A vector tells you how to move, not merely how far. Moving from A to B is the reverse of moving from B to A. Two journeys can have equal lengths and different directions. Two vector expressions can describe the same destination by different routes. These distinctions are the foundation of Secondary 4 vector questions, including the longer proofs involving ratios and collinearity.
This eleventh Secondary 4 Mathematics Learning Guide teaches vectors through journeys with named starting and finishing points. The aim is to make each plus sign, minus sign and fractional multiplier explainable. The guide belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
Scope: the principal references are the two-dimensional vector sections of 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310. This is not a claim of identical coverage across all subject levels. The examples do not require dot products, cross products or three-dimensional vector equations.
Notation: keep points, vectors and lengths separate
In this text, bold lower-case letters such as a and b denote vectors. Bold paired letters such as AB denote the directed vector from A to B. In handwritten work, draw the usual arrow above the paired letters. Plain AB denotes the length of the segment when we are discussing length, and |AB| denotes its magnitude.
For accessible reading on a phone, a two-component vector is sometimes written as (horizontal component, vertical component), such as (3, −2). In school working this is commonly displayed as a column vector. The meaning here is three units to the right and two units down. It is a displacement, not the name of a point, unless we explicitly identify it as a position vector from the origin.
A scalar is a number. Multiplying a vector by 3 triples its magnitude while retaining its direction. Multiplying by −3 triples its magnitude and reverses its direction. The magnitude is multiplied by the absolute value of the scalar; a length does not become negative.
The most useful sentence in a vector solution
To travel from A to C, I can travel from A to B and then from B to C. In vector language, AC = AB + BC. The intermediate point B joins the two journeys. Reading the endpoints is a reliable way to check that a proposed sum actually forms a connected route.
By contrast, AB + CB does not follow A to B to C. The second vector points from C to B, so its direction is reversed from the required route. To use it, write AC = AB − CB. A sign is not merely a symbol to manipulate; it records a direction.
Follow this guide through position vectors, ratios on a line, collinearity, intersecting routes and independent practice. Each stage adds a decision rather than a disconnected formula.
Worked example 1: direction and magnitude from coordinates
A has coordinates (−2, 5) and B has coordinates (4, −3). Find the vector from A to B and its magnitude. Subtract the starting coordinates from the finishing coordinates:
AB = (4 − (−2), −3 − 5) = (6, −8).
|AB| = √(6² + (−8)²) = √100 = 10.
The vector says six units right and eight units down. The magnitude says the straight-line distance is ten units. The reverse vector BA is (−6, 8), but its magnitude is still ten. A solution that gives only ten when asked for the vector has lost the direction.
Check by adding the displacement to A: (−2, 5) + (6, −8) = (4, −3), which is B. This is a direct return to the geometric meaning of the calculation.
Worked example 2: add movements component by component
Let u = (3, 4) and v = (−3, 4), measured in kilometres. Find the resultant displacement and compare it with the total distance travelled along the two straight legs. Adding components gives u + v = (0, 8). The resultant is eight kilometres north if the axes represent east and north.
Each leg has length √(3² + 4²) = 5 km. The travelled distance is therefore 10 km, while the magnitude of the resultant displacement is 8 km. The two horizontal movements cancel, but the traveller still covered both legs. This is why adding magnitudes is not generally the same as finding the magnitude of a sum.
There is no contradiction between the answers. They measure different things: the length of the actual route and the straight-line separation of its endpoints. Before calculating a “distance” in a word problem, decide which of those quantities the question means.
Equal vectors are not merely equal lengths
Vectors (3, 4) and (4, 3) both have magnitude five, but they point in different directions. They are not equal vectors. Vectors (3, 4) and (−3, −4) also have equal magnitudes, but point in opposite directions. Equality of vectors requires matching both components, or equivalently the same magnitude and direction.
Two equal directed segments can be drawn in different positions. A vector represents displacement and can be translated without changing that displacement. A position vector is more specific because its starting point has been fixed at the chosen origin.
