Vectors | Ronny Best Mathematics

Section 01 · Foundations

What is a Vector?

Junior Secondary

In everyday life, we distinguish between quantities that only need a size — like temperature or mass — and quantities that also need a direction — like force or wind. Understanding this distinction is the starting point of all vector mathematics.

Scalars and Vectors Compared

A scalar is a quantity described completely by its magnitude (size) alone. Temperature, time, mass, distance, and speed are all scalars. A vector, however, requires both magnitude and direction to be fully described. Displacement, velocity, acceleration, force, and momentum are all vectors.

The distinction matters enormously in real life. Saying "run 5 km" is incomplete — you must say "run 5 km north." That direction makes it a displacement vector, not merely a distance scalar. Similarly, a force of 10 N acting downward is very different from 10 N acting sideways, even though their magnitudes are equal.

✓ Scalars (magnitude only)

Mass: 60 kg

Temperature: 37 °C

Speed: 80 km/h

Time: 2 hours

⃗ Vectors (magnitude + direction)

Displacement: 5 km North

Velocity: 80 km/h East

Force: 100 N downward

Acceleration: 9.8 m/s² down

Why Vectors Matter in the CBC Curriculum

The KICD CBC Pure Mathematics curriculum introduces vectors because they provide the mathematical language for describing physical reality — from the trajectory of a projectile to the stress forces in a building. In the STEM pathway, vectors underpin physics, engineering, and computer graphics. Mastery of vectors is also essential for calculus-based work at university level, where vector fields describe phenomena such as gravitational and electromagnetic force distributions.

KCSE Exam Note: KCSE Paper 2 regularly tests vectors in the context of geometry problems — finding midpoints, dividing lines in given ratios, and proving collinearity. These problems combine position vectors with ratio concepts and are worth significant marks. Always show your vector working fully, step by step.

Section 02 · Language of Vectors

Vector Notation and Representation

Junior Secondary

Before working with vectors algebraically, you must be fluent in the different ways they are written and drawn. Different textbooks and examination papers use different conventions — understanding all of them prevents confusion.

Standard Notations for a Vector

Bold (print)

a, b, v — used in textbooks

Underline (handwriting)

a̲, b̲, v̲ — used in handwritten work and KCSE exams

Arrow over letters

AB⃗ — vector from point A to point B

Column vector (2D)

(x, y) or written as column: [x / y]

Column vector (3D)

(x, y, z) — three-dimensional space

Component form

xi + yj + zk — using unit vectors i, j, k

The i, j, k Unit Vector System

The most fundamental vectors in 3D space are the three unit vectors along the coordinate axes:

Standard Basis Unit Vectors

i (or î)

= (1, 0, 0) — one unit in the x-direction

j (or ĵ)

= (0, 1, 0) — one unit in the y-direction

k (or k̂)

= (0, 0, 1) — one unit in the z-direction

General vector

v = xi + yj + zk = (x, y, z)

Worked Examples — Vector Notation

Example 1 Write the vector (−3, 5, 2) in component form using i, j, k.

Step 1 Identify the x, y, z components: x = −3, y = 5, z = 2.

Step 2 Write each component multiplied by its unit vector: −3i + 5j + 2k.

v = −3i + 5j + 2k

Example 2 If A = (2, 1) and B = (7, 4), write the vector AB⃗ as a column vector.

Step 1 The vector AB⃗ goes from A to B. Use the rule: AB⃗ = B − A (subtract coordinates of the starting point).

Step 2 AB⃗ = (7 − 2, 4 − 1) = (5, 3).

Check Note: BA⃗ = A − B = (−5, −3). This is the reverse direction — same magnitude, opposite direction.

AB⃗ = (5, 3)

Section 03 · Core Concept

Position Vectors and Displacement

Senior Secondary

A position vector locates a point relative to a fixed origin O. This idea underpins nearly all KCSE vector geometry problems, including ratio division, midpoints, and proving collinearity.

