What you'll learn
Vectors form a critical component of the CXC CSEC Mathematics syllabus, testing your ability to represent and manipulate quantities that have both magnitude and direction. This guide covers position vectors, displacement vectors, and their geometric applications—concepts that appear consistently in Section III (Vectors and Matrices) of Paper 2. Understanding vectors enables you to solve problems involving navigation, forces, and spatial relationships using precise mathematical methods.
Key terms and definitions
Vector — a quantity with both magnitude (size) and direction, typically represented by a directed line segment or in component form such as $\begin{pmatrix} x \ y \end{pmatrix}$.
Position vector — a vector that describes the position of a point relative to a fixed origin O, usually denoted by lowercase bold letters or with an arrow notation (e.g., $\vec{OA}$ or a).
Displacement vector — a vector representing the change in position from one point to another, independent of the starting position; for example, the displacement from A to B is written as $\vec{AB}$.
Magnitude of a vector — the length or size of a vector; for a vector $\begin{pmatrix} x \ y \end{pmatrix}$, the magnitude is $\sqrt{x^2 + y^2}$.
Unit vector — a vector with magnitude equal to 1, often used to indicate direction; standard unit vectors are i (in the x-direction) and j (in the y-direction).
Resultant vector — the single vector obtained by adding two or more vectors together, representing the combined effect.
Parallel vectors — vectors that have the same or opposite direction; if a and b are parallel, then b = ka for some scalar k.
Collinear points — three or more points that lie on the same straight line, verified when their displacement vectors are parallel.
Core concepts
Representing vectors
Vectors can be represented in multiple forms, and you must be comfortable converting between them:
Column vector notation: $\begin{pmatrix} x \ y \end{pmatrix}$ where x represents the horizontal component and y the vertical component.
Component form using unit vectors: $x\mathbf{i} + y\mathbf{j}$, where i = $\begin{pmatrix} 1 \ 0 \end{pmatrix}$ and j = $\begin{pmatrix} 0 \ 1 \end{pmatrix}$.
Directed line segment: An arrow drawn from one point to another, where the length represents magnitude and the arrowhead indicates direction.
When sketching vectors on the Cartesian plane, always use a ruler and clearly mark the arrowhead. The vector $\begin{pmatrix} 3 \ 2 \end{pmatrix}$ means "move 3 units right and 2 units up from the starting point."
Position vectors and coordinates
The position vector of a point A with coordinates (x, y) relative to origin O is:
$$\vec{OA} = \begin{pmatrix} x \ y \end{pmatrix}$$
This establishes a direct link between coordinate geometry and vectors. If point B has position vector b = $\begin{pmatrix} 5 \ 3 \end{pmatrix}$, then B has coordinates (5, 3).
For Caribbean navigation contexts, if a lighthouse at origin O has a ship at position (8, 6) km, the position vector of the ship is $\begin{pmatrix} 8 \ 6 \end{pmatrix}$ km.
Displacement vectors and the fundamental relationship
The displacement from point A to point B is found using the triangle rule:
$$\vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB} = \vec{OB} - \vec{OA}$$
More simply: Displacement = Final position − Initial position
If A has position vector a = $\begin{pmatrix} 2 \ 1 \end{pmatrix}$ and B has position vector b = $\begin{pmatrix} 7 \ 4 \end{pmatrix}$, then:
$$\vec{AB} = \begin{pmatrix} 7 \ 4 \end{pmatrix} - \begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 3 \end{pmatrix}$$
Notice that $\vec{BA} = -\vec{AB}$, meaning the displacement from B to A is in the opposite direction.
Vector operations
Addition: Add corresponding components: $$\begin{pmatrix} x_1 \ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \ y_1 + y_2 \end{pmatrix}$$
Geometrically, this follows the triangle rule or parallelogram rule. If a boat travels with displacement d₁ = $\begin{pmatrix} 4 \ 3 \end{pmatrix}$ km then d₂ = $\begin{pmatrix} 2 \ -1 \end{pmatrix}$ km, the total displacement is $\begin{pmatrix} 6 \ 2 \end{pmatrix}$ km.
Subtraction: Subtract corresponding components: $$\begin{pmatrix} x_1 \ y_1 \end{pmatrix} - \begin{pmatrix} x_2 \ y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \ y_1 - y_2 \end{pmatrix}$$
Scalar multiplication: Multiply each component by the scalar: $$k\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} kx \ ky \end{pmatrix}$$
Scalar multiplication changes magnitude but preserves (for k > 0) or reverses (for k < 0) direction.
Magnitude and distance
The magnitude of vector v = $\begin{pmatrix} x \ y \end{pmatrix}$ is:
$$|\mathbf{v}| = \sqrt{x^2 + y^2}$$
This formula derives directly from Pythagoras' theorem. The magnitude represents the distance between two points when the vector is a displacement.
