SPEC 10.5 • VECTORS
Geometric Problem-Solving with Vectors
Using vectors to prove geometric properties and solve problems
Why Use Vectors for Geometry?
Vectors provide a powerful algebraic approach to geometric proofs.
Key Principle
Two vectors are equal if they have the same magnitude AND direction.
Two vectors are parallel if one is a scalar multiple of the other: a ∥ b means a = kb
What We Can Prove
Points are collinear (lie on the same line)
Lines are parallel • Describe quadrilaterals
A point divides a line in a given ratio
Find intersection points and angles
Parallel Vectors
Definition
a ∥ b ⟺ a = kb for some scalar k ≠ 0
Same Direction (k > 0)
If PQ = 2RS, then PQ ∥ RS and |PQ| = 2|RS|
Opposite Direction (k < 0)
If PQ = −3RS, then PQ ∥ RS but opposite
Describing Quadrilaterals
ShapeVector ConditionParallelogramAB = DC (equal and parallel)TrapeziumAB = kDC where k ≠ 1RhombusParallelogram with |AB| = |BC|
Worked Example01
A, B, C, D have coordinates (2,−5,−8), (1,−7,−3), (0,15,−10), (2,19,−20). Show AB ∥ DC and find |AB|:|DC|. What type of quadrilateral is ABCD?
1
AB = (1−2, −7−(−5), −3−(−8)) = (−1, −2, 5)
2
DC = (0−2, 15−19, −10−(−20)) = (−2, −4, 10)
3
DC = 2(−1, −2, 5) = 2AB ✓ So AB ∥ DC
4
|DC| = 2|AB|, so |AB|:|DC| = 1:2
ABCD is a trapezium (one pair of parallel sides with different lengths). ✓
Proving Collinearity
Three points A, B, C are collinear if they lie on the same straight line.
Collinearity Test
Points A, B, C are collinear if and only if:AB = kAC for some scalar kOR equivalently: AB and BC are parallel (and share point B)
Important
The vectors must share a common point! AB and CD being parallel does NOT prove A, B, C, D are collinear.
Worked Example02
Prove that A(1, 2), B(4, 5), and C(7, 8) are collinear.
1
AB = (4−1, 5−2) = (3, 3)
2
AC = (7−1, 8−2) = (6, 6)
3
AC = 2AB since (6, 6) = 2(3, 3) ✓
4
Since AC = 2AB, the vectors are parallel and share point A.
Therefore, A, B, and C are collinear. ✓
Point Dividing a Line in a Ratio
Key Formula
If P divides line segment AB in ratio λ : μ (from A), then:OP = OA + ^λ/_λ+μ AB
Common Ratios
Ratio λ:μFractionAP1:1 (midpoint)½½AB1:2⅓⅓AB2:1⅔⅔AB
Worked Example03
Points A and B have position vectors a and b. Point P divides AB in the ratio 1:2. Find the position vector of P.
1
Ratio 1:2 means λ = 1, μ = 2, so P is ¹/₃ of the way from A to B
2
OP = OA + ¹/₃AB
3
AB = b − a
4
OP = a + ¹/₃(b − a) = a + ¹/₃b − ¹/₃a = ²/₃a + ¹/₃b
OP = ²/₃ a + ¹/₃b ✓
Equating Coefficients
Key Principle
If a and b are non-parallel vectors, and:pa + qb = ra + sbThen p = r and q = s
How to Use
Write two expressions for the same vector using different routes
Collect terms in the form (coefficient of a) + (coefficient of b)
Equate coefficients of a and coefficients of b
Solve the simultaneous equations
04Worked Example
OABC is a parallelogram with OA = a and OC = c. Prove that the diagonals OB and AC bisect each other.
1
OB = a + c. Let P lie on OB: OP = λ(a + c)
2
Let P also lie on AC: OP = a + μ(c − a) = (1−μ)a + μc
3
Equating: λa + λc = (1−μ)a + μc
4
Coefficients of a: λ = 1−μ. Coefficients of c: λ = μ
5
From λ = μ and λ = 1−μ: μ = 1−μ, so 2μ = 1, μ = ½, λ = ½
P is the midpoint of both diagonals (λ = μ = ½). ✓
05Worked Example
In triangle OAB, OA = a and OB = b. Point M divides OA in ratio 2:1 and point N divides OB in ratio 1:3. Lines AN and BM intersect at P. Find OP.
1
OM = ²/₃a, ON = ¹/₄b
2
P on AN: OP = a + t(¹/₄b − a) = (1−t)a + ^t/₄b
3
P on BM: OP = b + s(²/₃a − b) = ^2s/₃a + (1−s)b
4
Equate a: 1−t = ^2s/₃. Equate b: ^t/₄ = 1−s
5
From b: t = 4−4s. Substitute: 1−(4−4s) = ^2s/₃ → −3+4s = ^2s/₃ → s = ⁹/₁₀
6
t = 4−4×⁹/₁₀ = ²/₅
OP = ³/₅ a + ¹/₁₀b ✓
06Worked Example
In triangle OAB, C lies on AB with AC:CB = 1:3, and D lies on OB with OD:DB = 2:1. Lines OC and AD intersect at P. Find OP .