Key Position Vector Relationships

Position of A from O

OA⃗ = a = (a₁, a₂, a₃)

Vector from A to B

AB⃗ = OB⃗ − OA⃗ = b − a

Midpoint M of AB

OM⃗ = ½(a + b)

Point dividing AB in ratio m:n (internally)

OP⃗ = (n·a + m·b) / (m + n)

Collinearity condition

Points A, B, C collinear if AB⃗ = k · AC⃗ for some scalar k

Worked Examples — Position Vectors (KCSE Style)

Example 1 — Midpoint O is the origin. OA⃗ = 2i + 3j and OB⃗ = 6i + j. Find the position vector of M, the midpoint of AB.

Step 1 Write out: a = (2, 3) and b = (6, 1).

Step 2 Midpoint formula: OM⃗ = ½(a + b) = ½[(2, 3) + (6, 1)] = ½(8, 4) = (4, 2).

Check Verify: AM⃗ = (4−2, 2−3) = (2, −1). MB⃗ = (6−4, 1−2) = (2, −1). Equal vectors confirm M is the midpoint. ✔

OM⃗ = 4i + 2j

Example 2 — Internal Division A has position vector a = (1, 2) and B has position vector b = (7, 8). Point P divides AB internally in the ratio 2:1. Find the position vector of P.

Step 1 Ratio m:n = 2:1, so m = 2, n = 1. Total parts = 3.

Step 2 Apply the section formula: OP⃗ = (n·a + m·b)/(m + n) = (1·(1,2) + 2·(7,8)) / 3.

Step 3 = [(1, 2) + (14, 16)] / 3 = (15, 18) / 3 = (5, 6).

OP⃗ = 5i + 6j

Example 3 — Collinearity (KCSE-style) OA⃗ = 2a, OB⃗ = 3b, OC⃗ = 4a + 3b. Show that A, B, C are not collinear, OR find the value of a scalar if they were. [Typical KCSE phrasing]

Step 1 Find AB⃗ = OB⃗ − OA⃗ = 3b − 2a.

Step 2 Find AC⃗ = OC⃗ − OA⃗ = (4a + 3b) − 2a = 2a + 3b.

Step 3 For collinearity, AB⃗ = k · AC⃗ for some scalar k. So: 3b − 2a = k(2a + 3b) = 2ka + 3kb. Comparing a-components: −2 = 2k → k = −1. Comparing b-components: 3 = 3k → k = 1. Since k = −1 and k = 1 are contradictory, A, B, C are not collinear.

A, B, C are not collinear (no single value of k satisfies both conditions).

Section 04 · Core Algebra

Vector Operations

Senior Secondary

Vectors can be added, subtracted, and scaled (multiplied by a scalar). These operations follow clear rules that are consistent across 2D and 3D space. Understanding these operations allows you to solve force problems, velocity problems, and geometric proofs.

Vector Operation Rules

Addition

(a₁, a₂, a₃) + (b₁, b₂, b₃) = (a₁+b₁, a₂+b₂, a₃+b₃)

Subtraction

(a₁, a₂, a₃) − (b₁, b₂, b₃) = (a₁−b₁, a₂−b₂, a₃−b₃)

Scalar multiplication

k(a₁, a₂, a₃) = (ka₁, ka₂, ka₃)

Zero vector

0 = (0, 0, 0) — the additive identity

Negative vector

−a = (−a₁, −a₂, −a₃) — same magnitude, opposite direction

Triangle Law

AB⃗ + BC⃗ = AC⃗ (vectors head-to-tail)

Parallelogram Law

Resultant of a and b = diagonal of parallelogram formed by a and b

Properties of Vector Addition

Vector addition is both commutative (a + b = b + a) and associative ((a + b) + c = a + (b + c)). Scalar multiplication distributes over vector addition: k(a + b) = ka + kb. These properties mirror those of real number arithmetic and are used constantly in proofs.

Worked Examples — Vector Operations

Example 1 a = (3, −1, 4), b = (−2, 5, 1). Find: (a) a + b, (b) 2a − 3b.

Part a a + b = (3+(−2), −1+5, 4+1) = (1, 4, 5).

Part b First: 2a = (6, −2, 8). Then: 3b = (−6, 15, 3). So: 2a − 3b = (6−(−6), −2−15, 8−3) = (12, −17, 5).

(a) (1, 4, 5) (b) (12, −17, 5)

Example 2 — Applied: Resultant Force Three forces act on a particle: F₁ = (5, 3) N, F₂ = (−2, 7) N, F₃ = (1, −4) N. Find the resultant force.