For a hurricane tracking problem where a storm moves with displacement $\begin{pmatrix} -12 \ 5 \end{pmatrix}$ km from its previous position, the distance traveled is:
$$\sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ km}$$
Midpoint using vectors
The position vector of the midpoint M of line segment AB is:
$$\vec{OM} = \frac{\vec{OA} + \vec{OB}}{2} = \frac{\mathbf{a} + \mathbf{b}}{2}$$
This formula appears frequently in coordinate geometry questions reframed using vector notation.
Ratio division and section formula
If point P divides the line segment AB in the ratio m:n, then:
$$\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m + n}$$
For a 2:1 ratio (P closer to B), with m = 2 and n = 1:
$$\vec{OP} = \frac{1 \cdot \vec{OA} + 2 \cdot \vec{OB}}{3} = \frac{\mathbf{a} + 2\mathbf{b}}{3}$$
Parallel vectors and collinearity
Two vectors are parallel if one is a scalar multiple of the other: b = ka.
Three points A, B, C are collinear if $\vec{AB}$ and $\vec{AC}$ are parallel, which requires:
$$\vec{AC} = k \cdot \vec{AB}$$
for some scalar k. This test is essential for proving geometric properties.
Example: Points A, B, C have position vectors $\begin{pmatrix} 1 \ 2 \end{pmatrix}$, $\begin{pmatrix} 4 \ 5 \end{pmatrix}$, $\begin{pmatrix} 7 \ 8 \end{pmatrix}$ respectively.
$$\vec{AB} = \begin{pmatrix} 3 \ 3 \end{pmatrix}, \quad \vec{AC} = \begin{pmatrix} 6 \ 6 \end{pmatrix} = 2\begin{pmatrix} 3 \ 3 \end{pmatrix} = 2\vec{AB}$$
Since $\vec{AC} = 2\vec{AB}$, the points are collinear.
Geometric applications
Vectors provide powerful tools for proving geometric properties:
Proving a quadrilateral is a parallelogram: Show that opposite sides are equal vectors ($\vec{AB} = \vec{DC}$ and $\vec{AD} = \vec{BC}$).
Proving a triangle: Use the triangle rule $\vec{AB} + \vec{BC} + \vec{CA} = \mathbf{0}$ to verify three points form a triangle.
Finding vertices: Given three vertices of a parallelogram, use the relationship that diagonals bisect each other to find the fourth vertex.
Worked examples
Example 1: Position and displacement vectors
Question: The position vectors of points P and Q are p = $\begin{pmatrix} -3 \ 5 \end{pmatrix}$ and q = $\begin{pmatrix} 4 \ 1 \end{pmatrix}$ respectively.
(a) Find $\vec{PQ}$. (2 marks)
(b) Calculate the distance PQ. (2 marks)
(c) Find the position vector of the midpoint M of PQ. (2 marks)
Solution:
(a) Using the displacement formula: $$\vec{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 4 \ 1 \end{pmatrix} - \begin{pmatrix} -3 \ 5 \end{pmatrix} = \begin{pmatrix} 7 \ -4 \end{pmatrix}$$
(b) The distance PQ is the magnitude of $\vec{PQ}$: $$|\vec{PQ}| = \sqrt{7^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65} = 8.06 \text{ units (to 2 d.p.)}$$
(c) Using the midpoint formula: $$\vec{OM} = \frac{\mathbf{p} + \mathbf{q}}{2} = \frac{1}{2}\left[\begin{pmatrix} -3 \ 5 \end{pmatrix} + \begin{pmatrix} 4 \ 1 \end{pmatrix}\right] = \frac{1}{2}\begin{pmatrix} 1 \ 6 \end{pmatrix} = \begin{pmatrix} 0.5 \ 3 \end{pmatrix}$$
Example 2: Collinearity
Question: Points A, B and C have position vectors $\begin{pmatrix} 2 \ -1 \end{pmatrix}$, $\begin{pmatrix} 5 \ 2 \end{pmatrix}$ and $\begin{pmatrix} 11 \ 8 \end{pmatrix}$ respectively.
Show that A, B and C are collinear. (3 marks)
Solution:
Find $\vec{AB}$: $$\vec{AB} = \begin{pmatrix} 5 \ 2 \end{pmatrix} - \begin{pmatrix} 2 \ -1 \end{pmatrix} = \begin{pmatrix} 3 \ 3 \end{pmatrix}$$
Find $\vec{AC}$: $$\vec{AC} = \begin{pmatrix} 11 \ 8 \end{pmatrix} - \begin{pmatrix} 2 \ -1 \end{pmatrix} = \begin{pmatrix} 9 \ 9 \end{pmatrix}$$
Check for scalar relationship: $$\vec{AC} = \begin{pmatrix} 9 \ 9 \end{pmatrix} = 3\begin{pmatrix} 3 \ 3 \end{pmatrix} = 3\vec{AB}$$
Since $\vec{AC} = 3\vec{AB}$, the vectors are parallel and share point A. Therefore A, B and C are collinear.
