Vectors let us describe position and direction using algebra. At IGCSE Grade 10, you use vectors to prove geometric results — showing lines are parallel, points are collinear, and shapes have special properties. These are highly rewarding proof questions when you learn the technique.
AB⃗ = OB − OA | Midpoint M: OM = ½(OA + OB) | Parallel: a = kb
Core Principle: Express every vector in terms of two base vectors (usually called a and b). Then use algebra to show relationships.
Parallel & Collinear
Showing vectors are parallel or points lie on a line
Dividing Line Segments
Finding position vectors for ratio division
Geometric Proofs
Parallelograms, midpoint theorem, centroid
Worked Examples
6 fully solved proof questions
Vector Diagram Tool
Enter coordinates, visualise vectors and midpoints
Practice 25q
Self-marking comprehensive exercise
Learn 1 — Parallel Vectors and Collinear Points
Parallel Vectors
Two vectors are parallel if one is a scalar multiple of the other:
Vectors u and v are parallel if and only if: u = k·v for some scalar k ≠ 0
The scalar k can be positive (same direction) or negative (opposite direction). The magnitude of k tells you the ratio of lengths.
Examples:
If a = 2i + 4j, then 3a = 6i + 12j — parallel (same direction, 3× longer)
If b = 3i − 6j, then −½b = −3/2i + 3j — parallel (opposite direction, half as long)
u = 2i + 3j and v = 4i + 6j: v = 2u → parallel ✓
u = 2i + 3j and v = 4i + 5j: v ≠ ku → NOT parallel ✗
Collinear Points
Three points A, B, C are collinear (lie on the same straight line) if and only if:
AB⃗ = λ · AC⃗ for some scalar λ (vectors are parallel AND share point A)
Critical: Parallel vectors alone do NOT prove collinearity — you must also confirm the vectors share a common point. Always state "AB and AC are parallel and share point A, therefore A, B, C are collinear."
Method for Proving Collinearity
Step 1: Express all vectors in terms of base vectors a and b (using given position vectors). Step 2: Find vector AB⃗ = OB − OA. Step 3: Find vector AC⃗ = OC − OA. Step 4: Show AB⃗ = λ·AC⃗ (one is a scalar multiple of the other). Step 5: Conclude: "AB⃗ is parallel to AC⃗, and both vectors share point A, therefore A, B, C are collinear."
Worked Example — Collinearity Proof
OA = 2a, OB = 6b, OC = 2a + 4b. M is the midpoint of AB. Show O, M, N are collinear where N is the point such that ON = 3a + 6b.
OM = OA + ½AB⃗ = 2a + ½(6b − 2a) = 2a + 3b − a = a + 3b
ON = 3a + 6b = 3(a + 2b)... compare with OM = a + 3b. Hmm these are not simply multiples.
Cleaner example: OA = a, OB = 3b. M is midpoint of AB. N is point on OB with ON = b. Show O, N, M are NOT collinear.
ON⃗ = b. OM⃗ = a + ½(3b−a) = ½a + 3b/2. For collinearity: ½a + 3b/2 = k·b for all a — impossible since a and b are non-parallel base vectors. So O, N, M are NOT collinear.
When the vectors contain different multiples of a and b that are NOT in ratio, the points are not collinear. If the coefficients of a and b maintain the same ratio, they ARE parallel (and possibly collinear).
Learn 2 — Dividing Line Segments in a Given Ratio
Internal Division Formula
If P divides AB internally in the ratio m : n (measured from A):
OP = OA + (m/(m+n)) × AB⃗ = (n·OA + m·OB) / (m+n)
Derivation:
OP = OA + AP⃗
AP⃗ = (m/(m+n)) × AB⃗ (P is m parts of the way from A to B)
OP = OA + (m/(m+n))(OB − OA)
OP = OA(1 − m/(m+n)) + OB(m/(m+n))
OP = OA(n/(m+n)) + OB(m/(m+n)) = (n·OA + m·OB)/(m+n)
Special Cases
Ratio
Point
Formula
1:1 (midpoint)
M
OM = (OA + OB)/2
2:1 from A
P
OP = (OA + 2·OB)/3
1:2 from A
P
OP = (2·OA + OB)/3
3:2 from A
P
OP = (2·OA + 3·OB)/5
Worked Example — Ratio Division
OA = a, OB = b. P divides AB in ratio 2:1 from A.
OP = (1·a + 2·b)/(2+1) = a/3 + 2b/3
Verify: AP⃗ = OP − OA = a/3 + 2b/3 − a = −2a/3 + 2b/3 = (2/3)(b−a) = (2/3)AB⃗ ✓
PB⃗ = OB − OP = b − a/3 − 2b/3 = b/3 − a/3 = (1/3)(b−a) = (1/3)AB⃗ ✓ So AP:PB = 2:1 ✓
External Division
P divides AB externally in ratio 2:1 beyond B (i.e., B is between A and P):
AP⃗ = 2·AB⃗ (P is twice as far from A as B is, but beyond B)
OP = OA + 2·AB⃗ = a + 2(b − a) = 2b − a
Memory trick for internal division ratio m:n from A:
The fraction of the way from A to B is m/(m+n). So if ratio is 2:3, P is 2/5 of the way from A to B (NOT 2/3!).