Step 1 Resultant = F₁ + F₂ + F₃ (vector addition is done component by component).

Step 2 x-components: 5 + (−2) + 1 = 4. y-components: 3 + 7 + (−4) = 6. Resultant = (4, 6) N.

Step 3 Magnitude = √(4² + 6²) = √(16 + 36) = √52 = 2√13 ≈ 7.21 N.

Resultant force = (4, 6) N, magnitude = 2√13 N ≈ 7.21 N

Section 05 · Measurement

Magnitude and Unit Vectors

Senior Secondary

The magnitude of a vector is its length — a scalar value obtained from the Pythagorean theorem extended to multiple dimensions. A unit vector has magnitude exactly 1 and is used to indicate pure direction.

Magnitude and Unit Vector Formulas

Magnitude in 2D

|a| = √(a₁² + a₂²)

Magnitude in 3D

|a| = √(a₁² + a₂² + a₃²)

Unit vector in direction a

â = a / |a|

Verification

|â| = 1 always

Vector of given length

If |v| = L in direction a: v = L · â = L · (a/|a|)

Worked Examples — Magnitude and Unit Vectors

Example 1 Find the magnitude and unit vector of v = (3, −4, 0).

Step 1 |v| = √(3² + (−4)² + 0²) = √(9 + 16 + 0) = √25 = 5.

Step 2 Unit vector: v̂ = v / |v| = (3, −4, 0) / 5 = (3/5, −4/5, 0).

Verify |v̂| = √((3/5)² + (−4/5)²) = √(9/25 + 16/25) = √(25/25) = 1 ✔

|v| = 5 · v̂ = (3/5, −4/5, 0)

Example 2 Find a vector of magnitude 6 in the direction of a = (1, 2, 2).

Step 1 Find |a| = √(1 + 4 + 4) = √9 = 3.

Step 2 Unit vector: â = (1, 2, 2)/3 = (1/3, 2/3, 2/3).

Step 3 Required vector = 6 × (1/3, 2/3, 2/3) = (2, 4, 4).

Required vector = (2, 4, 4), with magnitude 6 ✔

Example 3 — KCSE Style Given p = 3i − 4j and q = 5i + 12j, find: (a) |p|, (b) |q|, (c) a unit vector in the direction of p + q.

Part a |p| = √(3² + (−4)²) = √(9 + 16) = √25 = 5.

Part b |q| = √(5² + 12²) = √(25 + 144) = √169 = 13.

Part c p + q = (3+5, −4+12) = (8, 8). |p + q| = √(64 + 64) = √128 = 8√2. Unit vector = (8, 8)/(8√2) = (1/√2, 1/√2) = (√2/2, √2/2).

|p| = 5 · |q| = 13 · unit vector in direction (p+q) = (√2/2, √2/2)

Section 06 · Scalar Multiplication

Dot (Scalar) Product

Senior Secondary / A-Level

The dot product combines two vectors to produce a scalar (ordinary number). It is the key tool for measuring the angle between two vectors and for testing perpendicularity — both crucial in KCSE geometry and physics applications.

Dot Product — Two Equivalent Definitions

Geometric definition

a · b = |a| |b| cos θ

Component definition

a · b = a₁b₁ + a₂b₂ + a₃b₃

Angle between vectors

cos θ = (a · b) / (|a| |b|)

Perpendicular vectors

a ⊥ b ⟺ a · b = 0 (θ = 90°)

Parallel vectors (same direction)

a · b = |a| |b| (θ = 0°, cos 0° = 1)

Scalar projection of a onto b

comp_b(a) = (a · b) / |b|

Vector projection of a onto b

proj_b(a) = [(a · b) / |b|²] b

Key Properties of the Dot Product

The dot product is commutative: a · b = b · a. It distributes over addition: a · (b + c) = a · b + a · c. The dot product of a vector with itself gives the square of its magnitude: a · a = |a|². The standard unit vectors satisfy: i · i = j · j = k · k = 1, and i · j = j · k = i · k = 0.

Worked Examples — Dot Product

Example 1 — Find the angle Find the angle between a = (1, 2, 2) and b = (2, 1, −2).