Example 3: Parallelogram problem
Question: OABC is a parallelogram where O is the origin. The position vectors of A and C are $\begin{pmatrix} 6 \ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \ 5 \end{pmatrix}$ respectively.
(a) Find the position vector of B. (3 marks)
(b) A cargo ship at origin O needs to travel to point B via point A, then directly back to O. Calculate the total distance traveled. (3 marks)
Solution:
(a) In a parallelogram, $\vec{OB} = \vec{OA} + \vec{OC}$ (using the parallelogram rule): $$\vec{OB} = \begin{pmatrix} 6 \ 2 \end{pmatrix} + \begin{pmatrix} 3 \ 5 \end{pmatrix} = \begin{pmatrix} 9 \ 7 \end{pmatrix}$$
(b) Distance OA: $$|\vec{OA}| = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 6.32 \text{ units}$$
Distance AB (where $\vec{AB} = \vec{OC}$ in a parallelogram): $$|\vec{AB}| = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} = 5.83 \text{ units}$$
Distance BO: $$|\vec{BO}| = |\vec{OB}| = \sqrt{9^2 + 7^2} = \sqrt{81 + 49} = \sqrt{130} = 11.40 \text{ units}$$
Total distance = 6.32 + 5.83 + 11.40 = 23.55 units (or 23.6 units to 1 d.p.)
Common mistakes and how to avoid them
• Confusing position and displacement vectors: Position vectors describe location relative to the origin; displacement vectors describe movement between two points. Always use $\vec{AB} = \mathbf{b} - \mathbf{a}$, not $\mathbf{a} - \mathbf{b}$, when finding displacement from A to B.
• Incorrect magnitude calculation: Students often forget to square root after adding the squared components. The magnitude of $\begin{pmatrix} 3 \ 4 \end{pmatrix}$ is $\sqrt{3^2 + 4^2} = 5$, not $3^2 + 4^2 = 25$.
• Sign errors in component operations: When subtracting vectors, carefully handle negative signs. $\begin{pmatrix} 2 \ -3 \end{pmatrix} - \begin{pmatrix} -1 \ 4 \end{pmatrix} = \begin{pmatrix} 3 \ -7 \end{pmatrix}$, not $\begin{pmatrix} 1 \ -7 \end{pmatrix}$.
• Proving collinearity incompletely: Showing that $\vec{AB}$ and $\vec{AC}$ are parallel is insufficient. You must also state that the vectors share a common point (A) to conclude collinearity. Write: "Since $\vec{AC} = k\vec{AB}$ and they share point A, the points are collinear."
• Midpoint formula errors: The midpoint formula requires addition in the numerator before dividing by 2: $\frac{\mathbf{a} + \mathbf{b}}{2}$, not $\frac{\mathbf{a}}{2} + \frac{\mathbf{b}}{2}$ (though algebraically equivalent, this shows proper method).
• Omitting units in applied problems: When vectors represent physical quantities (kilometers, meters, newtons), always include units in your final answer, especially for magnitude calculations.
Exam technique for Vectors: Position vectors, displacement vectors and geometric applications
• Command word "Find": Present your answer clearly in the required form. If finding a position vector, write it as a column vector unless instructed otherwise. Show intermediate working—marks are often awarded for method even if the final answer contains an arithmetic error.
• Command word "Show" or "Prove": You must demonstrate the result through logical steps. For collinearity, explicitly show the scalar relationship between vectors and state the shared point. For parallel vectors, write the scalar equation clearly.
• Structured working for multi-part questions: Questions often build sequentially—part (b) may use the answer from part (a). If you cannot complete part (a), use a letter to represent that result and continue. Partial marks are available for correct method with incorrect values.
• Mark allocation guides effort: A 2-mark question typically requires a formula and one calculation. A 4-mark question might require finding two displacement vectors, proving they're parallel, and stating a conclusion. Budget approximately 1.5 minutes per mark.
Quick revision summary
Vectors have magnitude and direction. Position vectors locate points relative to origin O; displacement vectors describe movement between points using $\vec{AB} = \mathbf{b} - \mathbf{a}$. Calculate magnitude using $\sqrt{x^2 + y^2}$. Add and subtract vectors by combining corresponding components; scalar multiplication scales magnitude. Midpoint of AB: $\frac{\mathbf{a} + \mathbf{b}}{2}$. Points are collinear when displacement vectors between them are parallel (scalar multiples). Master these fundamentals and practice showing working clearly for maximum marks in CXC CSEC Mathematics examinations.