1
OC = a + ¹/₄(b−a) = ³/₄a + ¹/₄b, OD = ²/₃b
2
P on OC: OP = t(³/₄a + ¹/₄b) = ^3t/₄a + ^t/₄b
3
P on AD: OP = a + s(²/₃b − a) = (1−s)a + ^2s/₃b
4
Equate a: ^3t/₄ = 1−s. Equate b: ^t/₄ = ^2s/₃ → t = ^8s/₃
5
Substitute: ³/₄ × ^8s/₃ = 1−s → 2s = 1−s → s = ¹/₃, t = ⁸/₉
OP = ²/₃ a + ²/₉b ✓
Worked Example07
P, Q, R have coordinates (4, −9, −3), (7, −7, −7), (8, −2, 0). Find S such that PQRS is a parallelogram.
1
For a parallelogram PQRS: QP = RS
2
QP = (4−7, −9−(−7), −3−(−7)) = (−3, −2, 4)
3
OS = OR + RS = (8, −2, 0) + (−3, −2, 4)
4
OS = (5, −4, 4)
S has coordinates (5, −4, 4) ✓
Finding Angles Using Cosine Rule
Cosine Rule Method
To find angle B in triangle ABC, use the cosine rule:
Method
Find |BA| and |BC| (the two sides meeting at B)
Find |AC| (the side opposite to angle B)
Apply the cosine rule formula
Use inverse cosine to find the angle
Worked Example08
In triangle ABC, AB = 3 i − 2j and AC = i − 5j. Find angle BAC.
1
|AB| = √(9 + 4) = √13
2
|AC| = √(1 + 25) = √26
3
BC = AC − AB = (1−3, −5−(−2)) = (−2, −3)
4
|BC| = √(4 + 9) = √13
5
cos A = ^13 + 26 − 13/_2√13 × √26 = ²⁶/_2√338 = ²⁶/_2 × 13√2 = ¹/_√2
∠BAC = cos⁻¹(¹/_√2) = 45° ✓
3D Shapes: Parallelepiped
Parallelepiped
A 3D figure with 6 parallelogram faces. With OA = a, OB = b, OC = c:
OD = a + b
OG = a + b + c
OF = b + c
Space Diagonals
The four space diagonals (like OG) all pass through the centre and bisect each other.
09Worked Example
Prove that the space diagonals OG and AF of a parallelepiped bisect each other.
1
OG = a + b + c. Let H lie on OG: OH = λ(a + b + c)
2
OF = b + c, so AF = OF − OA = −a + b + c
3
H on AF: OH = a + μ(−a + b + c) = (1−μ)a + μb + μc
4
Equate: λ = 1−μ and λ = μ. So μ = 1−μ, giving μ = ½, λ = ½
H is the midpoint of both diagonals. ✓
10
Worked Example
In a parallelepiped with OA = a, OB = b, OC = c, point M lies on the space diagonal OG such that OM:MG = 1:3. Find OM and AM.
1
OG = a + b + c
2
OM:MG = 1:3 means M is ¹/₄ of the way from O to G
3
OM = ¹/₄OG = ¹/₄(a + b + c)
4
AM = OM − OA = ¹/₄(a + b + c) − a
5
AM = −³/₄a + ¹/₄b + ¹/₄c
OM = ¹/₄ a + ¹/₄b + ¹/₄c, AM = −³/₄a + ¹/₄b + ¹/₄c ✓
Practice
Question 1: Points A, B, C have position vectors (1, −4, 8), (4, 4, 7), (10, 0, 30). Show that |OA| = |OB| and |AC| = |BC|. Hence describe the quadrilateral OACB.
|OA| = √(1 + 16 + 64) = √81 = 9
|OB| = √(16 + 16 + 49) = √81 = 9 ✓
AC = (9, 4, 22), |AC| = √(81 + 16 + 484) = √581
BC = (6, −4, 23), |BC| = √(36 + 16 + 529) = √581 ✓
OACB is a kite (two pairs of adjacent equal sides).
Practice
Question 2: In triangle OAB, point M divides OA in the ratio 2:1. Point N lies on AB such that MN is parallel to OB. Find the position vector of N in terms of a and b.
OM = ²/₃a
Since MN ∥ OB, by similar triangles, N divides AB in the ratio 2:1.
ON = OA + ²/₃AB = a + ²/₃(b − a)
ON = ¹/₃ a + ²/₃b ✓
Practice
Question 3: A(7, 12, −1), B(11, 2, −9), C(14, −14, 3), D(8, 1, 15). Show that AB ∥ CD and find the ratio |AB|:|CD|.
AB = (11−7, 2−12, −9−(−1)) = (4, −10, −8)
CD = (8−14, 1−(−14), 15−3) = (−6, 15, 12)
CD = −1.5(4, −10, −8) = −1.5AB ✓
|CD| = 1.5|AB|, so |AB|:|CD| = 1:1.5 = 2:3
ABCD is a trapezium.
Summary
Key Results
To Prove/FindMethod
Parallel linesAB = k CD
Collinear points A, B, CAB = k AC (share point A)
P divides AB in ratio λ:μOP = OA + ^λ/_λ+μ AB
Intersection pointEquate coefficients of a and b
Angle between vectorsCosine rule
Exam Techniques
For parallelogram: AB = DC
For trapezium: AB = kDC where k ≠ 1
Equating coefficients: pa + qb = ra + sb ⟹ p = r, q = s
Next: Why Use Vectors for Geometry?