How to Solve Vectors Questions in ECZ Mathematics

Mathematics (Grade 12 & GCE)  |  Updated March 2026  |  12 min read

Vectors are tested in both Paper 1 (short 2-mark or 3-mark column vector questions) and Paper 2 (Section A 5-to-6 mark vector geometry problems) in ECZ Grade 12 and GCE Mathematics.

Senior Secondary Tip

Always remember the vector route rule: to get from point $A$ to point $B$ via point $O$, write $\vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB}$!

1. Column Vectors and Position Vectors

A vector has both magnitude (length) and direction:

\[ \vec{v} = \begin{pmatrix} x \\ y \end{pmatrix} \]
  • Position Vector: The vector from origin $O(0,0)$ to point $A(x_1, y_1)$ is $\vec{OA} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$.
  • Vector Between Two Points: $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$.

2. Magnitude of a Vector

The magnitude $|\vec{AB}|$ represents the geometric distance between points $A$ and $B$:

\[ |\vec{AB}| = \sqrt{x^2 + y^2} \]

ECZ Past Paper Style Problem:

In triangle $OAB$, $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Point $C$ lies on $AB$ such that $AC : CB = 1 : 2$.

  1. Express $\vec{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
  2. Express $\vec{AC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
  3. Hence, find position vector $\vec{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ in simplest form.

Solution Part (i):

\[ \vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a} \]

Solution Part (ii):

Since ratio $AC : CB = 1 : 2$, $AC$ is $\frac{1}{3}$ of total length $AB$:

\[ \vec{AC} = \frac{1}{3} \vec{AB} = \frac{1}{3}(\mathbf{b} - \mathbf{a}) \]

Solution Part (iii):

\[ \vec{OC} = \vec{OA} + \vec{AC} = \mathbf{a} + \frac{1}{3}(\mathbf{b} - \mathbf{a}) \] \[ \vec{OC} = \mathbf{a} + \frac{1}{3}\mathbf{b} - \frac{1}{3}\mathbf{a} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{3}(2\mathbf{a} + \mathbf{b}) \]
Collinear Points Proof

To prove points $A, B, C$ are collinear: 1. Show $\vec{AB} = k \vec{BC}$ (parallel vectors), and 2. State that point $B$ is a common point to both vectors!

Teacher Strategy

Remind learners to draw arrows on vector diagrams during working out so they don't accidentally reverse signs ($-\mathbf{a}$ vs $+\mathbf{a}$).

Frequently Asked Questions

Q1: What is a column vector and how is position vector calculated?

A column vector is written as [[x], [y]], representing horizontal displacement x and vertical displacement y. If points A(x1, y1) and B(x2, y2) have origin O, vector AB = OB - OA = [[x2 - x1], [y2 - y1]].

Q2: How do you calculate the magnitude of a vector?

The magnitude (length) of vector v = [[x], [y]] is calculated using Pythagoras' theorem: |v| = √(x² + y²).

Q3: What condition proves that two vectors are parallel or collinear?

Two vectors u and v are parallel if u = k * v (where k is a scalar constant). If they share a common point (e.g., AB = k * BC), points A, B, and C are collinear (lying on the same straight line).

Ace ECZ Mathematics Vector Geometry

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