Step 1 Compute a · b = (1)(2) + (2)(1) + (2)(−2) = 2 + 2 − 4 = 0.

Step 2 Since a · b = 0, cos θ = 0, therefore θ = 90°.

Conclusion The vectors are perpendicular. This result holds regardless of their magnitudes.

θ = 90° — the vectors are perpendicular.

Example 2 — Exact angle Find the angle between p = (3, 0, 4) and q = (1, 1, 0), giving your answer to 1 decimal place.

Step 1 p · q = (3)(1) + (0)(1) + (4)(0) = 3 + 0 + 0 = 3.

Step 2 |p| = √(9 + 0 + 16) = √25 = 5. |q| = √(1 + 1 + 0) = √2.

Step 3 cos θ = (p · q) / (|p| |q|) = 3 / (5√2) = 3/(5√2) ≈ 3/7.071 ≈ 0.4243.

Step 4 θ = arccos(0.4243) ≈ 64.9°.

θ ≈ 64.9°

Example 3 — Finding k for perpendicularity Vectors a = (k, 3, −1) and b = (2, k, 5) are perpendicular. Find k.

Step 1 For perpendicularity: a · b = 0.

Step 2 a · b = (k)(2) + (3)(k) + (−1)(5) = 2k + 3k − 5 = 5k − 5.

Step 3 Set equal to zero: 5k − 5 = 0 → 5k = 5 → k = 1.

k = 1

Example 4 — Projection Find the scalar projection of a = (4, 1, 3) onto b = (2, −2, 1).

Step 1 a · b = (4)(2) + (1)(−2) + (3)(1) = 8 − 2 + 3 = 9.

Step 2 |b| = √(4 + 4 + 1) = √9 = 3.

Step 3 Scalar projection = (a · b)/|b| = 9/3 = 3.

Scalar projection = 3

Section 07 · Vector Multiplication

Cross (Vector) Product

A-Level / University

Unlike the dot product, the cross product of two vectors produces another vector — one that is perpendicular to both. It is defined only in 3D space and is fundamental in physics (torque, angular momentum) and finding normal vectors to planes.

Cross Product — Definition and Formula

Geometric form

a × b = |a| |b| sin θ n̂

where n̂

is the unit vector perpendicular to both a and b (right-hand rule)

Determinant form

a × b = | i  j  k |
                                                            | a₁ a₂ a₃|
                                                            | b₁ b₂ b₃|

Expanded form

= (a₂b₃ − a₃b₂)i − (a₁b₃ − a₃b₁)j + (a₁b₂ − a₂b₁)k

Magnitude = Area

|a × b| = area of parallelogram with sides a and b

Triangle area

Area of triangle = ½|a × b|

Parallel test

a ∥ b ⟺ a × b = 0 (sin 0° = 0)

Critical Warning — Order Matters! The cross product is anti-commutative: a × b = −(b × a). Unlike the dot product, reversing the order gives the opposite vector (same magnitude, opposite direction). Always be careful about order.

The Right-Hand Rule

To determine the direction of a × b without calculation: point your right-hand fingers in the direction of a, curl them toward b, and your extended thumb points in the direction of a × b. This physical intuition helps you verify your calculated results and is essential in physics applications involving torque and magnetic force.

Worked Examples — Cross Product

Example 1 — Standard computation Find a × b where a = (1, 2, 3) and b = (4, 5, 6).

Step 1 Apply the formula, i-component: a₂b₃ − a₃b₂ = (2)(6) − (3)(5) = 12 − 15 = −3.

Step 2 j-component (note the minus sign in front): −(a₁b₃ − a₃b₁) = −[(1)(6) − (3)(4)] = −[6 − 12] = −(−6) = 6.

Step 3 k-component: a₁b₂ − a₂b₁ = (1)(5) − (2)(4) = 5 − 8 = −3.

Verify Check perpendicularity: (−3, 6, −3)·(1, 2, 3) = −3 + 12 − 9 = 0 ✔ and (−3, 6, −3)·(4, 5, 6) = −12 + 30 − 18 = 0 ✔

a × b = (−3, 6, −3) = −3i + 6j − 3k

Example 2 — Area of a triangle Vertices A(1, 0, 0), B(0, 2, 0), C(0, 0, 3). Find the area of triangle ABC.