Position vectors: end minus start
Let O be an origin, with OA = a and OB = b. To travel from A to B, go from A back to O and then from O to B. Therefore AB = −a + b = b − a.
This is the meaning behind “end minus start”. It is not a mnemonic detached from the diagram. If the requested direction is B to A, the answer is a − b. Read the arrow before choosing the order.
Changing the origin changes individual position vectors but not a displacement between two fixed points. Both endpoint position vectors shift by the same amount, and that common shift cancels in the difference. This explains why the geometry of AB should not depend on where you choose to label the origin.
Worked example 3: the same route in a triangle
O, A and B form a triangle, with OA = a and OB = b as vectors. Find AB, BA and the sum OA + AB + BO. The directed vectors are AB = b − a, BA = a − b, and BO = −b.
Thus OA + AB + BO = a + (b − a) − b = 0. The zero vector is expected because the journey finishes where it started. A non-zero result for a closed route is a signal to inspect a reversed arrow or missing segment.
Ratios: a fraction of the whole segment
If P lies between A and B and AP:PB = 2:3, the full segment AB contains five equal parts. Therefore AP = (2/5)AB, not (2/3)AB. The ratio compares one part with another part; the fraction 2/5 compares the first part with the whole.
Direction still matters. PB = (3/5)AB, while BP = −(3/5)AB. A correct fraction with the wrong direction is a wrong vector. Draw a small arrow on each segment before writing the multiplier.
Worked example 4: obtain a position vector from an internal ratio
OA = a and OB = b. P lies on AB with AP:PB = 2:3. Find OP. Start with a connected route: OP = OA + AP. We already know AP = (2/5)(b − a).
OP = a + (2/5)(b − a)
= (3/5)a + (2/5)b.
The coefficients might initially seem reversed relative to the ratio 2:3. The route explains them: you begin at A and move only two fifths of the way towards B. The position therefore remains weighted more towards A. Rather than memorising a section formula with unexplained cross-multiplication, reconstruct the journey.
For a numerical check, choose A = (5, 0) and B = (0, 10). The formula gives P = (3, 4). Moving from A to P gives (−2, 4), which is two fifths of AB = (−5, 10). The abstract expression and the coordinate picture agree.
The endpoint test checks a ratio formula
A general point on the line through A and B can be written as OP = a + t(b − a). If t = 0, P is A. If t = 1, P is B. If 0 < t < 1, P lies between them. If t > 1, P lies beyond B in the direction from A to B; if t < 0, it lies beyond A in the reverse direction.
Expanding gives (1 − t)a + tb. The coefficients sum to one because this is a position on the line through those two endpoint positions. This is a useful structural check for this particular representation, not a rule that the coefficients of every position vector in a triangle must sum to one.
Worked example 5: a point beyond the segment
Q lies on AB extended beyond B, with AQ = (5/3)AB as a directed vector. Find OQ. The same connected route works:
OQ = a + (5/3)(b − a)
= −(2/3)a + (5/3)b.
The negative coefficient does not mean the position is impossible. It reflects the extension beyond B. The coefficients still sum to one, and the parameter 5/3 is greater than one. A ratio formula should be checked against where the point is supposed to lie, not against a blanket rule that coefficients must always be positive.
Parallel vectors: the same scalar must multiply every component
Non-zero vectors are parallel when one is a scalar multiple of the other. A positive multiple gives the same direction, while a negative multiple gives the opposite direction. The relationship must work for the whole vector. Matching one component or one coefficient is insufficient.
Worked example 6: determine an unknown component
u = (2, −3) and v = (k, 12) are parallel. Find k. Write v = λu. The vertical component gives 12 = −3λ, so λ = −4. The horizontal component must use the same multiplier: k = −4 × 2 = −8.
Therefore v = (−8, 12) = −4u. It points in the opposite direction and has four times the magnitude. If you obtain k = 8, the vertical and horizontal components require different multipliers; the vectors are then not parallel.
