PGDDSA Study · Semester 1

Core Titles

Key headlines and terms for quick recall

Distance between two points: $\sqrt{\sum (x_i - y_i)^2}$
Point to plane: $\dfrac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$
Point to line: $\dfrac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}$
Distance between parallel lines / parallel planes
Skew lines — shortest distance via $\dfrac{|(\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{\|\vec{d}_1 \times \vec{d}_2\|}$

Basic Idea

What it is, why it matters, how it works

Two-point distance

$d(A, B) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Point to plane

For plane $ax + by + cz + d = 0$ and point $P_0(x_0, y_0, z_0)$ : $d = \frac{|a x_0 + b y_0 + c z_0 + d|}{\sqrt{a^2 + b^2 + c^2}}.$ Idea: project $\vec{r}_0 - \vec{p}$ (where $\vec{p}$ is any point on the plane) onto the normal $\vec{n}$ .

Point to line

Line through $A$ with direction $\vec{d}$ . For point $P$ : $d(P, L) = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}.$ Idea: the cross product's magnitude is the area of the parallelogram with sides $\vec{AP}$ and $\vec{d}$ ; dividing by $\|\vec{d}\|$ gives the perpendicular height.

Two parallel planes

Take the constant difference; for $ax+by+cz = d_1$ and $ax+by+cz = d_2$ : $d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}.$

Skew lines (non-parallel, non-intersecting)

For lines $\vec{r} = \vec{r}_1 + t \vec{d}_1$ and $\vec{r} = \vec{r}_2 + s \vec{d}_2$ : $d = \frac{|(\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{\|\vec{d}_1 \times \vec{d}_2\|}.$ Idea: $\vec{d}_1 \times \vec{d}_2$ is perpendicular to both lines, so the projection of the joining vector $\vec{r}_2 - \vec{r}_1$ onto this normal is the shortest distance.

Why this matters in Data Science

Distance-to-hyperplane is central to SVM margin and any classification confidence measure. Anomaly detection often computes distance to subspaces.

Mind Map

Visual structure of the concept

DISTANCES IN SPACE
├── Point ↔ Point: √Σ(xᵢ − yᵢ)²
├── Point ↔ Plane:  |ax₀+by₀+cz₀+d| / √(a²+b²+c²)
├── Point ↔ Line:   ‖AP × d‖ / ‖d‖
├── Parallel planes:  |d₁ − d₂| / ‖n‖
└── Skew lines:  |(r₂−r₁)·(d₁×d₂)| / ‖d₁×d₂‖

Exam Q&A

Part A (2 marks) and Part B (20 marks) style questions

Part A (2 marks each)

Q1. Distance between points $(1,2,3)$ and $(4,6,3)$ . $\sqrt{9 + 16 + 0} = 5$ .

Q2. Write the formula for distance from point to plane. $d = \dfrac{|a x_0 + b y_0 + c z_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ .

Q3. Find the distance from $(1,1,1)$ to the plane $2x + y - 2z + 3 = 0$ . $d = \dfrac{|2 + 1 - 2 + 3|}{\sqrt{4 + 1 + 4}} = \dfrac{4}{3}$ .

Q4. What are skew lines? Two lines in 3D that are neither parallel nor intersecting.

Part B (20 marks)

Q. Derive the formula for the perpendicular distance from a point to a plane. Find the shortest distance between the skew lines $\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + t(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + s(2\hat{i} + 3\hat{j} + 6\hat{k})$ . (Note: the two lines are parallel — adjust the example.)

We treat the second line as $\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + s(\hat{i} + 2\hat{j} + 3\hat{k})$ so that the lines are truly skew.

Point-to-plane distance — derivation.

Let the plane be $\Pi : ax + by + cz + d = 0$ , with normal $\vec{n} = (a, b, c)$ . Let $P_0 = (x_0, y_0, z_0)$ and let $Q$ be the foot of perpendicular from $P_0$ on $\Pi$ .

Then $\vec{P_0 Q} = \lambda \vec{n}$ for some scalar $\lambda$ , so $Q = P_0 + \lambda \vec{n}$ . Since $Q$ lies on $\Pi$ : $a(x_0 + \lambda a) + b(y_0 + \lambda b) + c(z_0 + \lambda c) + d = 0$ $\Rightarrow (a x_0 + b y_0 + c z_0 + d) + \lambda (a^2 + b^2 + c^2) = 0$ $\Rightarrow \lambda = -\dfrac{a x_0 + b y_0 + c z_0 + d}{a^2 + b^2 + c^2}.$

The perpendicular distance is $d(P_0, \Pi) = |\lambda| \|\vec{n}\| = \frac{|a x_0 + b y_0 + c z_0 + d|}{\sqrt{a^2 + b^2 + c^2}}. \quad \blacksquare$

Shortest distance between skew lines.

Lines: $L_1 : \vec{r}_1 = (1, 2, -4) + t(2, 3, 6)$ , $\vec{d}_1 = (2, 3, 6)$ . $L_2 : \vec{r}_2 = (3, 3, -5) + s(1, 2, 3)$ , $\vec{d}_2 = (1, 2, 3)$ .

Step 1 — $\vec{d}_1 \times \vec{d}_2$ . $\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(9 - 12) - \hat{j}(6 - 6) + \hat{k}(4 - 3) = -3\hat{i} + 0\hat{j} + \hat{k}$ .

Magnitude: $\sqrt{9 + 0 + 1} = \sqrt{10}$ .

Step 2 — Joining vector. $\vec{r}_2 - \vec{r}_1 = (2, 1, -1)$ .

Step 3 — Dot with $\vec{d}_1 \times \vec{d}_2$ . $(2)(-3) + (1)(0) + (-1)(1) = -6 + 0 - 1 = -7$ .

Step 4 — Shortest distance. $d = \frac{|-7|}{\sqrt{10}} = \frac{7}{\sqrt{10}} = \frac{7\sqrt{10}}{10}.$

Distances in Space