Step 1 Find two edge vectors: AB⃗ = B − A = (−1, 2, 0). AC⃗ = C − A = (−1, 0, 3).

Step 2 AB⃗ × AC⃗: i-component = (2)(3)−(0)(0) = 6. j-component = −[(−1)(3)−(0)(−1)] = −[−3−0] = 3. k-component = (−1)(0)−(2)(−1) = 0+2 = 2.

Step 3 AB⃗ × AC⃗ = (6, 3, 2). |AB⃗ × AC⃗| = √(36 + 9 + 4) = √49 = 7.

Step 4 Area of triangle = ½ × 7 = 3.5 square units.

Area of △ABC = 3.5 square units

Example 3 — Parallel test Are a = (2, −1, 3) and b = (4, −2, 6) parallel?

Method 1 Check if b = ka: (4, −2, 6) = k(2, −1, 3). From x: k = 2. From y: k = 2. From z: k = 2. All equal → parallel. ✔

Method 2 Compute a × b. i: (−1)(6)−(3)(−2) = −6+6 = 0. j: −[(2)(6)−(3)(4)] = −[12−12] = 0. k: (2)(−2)−(−1)(4) = −4+4 = 0. Cross product = (0,0,0). ✔

Yes, a and b are parallel. (b = 2a)

Section 08 · Applications

Geometric Applications of Vectors

Senior Secondary

One of the most powerful uses of vectors in the CBC curriculum is in plane geometry — proving properties of shapes, finding unknown points, and demonstrating collinearity or parallelism. These problems appear heavily in KCSE Paper 2 and account for a significant proportion of available marks.

Proving Parallel Lines

Show AB⃗ = k · CD⃗ for some scalar k ≠ 0. Two vectors are parallel if one is a scalar multiple of the other. If they also share a point, the lines are collinear (same line).

Proving Collinearity

Show AB⃗ = k · AC⃗. Points A, B, C are collinear if the vectors between any two pairs of points are parallel and share a common point.

Intersection of Lines

Parametric equations: set equal and solve for parameters. Two lines in 2D will always intersect (unless parallel); in 3D, they may be skew.

Worked Examples — Geometry (KCSE Style)

Example 1 — KCSE Geometry Proof OABC is a trapezium where OA is parallel to CB and CB = ½OA. OA⃗ = a, OC⃗ = c. D is the midpoint of AB. Find OD⃗ in terms of a and c. Hence show that O, D, B are not collinear (or verify the vector equation of the line through O and D).

Step 1 Since CB = ½OA and CB ∥ OA: CB⃗ = ½a. So OB⃗ = OC⃗ + CB⃗ = c + ½a.

Step 2 OA⃗ = a (given). AB⃗ = OB⃗ − OA⃗ = (c + ½a) − a = c − ½a.

Step 3 D is midpoint of AB: OD⃗ = OA⃗ + ½AB⃗ = a + ½(c − ½a) = a + ½c − ¼a = ¾a + ½c.

Check OD⃗ = ¾a + ½c. OB⃗ = ½a + c. Is OD⃗ = k·OB⃗? If so: ¾a + ½c = k(½a + c). From a: k = 3/2. From c: k = 1/2. Contradiction — O, D, B are NOT collinear. ✔

OD⃗ = ¾a + ½c

Section 09 · Three Dimensions

Lines and Planes in 3D Space

A-Level / University

In three dimensions, vectors provide the most elegant and powerful way to describe lines and planes. This is fundamental at university level and appears in A-Level applied mathematics.

Vector Equations of Lines and Planes

Line through point a, direction d

r = a + t·d (t ∈ ℝ)

Parametric form

x = a₁ + td₁, y = a₂ + td₂, z = a₃ + td₃

Cartesian symmetric form

(x−a₁)/d₁ = (y−a₂)/d₂ = (z−a₃)/d₃

Plane: normal n, point a

n · (r − a) = 0 → n · r = n · a

Cartesian plane equation

n₁x + n₂y + n₃z = d (d = n · a)

Dist. from point P to plane ax+by+cz=d

Distance = |aP_x + bP_y + cP_z − d| / √(a²+b²+c²)

Angle between two planes

cos θ = |n₁ · n₂| / (|n₁| |n₂|)

Worked Examples — Lines and Planes

Example 1 — Vector equation of a line Find the vector equation of the line through A(2, 1, −3) with direction d = (1, 2, −1).

