A vector describes both how far and in which direction. That makes it fundamentally different from an ordinary number. The same distance travelled in two different directions produces two different vectors, and a reversed direction produces the negative of the original vector.
This Secondary 3 Mathematics Learning Guide develops vectors in two dimensions from geometric meaning. It covers directed line segments, column-vector notation, magnitude, position vectors, translation, addition and subtraction, scalar multiplication, and geometric problems in which several vector routes must represent the same displacement.
The official 2027 SEC G3 Mathematics syllabus K310 includes vector notation, directed line segments, translation by a vector, position vectors, vector magnitude, sums and differences of vectors, scalar multiplication and geometric problems involving vectors. This guide follows that scope and avoids introducing advanced vector products.
Use the Secondary Mathematics Hub for the wider route. Within this guide, move through diagnostic · notation · operations · position vectors · geometry · practice · answers.
Scalar Versus Vector
A scalar has magnitude only. Mass, temperature and ordinary length measurements can be treated as scalars in elementary contexts. A vector has magnitude and direction. Displacement is therefore a vector: 5 km east and 5 km north have the same magnitude but are not the same displacement.
Speed is usually treated as a scalar because it records how fast. Velocity carries direction as well. The distinction matters because vector addition follows geometry, not just arithmetic.
A Six-Question Diagnostic
Let a = (3, 4). Find |a|. Find −a. Find 2a. If b = (−1, 5), find a + b. If OA = a and OB = b, express AB in terms of a and b. Finally, what translation does the vector (4, −2) describe?
The answers are 5; (−3, −4); (6, 8); (2, 9); b − a; and four units right and two units down. These questions separate magnitude, direction, scalar multiplication, vector addition, position-vector reasoning and translation interpretation.
Directed Line Segments
The vector from A to B is written as AB with an arrow notation in formal typesetting. Direction matters: the vector from B to A is the negative of the vector from A to B.
If A is at (2, 1) and B is at (7, 4), then the displacement from A to B is (7−2, 4−1) = (5, 3). Reversing the direction gives BA = (−5, −3).
The point coordinates describe locations relative to the origin. The vector difference describes the movement needed to get from one location to another.
Column Vectors
A vector can be represented by a column such as (x, y), where the first component records horizontal displacement and the second vertical displacement. A positive x-component points right; a negative one points left. A positive y-component points up; a negative one points down.
Thus (−3, 2) means three units left and two units up. Its direction is not captured by the number √13 alone; √13 is only its magnitude.
Magnitude Comes From Pythagoras
For vector a = (x, y), its magnitude is |a| = √(x² + y²). This is the distance formula applied to a displacement from the origin.
For (6, 8), magnitude = √(36 + 64) = 10. For (−6, 8), the magnitude is also 10. Sign affects direction, while the squares remove sign when measuring length.
Worked Example 1: Displacement and Magnitude
Point P is (−2, 3) and Q is (4, −5). Find PQ and its magnitude. Subtract starting coordinates from ending coordinates: PQ = (4−(−2), −5−3) = (6, −8).
The magnitude is √(6² + (−8)²) = √100 = 10. The vector describes direction and displacement; the magnitude gives the straight-line distance between P and Q.
Vector Addition Means Following One Displacement With Another
If a = (3, 1) and b = (2, 4), then a + b = (5, 5). Geometrically, move according to a, then from that new point move according to b. The overall displacement is the vector from the original start to the final position.
This is sometimes called the head-to-tail interpretation. The horizontal components add because horizontal movements combine; the vertical components add for the same reason.
Vector Subtraction Means Add the Reverse
The vector a − b is a + (−b). If b = (4, −2), then −b = (−4, 2). Reversing a vector changes direction but preserves magnitude.
This interpretation becomes especially useful for position vectors. If OA = a and OB = b, then AB = OB − OA = b − a.
Scalar Multiplication Changes Length and Sometimes Direction
For scalar k, the vector ka multiplies every component of a by k. If k > 1, the vector becomes longer in the same direction. If 0 < k < 1, it becomes shorter in the same direction. If k < 0, direction reverses as well.