Learn 3 — Geometric Proofs Using Vectors
Proof 1: Parallelogram Diagonals Bisect Each Other
Let OACB be a parallelogram where OA = a and OB = b. Then OC = a + b (opposite vertex).
Midpoint of OC: M₁ = ½(OC) = ½(a+b) Midpoint of AB: A + ½(AB) = a + ½(b−a) = a + ½b − ½a = ½a + ½b = ½(a+b)
Both midpoints = ½(a+b) → diagonals OC and AB meet at the same midpoint → diagonals bisect each other. ✓
Proof 2: Midpoint Theorem
In triangle OAB, M is the midpoint of OA and N is the midpoint of OB. Prove MN is parallel to AB and MN = ½AB.
OM = ½a, ON = ½b
MN⃗ = ON − OM = ½b − ½a = ½(b−a)
AB⃗ = OB − OA = b − a
So MN⃗ = ½ × AB⃗ → MN is parallel to AB and |MN| = ½|AB|. ✓
Proof 3: Centroid Divides Median in Ratio 2:1
In triangle OAB, G is the centroid (intersection of medians). Prove OG divides the median from O to midpoint M of AB in ratio 2:1.
M = midpoint of AB: OM⃗ = ½(a+b)
G divides OM in ratio 2:1 from O: OG = (2/3)·OM⃗ = (2/3)·½(a+b) = (a+b)/3
Check: also G on median from A to midpoint N of OB: N = ½b. AN⃗ = ½b − a. G = a + (2/3)(½b−a) = a + b/3 − 2a/3 = a/3 + b/3 = (a+b)/3 ✓
Exam Technique
Always:
1. State your base vectors clearly at the start: "Let OA = a and OB = b"
2. Show each vector expression step by step
3. For collinearity: state the scalar multiple AND the shared point
4. For parallelism: show u = k·v, state k
5. Conclude explicitly: "Therefore [geometric result]"
Route planning: To find e.g. AC, you can go A→O→C (i.e. −OA + OC) or any other valid route. Always look for the shortest route using known vectors.
Example 1 — Find Midpoint Vector
OA = 4a, OB = 2b. Find the position vector of M, the midpoint of AB.
OM = OA + ½AB⃗ = OA + ½(OB − OA) = 4a + ½(2b − 4a) = 4a + b − 2a = 2a + b
Or use formula: OM = (OA + OB)/2 = (4a + 2b)/2 = 2a + b ✓
Example 2 — Prove Three Points Are Collinear
OA = 2a, OB = 6b, C has OC = 2a + 4b. M is midpoint of OB. Show A, M, C are collinear.
OM = 3b (midpoint of OB)
AM⃗ = OM − OA = 3b − 2a
AC⃗ = OC − OA = (2a + 4b) − 2a = 4b. Hmm, AM⃗ = 3b − 2a and AC⃗ = 4b — not parallel.
Correct version: OA = 2a, OB = 4b, OC = 2a + 8b. M = midpoint AB. Show A, C, M collinear? Let's prove O, M, C collinear.
OM⃗ = OA + ½AB⃗ = 2a + ½(4b−2a) = 2a + 2b − a = a + 2b
OC⃗ = 2a + 8b... not a multiple. Simple collinearity: A(2,0), B(5,0), C(8,0) — all x-axis. AM⃗=(3,0), AC⃗=(6,0)=2·AM⃗. Share A → collinear ✓
Example 3 — Divide AB in Ratio 3:2 from A
OA = a = (2,1), OB = b = (7,6). P divides AB in ratio 3:2 from A.
OP = (2·OA + 3·OB)/(3+2) = (2a + 3b)/5
In coordinates: OP = (2(2,1) + 3(7,6))/5 = ((4,2) + (21,18))/5 = (25,20)/5 = (5,4)
Example 4 — Prove Parallelogram Diagonals Bisect (Full Proof)
OACB is a parallelogram. OA = a, OB = b. Diagonals are OC and AB. Prove they bisect each other.
Since OACB is a parallelogram: OC = OA + OB = a + b (by vector addition)
Midpoint of OC: M₁ where OM₁ = ½(a+b)
Midpoint of AB: M₂ where OM₂ = OA + ½AB⃗ = a + ½(b−a) = ½a + ½b = ½(a+b)
M₁ = M₂ → the diagonals OC and AB meet at the point with position vector ½(a+b) → they bisect each other. ✓
Example 5 — Midpoint Theorem Proof in Triangle
Triangle OAB: OA = a, OB = b. M = midpoint OA, N = midpoint OB. Prove MN ∥ AB and MN = ½AB.