Step 1 Use the formula r = a + td. Here a = (2, 1, −3) and d = (1, 2, −1).

Step 2 r = (2, 1, −3) + t(1, 2, −1), which in parametric form is: x = 2 + t, y = 1 + 2t, z = −3 − t.

r = (2, 1, −3) + t(1, 2, −1), t ∈ ℝ

Example 2 — Equation of a plane through 3 points Find the equation of the plane through A(1, 0, 0), B(0, 2, 0), C(0, 0, 3).

Step 1 Find two vectors in the plane: AB⃗ = (−1, 2, 0) and AC⃗ = (−1, 0, 3).

Step 2 Normal vector n = AB⃗ × AC⃗. i: (2)(3)−(0)(0) = 6. j: −[(−1)(3)−(0)(−1)] = 3. k: (−1)(0)−(2)(−1) = 2. n = (6, 3, 2).

Step 3 Plane equation: n · r = n · a = (6)(1)+(3)(0)+(2)(0) = 6.

Step 4 Full equation: 6x + 3y + 2z = 6.

Verify A(1,0,0): 6(1)+3(0)+2(0)=6 ✔ B(0,2,0): 6(0)+3(2)+2(0)=6 ✔ C(0,0,3): 6(0)+3(0)+2(3)=6 ✔

6x + 3y + 2z = 6 (or equivalently: 6x + 3y + 2z − 6 = 0)

Example 3 — Distance from a point to a plane Find the perpendicular distance from P(1, 2, 3) to the plane 2x − y + 2z = 9.

Step 1 Identify: a = 2, b = −1, c = 2, d = 9. Point P = (1, 2, 3).

Step 2 Distance = |a·Px + b·Py + c·Pz − d| / √(a² + b² + c²) = |2(1) + (−1)(2) + 2(3) − 9| / √(4 + 1 + 4).

Step 3 = |2 − 2 + 6 − 9| / √9 = |−3| / 3 = 3/3 = 1 unit.

Perpendicular distance = 1 unit

Section 10 · University Level

Vector Calculus

University

Vector calculus extends differentiation and integration to vector-valued functions and multi-dimensional scalar fields. It is the mathematical language of physics — describing gravitational, electric, and fluid flow fields. It underpins university-level STEM courses across all Kenyan universities.

Vector Calculus Operators

Gradient (grad f)

∇f = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

What gradient means

Points in the direction of steepest increase of f; its magnitude is the rate of change in that direction.

Divergence (div F)

∇·F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z (scalar)

What divergence means

Measures how much a vector field "spreads out" from a point. Positive = source; Negative = sink.

Curl (curl F)

∇×F = |i j k | / |∂/∂x ∂/∂y ∂/∂z| / |F₁ F₂ F₃|

What curl means

Measures the rotation (circulation) of a vector field around a point. Zero curl = irrotational field.

Laplacian (∇²f)

∇²f = ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z²

Directional derivative

D_u f = ∇f · û (rate of change of f in direction of unit vector û)

Worked Examples — Vector Calculus

Example 1 — Gradient Find ∇f at the point (1, 2, 1) where f(x, y, z) = x²y + yz².

Step 1 ∂f/∂x = 2xy. ∂f/∂y = x² + z². ∂f/∂z = 2yz.

Step 2 ∇f = (2xy, x² + z², 2yz).

Step 3 At (1, 2, 1): ∇f = (2(1)(2), 1² + 1², 2(2)(1)) = (4, 2, 4).

∇f at (1,2,1) = (4, 2, 4) = 4i + 2j + 4k

Example 2 — Divergence Find ∇·F where F = (x²y, yz, xz²).

Step 1 ∂F₁/∂x = ∂(x²y)/∂x = 2xy.

Step 2 ∂F₂/∂y = ∂(yz)/∂y = z.