For a = (2, 3), 3a = (6, 9), while −2a = (−4, −6). The magnitudes are multiplied by |k|.
Worked Example 2: Parallel Vector Relationships
a = (4, 6) and b = (−2, −3). Describe their relationship. Since b = −1/2 a, the vectors are parallel but point in opposite directions. The magnitude of b is half the magnitude of a.
Scalar-multiple reasoning is often more informative than comparing component ratios separately. If one vector is a scalar multiple of another, their directions are parallel or anti-parallel depending on the sign.
Position Vectors Anchor Points to the Origin
The position vector of point A is the vector OA from the origin O to A. If A has coordinates (5, −1), then OA = (5, −1).
Once position vectors are known, vectors between points can be obtained by subtraction. This turns geometric diagrams into algebraic relationships.
Worked Example 3: Position Vectors
OA = a and OB = b. Express AB and BA. Travelling from A to B is equivalent to travelling from A back to O and then from O to B. Thus AB = −a + b = b − a.
Similarly, BA = a − b = −(b − a). This makes the reversal property explicit.
Midpoints Through Vectors
If M is the midpoint of AB and OA = a, OB = b, then OM = (a + b)/2. The midpoint lies halfway along the displacement from A to B.
To see this, AB = b − a. Halfway from A gives OM = a + 1/2(b − a) = 1/2(a + b).
Worked Example 4: Midpoint Position Vector
OA = (2, 7) and OB = (8, −1). Find the midpoint position vector. OM = [(2,7) + (8,−1)]/2 = (10,6)/2 = (5,3).
The coordinate midpoint formula is therefore a vector average. This connects vectors directly to coordinate geometry.
Translations Are Vector Movements
A translation by vector (p, q) sends point (x, y) to (x+p, y+q). Every point moves by the same vector, so lengths, angles and shape are preserved.
For example, translating (3, −2) by (5, 4) gives (8, 2). Translating an entire polygon means adding the same vector to every vertex.
The companion Coordinate Geometry and Transformations guide develops this representation alongside lines, distances and scale transformations.
Different Routes Between the Same Points Must Give the Same Vector
This is one of the most powerful vector principles. If two routes begin at the same point and end at the same point, their total displacement vectors are equal.
In a parallelogram ABCD, AB + BC = AC and AD + DC = AC. Therefore the two routes around adjacent sides produce the same diagonal vector.
Worked Example 5: Parallelogram
AB = a and AD = b in parallelogram ABCD. Express AC, BC and DC. Opposite sides are equal and parallel as directed vectors, so BC = b and DC = a.
The diagonal AC can be reached by AB then BC, giving AC = a + b. It can also be reached by AD then DC, giving b + a, consistent with vector addition.
Dividing a Line Segment in a Ratio
If point P lies on AB with AP:PB = 2:3, then P is two fifths of the way from A to B. Therefore OP = OA + 2/5 AB.
Using OA = a and OB = b, AB = b − a, so OP = a + 2/5(b − a) = 3/5 a + 2/5 b.
Worked Example 6: Ratio Point
OA = a, OB = b and P divides AB internally in the ratio AP:PB = 1:2. Express OP. P lies one third of the way from A to B.
OP = a + 1/3(b − a) = 2/3 a + 1/3 b. The coefficients add to 1 because P is a weighted average of the two endpoint positions.
Collinearity Through Scalar Multiples
If AB is a scalar multiple of AC, then both vectors point along the same line through A. This can establish that A, B and C are collinear, provided the vectors share the same starting point.
For example, if AB = 2p and AC = 5p, then both displacements lie along the direction p. Therefore A, B and C are collinear.
Worked Example 7: Prove Collinearity
OA = a, OB = 3a + 2b and OC = 5a + 4b. Show that A, B and C are collinear.
AB = OB − OA = 2a + 2b = 2(a+b). AC = OC − OA = 4a + 4b = 4(a+b).
