Study Guide · Vectors

Vectors

Magnitude and direction together — from components to the dot product

6 sections 19 worked examples 7 checkpoints Intuition-first

What is a Vector?

Magnitude and direction together

Some quantities in nature are fully described by a single number (temperature: 22°C). Others need both a size AND a direction (velocity: 80 km/h North). Vectors capture both in one object.

💭

Speed vs velocity: A car going 80 km/h has speed = 80. If it's going 80 km/h North, that's velocity — a vector. The direction matters because North takes you somewhere completely different from South at the same speed.

Scalar (magnitude only)	Vector (magnitude + direction)
Mass (kg)	Force (N, with direction)
Speed (km/h)	Velocity (km/h, with direction)
Distance (km)	Displacement (km, with direction)
Time (s)	Acceleration (m/s², with direction)
Temperature (°C)	Weight (N, downward)

Key vocabulary

Magnitude — the length/size of the vector, written ‖v‖ or |v|, always ≥ 0
Direction — the angle the vector makes (usually measured from East, counter-clockwise)
Equivalent vectors — same magnitude AND direction; position doesn't matter
Opposite vector — same magnitude, reversed direction; written −v
Zero vector — magnitude 0, no defined direction; written 0
Unit vector — magnitude exactly 1; often used to indicate direction only

Basic

Classify as scalar or vector

Classify: (a) 35°C temperature (b) 12 N downward force (c) 5 km distance (d) 5 km southwest

▶

(a) 35°C temperature: No direction — scalar

(b) 12 N downward: Has magnitude (12 N) AND direction (downward) — vector

(c) 5 km distance: No direction — scalar

(d) 5 km southwest: Has magnitude AND direction — vector (displacement)

(a) scalar (b) vector (c) scalar (d) vector

Basic

Equal and opposite vectors

Vector u has magnitude 8 and direction 30°. Describe: (a) a vector equal to u, (b) −u

▶

(a) Equal vector: Any vector with magnitude 8 AND direction 30°. Position (where it's drawn) doesn't matter — same ‖·‖ and same angle

(b) −u: Same magnitude (8), but direction reversed: 30° + 180° = 210°

Equal to u: magnitude 8, direction 30°. Opposite −u: magnitude 8, direction 210°

Checkpoint 1 — Vector Basics

Give two examples of scalar quantities and two of vector quantities
Vector w has magnitude 15 and direction 120°. What is the magnitude and direction of −w?
True or false: Two vectors with the same magnitude are always equal

Scalars: mass, temperature, time, distance, speed. Vectors: force, velocity, displacement, acceleration, weight
Magnitude: 15 (unchanged); direction: 120° + 180° = 300°
False — equal vectors also require the same direction

Notation & the Compass

Writing vectors and reading directions

A vector from point A to point B is written AB⃗, or as a single letter in bold: v. In handwriting, draw an arrow above: v⃗. The magnitude (length) is written ‖v‖ or |v|.

Standard Angle Convention

Angles are measured from East = 0°, going counter-clockwise (same as unit circle)
North = 90°, West = 180°, South = 270°
Compass bearing "N 40° E" = start at North, rotate 40° toward East → standard angle = 90° − 40° = 50°
Compass bearing "W 50° S" = start at West (180°), rotate 50° toward South → standard angle = 180° + 50° = 230°
Always sketch the direction before computing!

Direction	Standard Angle	Notes
East	0°	Reference direction
North	90°	Quarter turn CCW
West	180°	Half turn
South	270°	Three-quarter turn CCW
N 40° E	50°	90° − 40° = 50°
S 30° W	240°	180° + 60° = 240° — start at South (270°), go 30° toward West: 270°−30°=240°
N 70° W	160°	Start at North (90°), go 70° toward West: 90°+70°=160°

⚠️

Read the bearing carefully: "40° South of East" means start at East, rotate 40° toward South → standard angle = 360°−40° = 320°. Always draw a quick sketch.

Basic

Convert compass bearing to standard angle

Convert to standard angle: (a) N 25° E (b) S 60° W (c) N 50° W

▶

(a) N 25° E: North = 90°. Rotating 25° toward East decreases the angle: 90° − 25° = 65°

(b) S 60° W: South = 270°. Rotating 60° toward West decreases from 270°: 270° − 60° = 210°

(c) N 50° W: North = 90°. Rotating 50° toward West increases the angle: 90° + 50° = 140°

(a) 65° (b) 210° (c) 140°

Medium

Convert standard angle to compass bearing

Express as a compass bearing: (a) 135° (b) 300°

▶

(a) 135°: This is between North (90°) and West (180°). Distance from North: 135°−90°=45°. Distance from West: 180°−135°=45°. Equidistant → N 45° W

(b) 300°: This is between West (180°) and North going around (360°/0°). Easier: 300° is between South (270°) and West reversed. 300°−270°=30° past South toward West → N 30° W ... let's check: 360°−300°=60° short of North, from West side. Best: between North and West: 360°−300°=60° from North toward West → N 60° W

(a) N 45° W (b) N 60° W

Checkpoint 2 — Compass Bearings

Convert to standard angle: S 45° E
Convert to standard angle: W 20° N
Express 225° as a compass bearing

South = 270°, go 45° toward East: 270°+45°= 315° → actually S 45° E: start from South toward East means decreasing from 270°... South-East quadrant: 270°+45°=315°. Wait, let me think: going from South toward East: East is at 0° (or 360°). From 270° toward 360°: 270°+45°=315°. So standard angle = 315°
West = 180°, go 20° toward North: 180°−20° = 160°
225° is 45° past West (180°), heading toward South → S 45° W

Components & Magnitude

Breaking a vector into horizontal and vertical parts

Every vector in 2D can be written as [x, y] — the horizontal (x) and vertical (y) components. This decomposition is the key tool for all vector calculations.