Collinearity needs a common line, not just parallel directions
Three points are collinear when they lie on one straight line. To prove A, P and Q collinear, a useful method is to show that AQ is a scalar multiple of AP, with AP non-zero. These two vectors share the starting point A. Their parallel directions must therefore lie on the same line through A.
Showing AB parallel to CD is different. The segments may lie on two separate parallel lines. Without a shared point or another condition identifying the same line, parallelism does not prove that A, B, C and D are all collinear. This distinction is small in wording and large in proof.
Worked example 7: prove collinearity by using a shared starting point
O, A and B form a non-degenerate triangle. OA = a, OB = b, OP = (2/3)a + (1/3)b and OQ = −(1/3)a + (4/3)b. Show that A, P and Q are collinear and find AP:PQ. Do not compare OP and OQ directly: those start from O, whereas the target line runs through A.
AP = OP − OA = (1/3)(b − a).
AQ = OQ − OA = (4/3)(b − a) = 4AP.
The vectors share A, so A, P and Q are collinear. Also PQ = AQ − AP = 3AP. Therefore AP:PQ = 1:3. The multiplier four compares AP with the whole AQ; it does not directly give the part-to-part ratio AP:PQ.
You can also locate the points on AB extended. P is one third of the way from A to B, while Q is four thirds of the way. The order is A, P, B, Q. This positional check confirms that the lengths AP and PQ are being compared in the intended directions.
Worked example 8: the midpoint theorem from vector routes
In triangle OAB, M and N are the midpoints of OA and OB respectively. Show that MN is parallel to AB and find the area ratio OMN:OAB. The position vectors are OM = a/2 and ON = b/2.
MN = ON − OM = (1/2)(b − a) = (1/2)AB.
Thus MN is parallel to AB and half its length. The triangles are similar with length scale factor 1/2, so their areas are in the ratio 1:4. The area ratio is not 1:2. The vector calculation establishes a length relationship; similarity determines how area changes.
Compare this with a point P dividing AB in the ratio 2:3. Triangles OAP and OPB share the same perpendicular height to line AB, so their areas are in the ratio 2:3, matching their bases. You do not square that ratio because these two triangles are not being established as similar scaled copies. The reason for the area comparison determines the operation.
A parallelogram is a useful vector organiser
Let O, A, C and B be consecutive vertices of a parallelogram, with OA = a and OB = b. The vector AC equals b, so OC = a + b. The diagonal AB is still b − a. Sums and differences describe different diagonals and directions.
If D is the midpoint of AC, then OD = a + b/2. From B to D, subtract the starting position: BD = a − b/2. This is a compact example of how shape properties, a midpoint and an endpoint difference combine without requiring a new formula.
Intersecting lines: write two routes to the same point
Longer vector questions often involve an intersection X. The central idea is simple: because X lies on two lines, its position can be expressed by travelling along either line. Write one expression for each route, then equate them. Unknown scalar parameters record how far along each line you travel.
When a and b are non-parallel vectors, equal vector expressions in terms of them have equal corresponding coefficients. The non-parallel condition matters. If b were a multiple of a, the same vector could have different-looking coefficient pairs, so comparing coefficients independently would not be justified.
Worked example 9: solve an intersection with two parameters
O, A and B form a triangle. P lies on OA with OP = (2/3)OA, and Q lies on OB with OQ = (3/4)OB. Lines AQ and BP meet at X. Find OX, AX:XQ and BX:XP. Set OA = a and OB = b. Because the triangle is non-degenerate, these vectors are non-parallel.
Travel first along AQ. Write AX = tAQ. Since AQ = (3/4)b − a,
OX = a + t[(3/4)b − a]
= (1 − t)a + (3t/4)b.
Now travel along BP. Write BX = sBP. Since BP = (2/3)a − b,
OX = b + s[(2/3)a − b]
= (2s/3)a + (1 − s)b.
