confusing between line’s direction and normal vectors.
- ❌ Substituting point incorrectly in the plane equation (sign errors).
🌟 Why It Matters
These formulas are central to:
- Shortest distances in geometry,
- Line-plane intersection and angles,
- Optimization (e.g., minimal path problems),
- Physics & Engineering (motion along or perpendicular to a field).
The beauty of 3D geometry is that algebra expresses space precisely.
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✅ Derive geometric results using vectors,
✅ Visualize perpendicular distances and projections,
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