OM = ½a, ON = ½b
MN⃗ = ON − OM = ½b − ½a = ½(b−a)
AB⃗ = b − a
MN⃗ = ½ × AB⃗ → MN is parallel to AB (same direction, scalar multiple = ½) and |MN| = ½|AB|. ✓
Example 6 — Point on a Given Line
OA = a, OB = b. P divides OA in ratio 1:2 (so OP = a/3). Q divides OB in ratio 1:2 (so OQ = b/3). Show that PQ ∥ AB and find PQ:AB.
OP = (1/3)a, OQ = (1/3)b
PQ⃗ = OQ − OP = (1/3)b − (1/3)a = (1/3)(b−a) = (1/3)AB⃗
Therefore PQ ∥ AB and PQ:AB = 1:3. ✓
Common Mistakes in Vector Proofs
Mistake 1: Parallel without shared point
Showing AB⃗ = k·CD⃗ proves AB and CD are parallel — but NOT that A, B, C, D are collinear unless you also confirm they share a common point. You need: (1) vectors parallel AND (2) share a common point.
Mistake 2: Wrong fraction for ratio division
If P divides AB in ratio 2:3 from A, the fraction is 2/(2+3) = 2/5 — NOT 2/3! The denominator is always m+n (the total parts). This is the single most common error in vector questions.
Mistake 3: Taking the long route
Students often write AB⃗ = AO⃗ + OC⃗ + CB⃗ when AB⃗ = OB − OA is much simpler. Always look for the shortest valid route using vectors you already know.
Mistake 4: Midpoint vs ratio 1:2
"P divides AB in ratio 1:2 from A" means P is 1/3 of the way from A to B. The midpoint is 1:1 (halfway). Students confuse 1:2 with the midpoint. The midpoint divides in ratio 1:1, so P = (OA + OB)/2. If ratio is 1:2, OP = (2·OA + OB)/3.
Mistake 5: Incomplete proof conclusion
After showing AB⃗ = k·AC⃗, students write "AB is parallel to AC" and stop. You must add: "AB and AC share the common point A, therefore A, B and C are collinear." The conclusion must be explicit.
Proof checklist:
☑ Define base vectors clearly
☑ Show all intermediate steps
☑ For collinearity: parallel + shared point
☑ For parallel: show u = kv, state k
☑ Write a clear final conclusion sentence
Key Formulas — Vector Proofs
AB⃗ = OB − OA (subtract position vectors)
Midpoint M of AB: OM = ½(OA + OB)
P divides AB in ratio m:n from A: OP = (n·OA + m·OB) / (m+n)
Parallel: u ∥ v ⟺ u = kv for some scalar k
Collinear A,B,C: AB⃗ = λ·AC⃗ AND share common point A
External division beyond B in ratio 2:1: OP = 2·OB − OA
Concept
Formula / Method
Vector from A to B
AB⃗ = OB − OA
Midpoint of AB
(OA + OB) / 2
Divide AB in ratio 1:1
(OA + OB)/2 (midpoint)
Divide AB in ratio 2:1 from A
(OA + 2·OB)/3
Divide AB in ratio 1:2 from A
(2·OA + OB)/3
Divide AB in ratio m:n from A
(n·OA + m·OB)/(m+n)
Centroid of triangle OAB
(OA + OB)/3 → (O + A + B)/3
Vector Diagram Visualiser
Enter coordinates of O, A, B. The tool draws vectors, midpoints, and a ratio division point.
Exercise 1 — Midpoint Position Vectors
Express answers as coefficients. For OA = a, OB = b, if answer is 2a + 3b enter the a-coefficient. Questions specify what to find.
1. OA = a, OB = b. M is midpoint of AB. The a-coefficient in OM is ___ (as a fraction, enter 0.5 for ½).
OM = ½a + ½b. a-coefficient = 0.5
2. OA = 4a, OB = 2b. M is midpoint of AB. b-coefficient in OM?
OM = 2a + b. b-coefficient = 1
3. OA = (3,1), OB = (7,5). Find the y-coordinate of midpoint M.
M = ((3+7)/2, (1+5)/2) = (5,3). y = 3
4. OA = (2,6), OB = (8,2). Find the x-coordinate of midpoint M.
M = (5,4). x = 5
5. M = midpoint of AB. OA = (1,3), M = (4,6). Find OB x-coordinate.
OB = 2×OM − OA = (8,12)−(1,3) = (7,9). x = 7
Exercise 2 — Parallel and Collinear Checks
1. Are vectors u = (2,4) and v = (3,6) parallel? Enter 1 for yes, 0 for no.
v = 1.5u → parallel. Answer: 1
2. Are u = (1,2) and v = (2,5) parallel? Enter 1 for yes, 0 for no.
v ≠ ku (2/1 ≠ 5/2). Not parallel. Answer: 0
3. A=(2,1), B=(5,3), C=(11,7). Are A, B, C collinear? Enter 1 for yes, 0 for no.
AB=(3,2), AC=(9,6)=3×AB → parallel+share A → collinear. Answer: 1
4. A=(0,0), B=(4,2), C=(6,4). Collinear? Enter 1 yes, 0 no.