Step 3 ∂F₃/∂z = ∂(xz²)/∂z = 2xz.

Step 4 ∇·F = 2xy + z + 2xz.

div F = 2xy + z + 2xz

Integral Theorems: Three major theorems connect vector calculus integrals: Green's Theorem relates a line integral around a closed 2D curve to a double integral over the enclosed region; Stokes' Theorem extends this to 3D, relating a surface integral of curl F to the line integral around the boundary curve; the Divergence Theorem (Gauss's Theorem) relates the volume integral of divergence to the surface integral over the boundary. These are the foundation of electromagnetism and fluid dynamics.

Section 11 · Quick Reference

Complete Vectors Formula Sheet

All Levels

Magnitude

2D: |a| = √(a₁² + a₂²)

3D: |a| = √(a₁² + a₂² + a₃²)

Unit vector: â = a / |a|

a · a = |a|²

Position Vectors

AB⃗ = b − a

Midpoint: OM⃗ = ½(a + b)

Ratio m:n: (na + mb)/(m+n)

Collinear: AB⃗ = k·AC⃗

Dot Product

a · b = a₁b₁ + a₂b₂ + a₃b₃

a · b = |a||b|cos θ

cos θ = (a·b)/(|a||b|)

Perp: a·b = 0

Proj: (a·b)/|b|² × b

Cross Product

a×b = |a||b|sin θ n̂

i: a₂b₃ − a₃b₂

j: −(a₁b₃ − a₃b₁)

k: a₁b₂ − a₂b₁

Area △ = ½|a×b|

Parallel: a×b = 0

Lines in 3D

r = a + td

x = a₁+td₁, y = a₂+td₂, z = a₃+td₃

Cartesian: (x−a₁)/d₁ = (y−a₂)/d₂ = (z−a₃)/d₃

Planes in 3D

n·(r−a) = 0

n₁x + n₂y + n₃z = d

Normal: n = AB⃗ × AC⃗

Dist = |ax₀+by₀+cz₀−d|/√(a²+b²+c²)

Vector Calculus

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)

div F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z

∇²f = ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z²

D_u f = ∇f · û

Key Identities

a × b = −(b × a)

a · b = b · a

i·i = j·j = k·k = 1

i·j = j·k = i·k = 0

i×j = k, j×k = i, k×i = j

Section 12 · Exam Preparation

Common Errors in Vectors

All Levels

These are the most frequently penalised mistakes in KCSE and Senior Secondary vector assessments. Learning to recognise them in your own work is as important as learning the correct methods.

✗ WrongAB⃗ = A − B = (1,2) − (4,6) = (−3,−4)

✓ RightAB⃗ = B − A = (4,6) − (1,2) = (3,4)

AB⃗ goes from A to B. Subtract the starting point (A) from the destination (B), not the other way around.

✗ Wronga · b = (a₁, a₂, a₃)(b₁, b₂, b₃) = (a₁b₁, a₂b₂, a₃b₃)

✓ Righta · b = a₁b₁ + a₂b₂ + a₃b₃ (a single scalar)

The dot product always produces a scalar (single number), not a vector. Add all component products together.

✗ Wronga × b = b × a

✓ Righta × b = −(b × a)

Unlike the dot product, the cross product is NOT commutative. Reversing the order negates the result — a very common sign error.

✗ WrongUnit vector of (3,4): û = (3,4)/7

✓ Right|(3,4)| = √(9+16) = 5, û = (3,4)/5 = (0.6, 0.8)

Always calculate the magnitude first before dividing. Using 7 (sum of components) instead of 5 (magnitude) is a very frequent error.

✗ Wrongj-component of cross product: a₁b₃ − a₃b₁

✓ Rightj-component: −(a₁b₃ − a₃b₁) — note the negative sign!

The j-component of the cross product has a minus sign in front. Forgetting this sign is the single most common cross product error.

✗ WrongFor collinearity, just show AB⃗ = k·CD⃗ for some k.

✓ RightAlso confirm that A, B, C, D share a common point, OR use the same base point (e.g., AC⃗ and AB⃗).

Parallel vectors (AB⃗ = k·CD⃗) means lines are parallel — not necessarily collinear. Collinearity requires shared points, shown through position vectors from the same origin.