Equating coefficients gives 1 − t = 2s/3 and 3t/4 = 1 − s. From the second equation, s = 1 − 3t/4. Substitute into the first: 1 − t = 2/3 − t/2. Therefore t = 2/3 and s = 1/2.
Substitute back to obtain OX = (1/3)a + (1/2)b. Since AX is two thirds of AQ, the remaining XQ is one third, giving AX:XQ = 2:1. Since BX is half of BP, BX:XP = 1:1.
The parameters are fractions of complete directed segments, not directly the requested part-to-part ratios. Translating t and s back into the geometry is the final stage, not an optional interpretation after the algebra has ended.
Numerical verification of the intersection
Use a concrete triangle to check the algebra: let O = (0, 0), A = (12, 0) and B = (0, 12). Then P = (8, 0), Q = (0, 9), and the vector result gives X = (4, 6).
From A to Q, the displacement is (−12, 9). Two thirds of it is (−8, 6), which takes A to (4, 6). From B to P, the displacement is (8, −12). Half of it is (4, −6), which also takes B to (4, 6). Both routes reach the same point with the claimed fractions.
This numerical check is not the proof for every triangle; the vector derivation supplied that. Its role is to expose a possible sign, fraction or coefficient error in a transparent special case. A check should complement the proof without being mistaken for it.
Write the geometric conclusion, not only the algebra
A line such as AQ = 4AP is powerful, but the requested conclusion may still need to be stated. Explain that the vectors have a shared point and parallel directions, so A, P and Q lie on one line. Then interpret the scalar if a ratio or point order is requested.
Similarly, MN = (1/2)AB proves parallelism and a length relationship. It does not prove that M, N, A and B are all collinear. Read precisely which geometric claim follows from the calculation. A vector proof fails when its final sentence claims more than its algebra established.
A short error diagnosis before more practice
| Visible mistake | The missing distinction | A focused repair |
|---|---|---|
| AB is written as a − b | Start and finish have been reversed. | Say the route A to O to B before simplifying. |
| AP:PB = 2:3 becomes AP = (2/3)AB | Part-to-part ratio is confused with part of the whole. | Mark five equal parts on the segment. |
| A magnitude is given when a vector is requested | Length is confused with directed displacement. | State components first, then magnitude separately. |
| Parallelism is called collinearity | A common point or line has not been established. | Compare vectors with the same starting point. |
| Only one coefficient is matched | The scalar must act on the entire vector. | Check every component with the same multiplier. |
| An intersection fraction becomes the wrong ratio | A part of a whole is confused with two parts. | Convert t into t:(1 − t). |
Independent practice
For the symbolic questions, use OA = a and OB = b, with O, A and B forming a non-degenerate triangle unless another shape is specified. Write the connected route before the simplified expression. The purpose is to practise the decision that determines the sign and fraction.
- A = (−1, 4) and B = (5, −4). Find AB as a vector and find its magnitude.
- u = (3, −2) and v = (−1, 5). Find 2u − v.
- The vectors (2, −3) and (k, 12) are parallel. Find k and describe their relative directions.
- P lies on AB with AP:PB = 3:2. Find OP.
- M and N are the midpoints of OA and OB. Express MN in terms of a and b and state the area ratio OMN:OAB.
- OP = (1/4)a + (3/4)b and OQ = −(1/2)a + (3/2)b. Show that P is the midpoint of AQ.
- AB = p and BC = q as vectors. Express CA in terms of p and q.
- Find the magnitude of −3u when u = (−3, 4). Explain why the answer is not negative.
- O, A, C and B are consecutive vertices of a parallelogram. D is the midpoint of AC. Find BD.
- P lies on OA with OP = (3/4)OA, and Q lies on OB with OQ = (2/3)OB. AQ and BP intersect at X. Find OX, AX:XQ and BX:XP.
