A vector is not merely an arrow with numbers beside it. It is a compact way to represent movement, displacement and geometric relationship. The same vector can appear as a directed line segment, a column vector, a position vector, a scalar multiple or a translation. The notation changes; the directed relationship remains.
This forty-ninth Secondary 4 Mathematics Learning Guide develops vectors as a representation system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
It deepens Vectors, Ratios and Collinearity by focusing on how one vector relationship is translated across several forms. The current SEC G3 Mathematics syllabus includes vector notation, position vectors, magnitude, scalar multiplication, vector addition and subtraction, and geometric applications.
A vector carries magnitude and direction
A scalar has magnitude only. A vector has magnitude and direction. A distance of 5 km is scalar. A displacement of 5 km east is vector.
Direction is part of the value.
This is why reversing a vector changes its sign. If vector AB is a, then vector BA is −a.
Worked Example 1 | Reverse a directed segment
If AB=(4,−3), find BA.
BA=(−4,3).
Every component reverses because the direction of travel has reversed.
Column vectors encode horizontal and vertical change
A column vector such as (3,−2) means move 3 units in the positive x-direction and 2 units in the negative y-direction.
The components are not coordinates of a point unless the context specifically defines them as a position vector from the origin.
Worked Example 2 | Translate a point
Point P(2,5) is translated by vector (4,−3). Find P′.
P′=(2+4,5−3)=(6,2).
The vector represents the same movement regardless of the starting point.
Position vectors locate points relative to the origin
If O is the origin and OP=p, then p is the position vector of P. It describes the displacement from O to P.
If OA=a and OB=b, then:
AB=OB−OA=b−a.
Worked Example 3 | Find a vector between two points
OA=(2,1) and OB=(7,9). Find AB.
AB=(7,9)−(2,1)=(5,8).
This is the same subtraction rule used with coordinates: destination minus start.
Vector addition represents consecutive movement
If AB=a and BC=b, then travelling A→B→C gives:
AC=a+b.
This is sometimes called the triangle law of vectors. The intermediate point disappears from the net displacement.
Worked Example 4 | Two-stage movement
AB=(3,4) and BC=(−1,6). Find AC.
AC=(3,4)+(−1,6)=(2,10).
Scalar multiples preserve or reverse direction
If b=ka, the vectors are parallel. If k>0, they point in the same direction. If k<0, they point in opposite directions. The magnitude is multiplied by |k|.
Worked Example 5 | Read a scalar multiple
Let a=(2,−5). Find −3a.
−3a=(−6,15).
The magnitude triples and the negative scalar reverses direction.
Magnitude turns a vector into a scalar length
For vector (x,y), magnitude is:
√(x²+y²).
This comes directly from Pythagoras. The magnitude has no direction; it is the length of the vector.
Worked Example 6 | Magnitude
Find the magnitude of vector (6,8).
√(6²+8²)=√100=10.
Equal vectors need equal direction and equal magnitude
Two vectors can be equal even when they are drawn in different places. What matters is that their components, direction and magnitude are the same.
This lets vectors describe repeated geometric structure such as opposite sides of a parallelogram.
Worked Example 7 | Parallelogram structure
In parallelogram ABCD, AB=p and AD=q. Express AC.
Travel A→B→C. Since BC=AD=q:
AC=p+q.
Ratios on a line create fractional vectors
If P divides AB internally in a known ratio, the vector AP is a corresponding fraction of AB.
Worked Example 8 | Divide a segment
P lies on AB with AP:PB=2:3. If AB=v, find AP.
AP=2/5 v.
The whole segment is five equal ratio parts; AP occupies two of them.
Position vector of a point on a segment
If OA=a and AB=v, then a point P with AP=λv has:
OP=a+λv.
This is a useful template because it separates starting position from movement along the segment.
Worked Example 9 | Position vector by ratio
OA=a, OB=b, and P divides AB so that AP:PB=1:3. Find OP.
AB=b−a. Since AP is one quarter of AB:
OP=a+1/4(b−a)=3/4 a+1/4 b.
Collinearity is a scalar-multiple statement
If AB=kAC for some scalar k, then AB and AC are parallel. If they also share the point A, the points A, B and C lie on the same straight line.
Worked Example 10 | Prove collinearity
AB=(4,6) and AC=(10,15).
AC=2.5AB.
Therefore AB and AC are parallel and share A, so A, B and C are collinear.
Translation is a vector transformation
A translation moves every point by the same vector. Shape, size, orientation and parallel relationships are preserved.
Worked Example 11 | Translate a triangle
A(1,2), B(4,2), C(2,6) are translated by (−3,5).
- A′=(−2,7)
- B′=(1,7)
- C′=(−1,11)
Every vertex receives the same displacement.
The route method for vector geometry
In complicated diagrams, write a route from start to destination. If you need AC, one route might be A→B→C. Another might be A→D→C. Both routes must produce the same net vector.
Equating two valid routes can reveal an unknown scalar or prove a geometric relationship.
Worked Example 12 | Equivalent routes
Suppose AB=a, BC=b, AD=c, and DC=d. Since both routes reach C:
a+b=c+d.
This equation is not a new vector rule. It is simply the statement that the final displacement from A to C is unique.
Vector representation audit
- What are the start and end points?
- Is this a position vector from the origin or a displacement between two points?
- Have I preserved direction?
- Can the route be written as a sum or difference?
- Does a scalar multiple imply parallelism or collinearity?
- Does the question need the vector itself or only its magnitude?
- Can a second route verify the result?
Common failure modes
| Failure | Cause | Repair |
|---|---|---|
| Uses AB and BA interchangeably | Direction ignored | Reverse sign when route reverses |
| Treats a vector as a point | Position and displacement mixed | State what the components represent |
| Finds magnitude by adding components | Pythagorean structure lost | Use √(x²+y²) |
| Says scalar multiples are always same direction | Negative scalar ignored | Use sign of scalar to determine direction |
| Proves parallel but not collinear | Shared point condition omitted | State both scalar multiple and common line point |
| Adds position vectors when a displacement is required | Destination-minus-start rule missed | Use AB=OB−OA |
Independent practice
- If AB=(5,−2), find BA.
- Translate P(−1,4) by vector (6,−3).
- OA=(2,5), OB=(9,1). Find AB.
- Find the magnitude of (8,15).
- If a=(3,−4), find 2a and −1/2a.
- P divides AB in the ratio AP:PB=3:2. Express AP in terms of AB.
- AB=(6,9), AC=(2,3). Explain whether A, B and C are collinear.
Explained answers
1. (−5,2).
2. P′=(5,1).
3. (9,1)−(2,5)=(7,−4).
4. √(64+225)=√289=17.
5. 2a=(6,−8); −1/2a=(−3/2,2).
6. AP=3/5 AB.
7. AB=3AC. The vectors are scalar multiples and share A, so A, B and C are collinear.
Final thought
Vectors become easier when they are treated as one relationship moving through several representations. Directed segments, position vectors, column vectors and translations all describe displacement with direction preserved.
Track the route, preserve the direction, and use the representation that makes the geometry easiest to see.
Return to the Secondary Mathematics Hub.
