PGDDSA Study · Semester 1

Core Titles

Key headlines and terms for quick recall

Line vector form $\vec{r} = \vec{r}_0 + t \vec{d}$
Line parametric form $x = x_0 + at,\; y = y_0 + bt,\; z = z_0 + ct$
Line symmetric form $\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c}$
Plane vector form $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$
Plane Cartesian $ax + by + cz = d$
Intersection of two planes is a line (or empty)

Basic Idea

What it is, why it matters, how it works

Line in space

A line is determined by a point $\vec{r}_0 = (x_0, y_0, z_0)$ on it and a direction vector $\vec{d} = (a, b, c)$ .

Vector form $\vec{r}(t) = \vec{r}_0 + t\vec{d}, \quad t \in \mathbb{R}.$

Parametric form $x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct.$

Symmetric form (when $a, b, c$ all non-zero) $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}.$

Line through two points $A, B$

Direction = $\vec{B} - \vec{A}$ . So $\vec{r}(t) = \vec{A} + t(\vec{B} - \vec{A})$ .

Plane in space

Determined by a point $\vec{r}_0$ and a normal vector $\vec{n} = (a, b, c)$ .

Vector form $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$

Cartesian form $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \quad \text{or} \quad ax + by + cz = d.$

Plane through three non-collinear points $A, B, C$

Take $\vec{n} = (\vec{B} - \vec{A}) \times (\vec{C} - \vec{A})$ , then use point-normal form.

Intersections

Two planes with non-parallel normals → a line.
Line and plane: substitute parametric line into plane equation, solve for $t$ .

Why this matters in Data Science

Lines/planes are the geometric backbone of classification (decision boundary), regression (best-fit plane), and dimensionality reduction.

Mind Map

Visual structure of the concept

LINES & PLANES IN SPACE
├── LINE
│   ├── Vector: r = r₀ + td
│   ├── Parametric: (x₀+at, y₀+bt, z₀+ct)
│   ├── Symmetric: (x−x₀)/a = (y−y₀)/b = (z−z₀)/c
│   └── Through A, B: d = B − A
└── PLANE
    ├── Vector: n·(r − r₀) = 0
    ├── Cartesian: ax + by + cz = d
    ├── Normal vector n
    └── Through 3 points: n = (B−A) × (C−A)

Exam Q&A

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

Part A (2 marks each)

Q1. Write the vector equation of a line. $\vec{r}(t) = \vec{r}_0 + t \vec{d}$ , where $\vec{r}_0$ is a point on the line and $\vec{d}$ its direction.

Q2. Write the equation of a plane in vector form. $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ .

Q3. Find the equation of the line through $(1, 0, 2)$ and $(3, 4, -1)$ . Direction $(2, 4, -3)$ . Equation: $\vec{r}(t) = (1, 0, 2) + t(2, 4, -3)$ .

Q4. What is the normal to the plane $2x - 3y + 4z = 5$ ? $\vec{n} = (2, -3, 4)$ .

Part B (20 marks)

Q. Derive the equation of a line and a plane in space. Find the equation of the plane passing through $A(1,1,1)$ , $B(2,3,4)$ , $C(0,1,2)$ , and the equation of the line through the origin perpendicular to this plane.

Equation of a line. Given a point $\vec{r}_0$ and direction $\vec{d}$ , every point on the line is $\vec{r}_0 + t\vec{d}$ for some scalar $t$ . Hence vector form $\vec{r}(t) = \vec{r}_0 + t\vec{d}$ ; in components, parametric equations.

Equation of a plane. Given point $\vec{r}_0$ on the plane and normal $\vec{n}$ , any point $\vec{r}$ on the plane satisfies $(\vec{r} - \vec{r}_0) \perp \vec{n}$ , i.e., $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ .

Plane through $A, B, C$ .

Step 1 — Two edge vectors. $\vec{AB} = (1, 2, 3), \quad \vec{AC} = (-1, 0, 1)$ .

Step 2 — Normal via cross product. $\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 0 & 1 \end{vmatrix} = \hat{i}(2 \cdot 1 - 3 \cdot 0) - \hat{j}(1 \cdot 1 - 3 \cdot -1) + \hat{k}(1 \cdot 0 - 2 \cdot -1)$ $= 2\hat{i} - 4\hat{j} + 2\hat{k} = (2, -4, 2)$ . Simplify: $(1, -2, 1)$ .

Step 3 — Cartesian form. Using point $A(1,1,1)$ : $1(x - 1) - 2(y - 1) + 1(z - 1) = 0 \Rightarrow x - 2y + z = 0$ .

Line through origin perpendicular to this plane. Direction = normal $\vec{n} = (1, -2, 1)$ , passes through origin: $\vec{r}(t) = (0,0,0) + t(1, -2, 1) = (t, -2t, t).$ Parametric: $x = t, \; y = -2t, \; z = t$ . Symmetric: $\dfrac{x}{1} = \dfrac{y}{-2} = \dfrac{z}{1}$ .

Lines and Planes in Space