Answers with the checks that matter
1. AB = (5 − (−1), −4 − 4) = (6, −8), and its magnitude is 10. Adding this displacement to A returns B.
2. 2u − v = (6, −4) − (−1, 5) = (7, −9). Subtract both components, including the negative horizontal component of v.
3. The vertical component requires multiplier −4. Hence k = −8. The vectors point in opposite directions, and the second has four times the magnitude of the first.
4. OP = a + (3/5)(b − a) = (2/5)a + (3/5)b. P is three fifths of the way from A to B, so it lies closer to B.
5. MN = (1/2)(b − a) = (1/2)AB. The length scale factor is one half, and the area ratio is 1:4.
6. AP = (3/4)(b − a), while AQ = (3/2)(b − a) = 2AP. Both start at A and point in the same direction, so P lies halfway from A to Q. Therefore P is the midpoint of AQ.
7. AC = p + q, so CA = −p − q. The negative sign reverses the whole journey, not only its first segment.
8. |u| = 5, so |−3u| = 15. The negative scalar reverses direction. It does not create a negative length.
9. OD = a + b/2. Subtracting OB gives BD = a − b/2. Starting from B and adding that vector reaches the midpoint of AC.
10. Write OX = (1 − t)a + (2t/3)b along AQ, and OX = (3s/4)a + (1 − s)b along BP. Equate coefficients: 1 − t = 3s/4 and 2t/3 = 1 − s. Solving gives t = 1/2 and s = 2/3. Therefore OX = a/2 + b/3, AX:XQ = 1:1, and BX:XP = 2:1.
How to practise without turning vectors into symbol guessing
Begin with route-only questions. Ask the learner to express AC through a named intermediate point, then reverse the requested direction. No numbers are necessary. A student who can explain the route but makes an arithmetic error needs a different repair from a student who cannot decide whether a vector should be added or subtracted.
Next, mix internal ratios, midpoints and extensions beyond a segment. Require the student to identify whether a fraction describes AP/AB or AP/PB. Use simple numerical endpoints to test the symbolic expression. The coordinate check should make the position visible, not replace the reasoning that produced it.
Finally, give a proof whose conclusion changes: first ask for parallelism, then for collinearity, then for a ratio. The learner must state what additional information each conclusion requires. This is a practical way to separate completing algebra from understanding what that algebra proves.
Four questions to settle before the next paper
Why is AB equal to b − a? Because travelling from A to B can be decomposed into A to O and O to B. Those displacements are −a and b. End minus start is a connected journey, not an arbitrary sign convention.
Can a zero vector be used to decide a direction? A zero vector has magnitude zero and no unique direction. Be careful with a proof that tries to infer a line direction from it. The parallelism and collinearity arguments here use non-zero directed segments where a direction is needed.
Why compare coefficients only when a and b are non-parallel? Then the two vectors provide independent directions. A combination equalling another combination forces the corresponding coefficients to match. With parallel vectors, one direction can be expressed using the other, so the coefficient representation is not unique.
Does a scalar multiple always give the requested ratio? It gives a comparison of the complete vectors in that equation. If AQ = 4AP, then AP:AQ is 1:4, but AP:PQ is 1:3 when the points are in the stated order. Read the requested segment pair before reporting the ratio.
Sources, boundaries and connected learning
For assessed scope, consult the official 2026 Mathematics 4052 syllabus and 2027 G3 Mathematics K310 syllabus. The worked routes and practice problems are original teaching material. A school may sequence them differently; mastery of this guide is not a claim that the entire examination syllabus has been covered.
Connect vectors to Coordinate Geometry and Transformations for numerical position checks, Ratio, Percentage and Rates for part-to-whole reasoning, and Circle Theorems and Geometrical Proof for another way to make geometric conclusions depend on explicit conditions.
A good vector solution can be read as a journey. You know where it starts, where it finishes, which direction each part takes and what proportion of the full segment has been travelled. Once those meanings are secure, the algebra becomes a record of the geometry rather than a substitute for understanding it.
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