Hence AC = 2AB. Since both vectors start at A and one is a scalar multiple of the other, the three points are collinear.
Worked Example 8: Find an Unknown Scalar
Vector a = (2, 5). Vector b = (6, 15). Find k such that b = ka. Compare components: 6 = 2k gives k = 3, and 15 = 5k confirms the same value.
Therefore b = 3a. Both component equations must agree. If they give different k-values, the vectors are not scalar multiples.
Common Errors
Subtracting in the wrong order: AB is found as A−B instead of B−A. Repair by saying “end minus start”.
Confusing vector and magnitude: the answer √13 is given when the question asks for a vector. Repair by identifying whether direction is required.
Ignoring direction in parallel vectors: a negative scalar multiple is treated as the same direction. Repair by separating parallel line relationship from vector direction.
Mixing position and displacement: OA and AB are treated as interchangeable. Repair by drawing the origin and writing the start and end of each vector.
Independent Practice
1. For a=(5,12), find |a|.
2. For b=(−3,4), find −2b.
3. If a=(2,−1), b=(4,5), find a+b and b−a.
4. A=(1,3), B=(7,−2). Find AB.
5. Find the magnitude of the vector in Question 4.
6. Translate P=(−2,5) by vector (6,−3).
7. If OA=a and OB=b, express BA.
8. If M is the midpoint of AB, express OM in terms of a and b.
9. If AP:PB=3:2, express OP in terms of a=OA and b=OB.
10. In parallelogram ABCD, AB=p and AD=q. Express AC and BD.
11. Show that vectors (4,6) and (−6,−9) are parallel.
12. A=(2,1), B=(6,5), C=(10,9). Show A, B and C are collinear.
13. OA=a, OB=2a+b, OC=5a+3b. Find AB and AC. Are they scalar multiples?
14. a=(x,6) and b=(4,12) with b=2a. Find x.
15. Explain why vectors with equal magnitudes need not be equal.
Explained Answers
1. √(25+144)=13.
2. −2b=(6,−8).
3. a+b=(6,4); b−a=(2,6).
4. AB=(7−1,−2−3)=(6,−5).
5. |AB|=√61.
6. P′=(4,2).
7. BA=a−b.
8. OM=(a+b)/2.
9. P lies 3/5 from A to B, so OP=a+3/5(b−a)=2/5a+3/5b.
10. AC=p+q. BD=q−p, because B to D equals BO+OD=−p+q.
11. (−6,−9)=−3/2(4,6), so one is a scalar multiple of the other.
12. AB=(4,4), AC=(8,8)=2AB, so the three points are collinear.
13. AB=a+b. AC=4a+3b. Without extra information relating a and b, these are not generally scalar multiples.
14. 2a=(2x,12)=(4,12), so x=2.
15. Equal magnitude means equal length only. Different directions produce different vectors.
A Reliable Vector Workflow
Mark every direction. Write known position vectors. Express unknown displacements as end minus start. Use one complete route between the required points. Simplify coefficients only after the route is correct. Finally, check whether an equivalent second route gives the same vector.
For collinearity, look for scalar multiples. For midpoint or ratio points, interpret the location as a fraction of the displacement from one endpoint to the other.
Teacher and Parent Prompts
Ask “Where does this vector start, and where does it end?” before correcting signs. Ask “Does the answer need direction or only magnitude?” For diagram problems, ask the learner to describe two possible routes between the same pair of points.
For extension, give a ratio point or midpoint and ask for its position vector in two different ways. Equivalent routes should simplify to the same result.
Continue the Secondary 3 Learning Route
Continue with Coordinate Geometry and Transformations for gradients, distances, equations and translations; Trigonometry, Bearings and Navigation for directed geometric modelling; and Probability and Statistical Reasoning for uncertainty and data.
Vector competence means preserving direction, magnitude and route at the same time. Return to the Secondary Mathematics Hub for the full S1–S4 capability map.
