Ultimate 2D Vector Masterclass

Master Vector Geometry

Welcome to the definitive reference for Add Math 2D Vectors. All study chapters, rigorous path walkthroughs, and exam practice modules are organized inside the hamburger navigation menu on the left.

Path Traversal & Rules

Deep dive into polygon vector navigation, direction reversal rules, and geometric proofs.

Interactive Playground

Test vectors $\mathbf{a}$ and $\mathbf{b}$ dynamically using live sliders and real-time canvas rendering.

40 Hard Questions

Test your mastery with high-difficulty exam questions complete with score tracking and diagrams.

1

Vector Notation & Core Basics

Foundational principles of magnitude, direction, and vector representation.

What is a Vector?

Unlike a scalar quantity (which has magnitude only, like mass or temperature), a vector possesses both magnitude (length) and a definite direction. In 2D coordinate geometry, vectors can be expressed algebraically or graphically.

Column Vector Form

Represents net horizontal displacement $x$ and vertical displacement $y$.

\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}

Unit Vector Form

Expressed in terms of standard orthogonal unit vectors $\mathbf{i}$ and $\mathbf{j}$.

\mathbf{a} = x\mathbf{i} + y\mathbf{j}

Fundamental Rules of Vector Algebra

Direction Reversal: Reversing the endpoints of a vector negates its value. $\vec{AB} = -\vec{BA}$
Scalar Multiplication: Multiplying a vector by scalar $k$ scales its magnitude by $|k|$. If $k > 0$, the direction remains parallel. If $k < 0$, the direction is reversed.

2

Magnitude & Unit Vectors

Calculating lengths and normalizing directional vectors.

Calculating Magnitude

The magnitude (or length) of a vector $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ is derived directly from Pythagoras' Theorem across the coordinate axes:

|\mathbf{v}| = \sqrt{x^2 + y^2}

Unit Vector Normalization

A unit vector $\hat{\mathbf{v}}$ is a vector pointing in the exact same direction as $\mathbf{v}$ but scaled down or up to have a magnitude of exactly 1.

\hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|} \mathbf{v} = \frac{1}{\sqrt{x^2 + y^2}} \begin{pmatrix} x \\ y \end{pmatrix}

Worked Example

Find the unit vector in the direction of $\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}$.

Calculate magnitude: $|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Divide vector by its magnitude: $\hat{\mathbf{v}} = \frac{1}{5}(3\mathbf{i} - 4\mathbf{j}) = 0.6\mathbf{i} - 0.8\mathbf{j}$.

3

Vector Paths & Geometric Routes

Advanced strategies for traversing polygons and expressing unknown vectors in terms of bases.

The Golden Rule of Path Traversal

In Add Math vector questions, you are almost always given base vectors (e.g., $\vec{OA} = \mathbf{a}$, $\vec{OB} = \mathbf{b}$) and asked to find complex routes like $\vec{PQ}$ or position vectors of intersection points.

\vec{AC} = \vec{AB} + \vec{BC}

Crucial Navigation Rule: To find a vector from point A to point C, you can take any continuous detour path through intermediate nodes (like B). If you travel against a defined arrow on a diagram, you must invert its sign (multiply by $-1$).

Step-by-Step Problem Solving Strategy

1 Identify Terminals: Clearly locate your start point and destination point on the given geometric figure.
2 Trace Known Routes: Look for a sequence of edges where the vectors are already defined or can be easily inferred from polygon properties (parallelogram, trapezium, midpoint ratios).
3 Simplify Algebra: Collect like terms of $\mathbf{a}$ and $\mathbf{b}$ carefully. Factorize where necessary.

Parallelogram Geometry Check

In parallelogram OACB: $\vec{OC} = \mathbf{a} + \mathbf{b}$ and $\vec{AB} = \mathbf{b} - \mathbf{a}$.

4

Master Formulas & Theorems

Essential formulas for position vectors, division of line segments, and ratios.

Position Vector

A vector relative to origin $O(0,0)$. The vector between two position vectors $\mathbf{a}$ and $\mathbf{b}$:

\vec{AB} = \mathbf{b} - \mathbf{a}

"Position of destination minus position of origin."

Ratio Theorem

If point $P$ divides line $AB$ in ratio $m:n$:

\vec{OP} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}

Crucial for dividing internal segments precisely.

Parallelism

Two vectors $\mathbf{p}$ and $\mathbf{q}$ are parallel if one is a scalar multiple of the other:

\mathbf{p} = k\mathbf{q}

Used extensively in collinearity proofs.

5

Collinearity & Intersecting Lines

Advanced proofs for straight-line points and simultaneous coefficient equating.

Proving Collinearity

Three points $A, B, C$ lie on the exact same straight line (collinear) if and only if:

The vectors connecting them are parallel: $\vec{AB} = k\vec{BC}$.
They share a common point (in this case, point $B$). Because line $AB$ and line $BC$ share $B$ and have identical slopes, they form one continuous straight line.

The "Equating Coefficients" Trick for Intersections

When two lines intersect at an unknown point $X$ (e.g., $AC$ and $OB$), exam questions require finding the ratio of intersection. The winning strategy is to express $\vec{OX}$ in two independent ways using parameters $\lambda$ and $\mu$, then equate the coefficients of $\mathbf{a}$ and $\mathbf{b}$.

// Route 1: OX via path O -> A -> C

$\vec{OX} = \vec{OA} + \lambda\vec{AC} = (1-\lambda)\mathbf{a} + 3\lambda\mathbf{b}$

// Route 2: OX via direct path

$\vec{OX} = \mu\vec{OB} = 2\mu\mathbf{a} + \mu\mathbf{b}$

// Equate coefficients of a and b to solve simultaneous equations for λ and μ.

6

Interactive Vector Playground

Visualize vector addition $\mathbf{a} + \mathbf{b}$ using live coordinate sliders.

Vector a

x (i): 4

y (j): 2

Vector b

x (i): -2

y (j): 5

Resultant $\mathbf{a} + \mathbf{b}$

$\begin{pmatrix} 2 \\ 7 \end{pmatrix}$

7

Worked Exam Examples

Detailed step-by-step solutions to typical Cambridge O Level / Add Math past paper problems.

Example: Triangle Ratio & Vector Paths

In $\triangle OAB$, $\vec{OA} = 6\mathbf{a}$ and $\vec{OB} = 9\mathbf{b}$. Point $P$ lies on $AB$ such that $AP:PB = 1:2$. Find $\vec{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Step 1: Find vector AB

\vec{AB} = \vec{AO} + \vec{OB} = -6\mathbf{a} + 9\mathbf{b}

Step 2: Express AP

Since $AP:PB = 1:2$, the segment $AP$ is exactly $\frac{1}{3}$ of the total vector length $AB$.

\vec{AP} = \frac{1}{3}\vec{AB} = \frac{1}{3}(-6\mathbf{a} + 9\mathbf{b}) = -2\mathbf{a} + 3\mathbf{b}

Step 3: Find OP via continuous path O → A → P

\vec{OP} = \vec{OA} + \vec{AP} = 6\mathbf{a} + (-2\mathbf{a} + 3\mathbf{b}) = 4\mathbf{a} + 3\mathbf{b}

The Ultimate Hard Quiz

40 Rigorous Exam-Style Questions covering magnitude, unit vectors, polygons, ratio theorems, and collinearity.

Q 1 / 40 Score: 0

Loading sequence...

Home Dashboard