Decompose into x (horizontal) and y (vertical). Reassemble with Pythagoras and arctan.

Component Formulas

x = ‖v‖ · cos β — horizontal component (β = standard angle)
y = ‖v‖ · sin β — vertical component
‖v‖ = √(x² + y²) — Pythagorean theorem on components
β = arctan(y/x) — but check the quadrant! arctan only gives Q1 or Q4
CAST rule: Q1 (all +), Q2 (sin +), Q3 (tan +), Q4 (cos +)

⚠️

Quadrant trap with arctan: If x < 0, add 180° to the arctan result. If x > 0 and y < 0, add 360°. Always sketch the vector to verify the quadrant.

Medium

Components from magnitude and direction

A vector has magnitude 50 and direction 127° (standard). Find the components.

▶

Step 1 — Apply formulas:
x = 50 · cos(127°)
y = 50 · sin(127°)

Step 2 — Calculate:
cos(127°) = −cos(53°) ≈ −0.602 → x ≈ −30.1
sin(127°) = sin(53°) ≈ 0.799 → y ≈ 39.9
In Q2: x is negative, y is positive — consistent with CAST ✓

v ≈ [−30.1, 39.9]

Basic

Magnitude and direction from components

Find the magnitude and direction of v = [−5, 12]

▶

Step 1 — Magnitude: ‖v‖ = √((−5)² + 12²) = √(25 + 144) = √169 = 13

Step 2 — Raw arctan: arctan(12/−5) = arctan(−2.4) ≈ −67.4°

Step 3 — Quadrant check: x = −5 < 0, y = 12 > 0 → Q2. Add 180°: −67.4° + 180° = 112.6°

‖v‖ = 13, direction ≈ 112.6°

Medium

Vector from a compass bearing

A force of 40 N acts in direction N 35° E. Find its components.

▶

Step 1 — Convert to standard angle: N 35° E → 90° − 35° = 55°

Step 2 — Find components:
x = 40 · cos(55°) ≈ 40 · 0.574 ≈ 22.9 N
y = 40 · sin(55°) ≈ 40 · 0.819 ≈ 32.8 N

Horizontal (East): ≈ 22.9 N, Vertical (North): ≈ 32.8 N

Checkpoint 3 — Components & Magnitude

Find components of a vector with magnitude 20 and direction 210°
Find magnitude and direction of v = [7, −7]
A 60 N force acts in direction S 50° W. Find its components.

x = 20·cos(210°) = 20·(−√3/2) ≈ −17.3; y = 20·sin(210°) = 20·(−1/2) = −10
‖v‖ = √(49+49) = 7√2 ≈ 9.9; arctan(−7/7) = −45°; x > 0, y < 0 → Q4: 360°−45° = 315°
S 50° W → standard angle: 270°−50°=220° (or: from South toward West). x = 60·cos(220°) ≈ −46.0 N; y = 60·sin(220°) ≈ −38.6 N

Vector Operations

Adding, subtracting, and scaling vectors

Vectors are added component by component. Geometrically this is the head-to-tail rule: place the tail of the second vector at the head of the first, then draw the resultant from start to end.

Head-to-tail rule geometrically; component addition algebraically. Both give the same result.

Operations Summary

u + v = [x₁+x₂, y₁+y₂] — add component by component
u − v = [x₁−x₂, y₁−y₂] — subtract component by component
k·v = [kx, ky] — scalar multiplication scales both components
−v = [−x, −y] — opposite vector
Two vectors are collinear (parallel) if v = k·u for some scalar k
A vector is a unit vector if ‖v‖ = 1; unit vector in direction of v: v̂ = v / ‖v‖

⚠️

Never add magnitudes: ‖u + v‖ ≠ ‖u‖ + ‖v‖ in general. Always add components first, then find the magnitude of the result.

Medium

Finding the resultant vector

Find u + v and ‖u + v‖ where u = [3, 4] and v = [5, −2]

▶

Step 1 — Add components: u + v = [3+5, 4+(−2)] = [8, 2]

Step 2 — Magnitude: ‖u + v‖ = √(8² + 2²) = √(64 + 4) = √68 ≈ 8.25

u + v = [8, 2], magnitude ≈ 8.25

Basic

Scalar multiplication and subtraction

Given u = [2, −3] and v = [4, 1], find: (a) 3u (b) u − 2v (c) ‖2u‖

▶

(a) 3u: [3·2, 3·(−3)] = [6, −9]

(b) u − 2v: [2−2·4, −3−2·1] = [2−8, −3−2] = [−6, −5]

(c) ‖2u‖: 2u = [4, −6]; ‖2u‖ = √(16+36) = √52 ≈ 7.21
Note: ‖2u‖ = 2·‖u‖ = 2·√13 ≈ 7.21 — scalar multiplication scales the magnitude too ✓

(a) [6, −9] (b) [−6, −5] (c) ≈ 7.21

Medium

Find a unit vector

Find the unit vector in the direction of v = [3, 4]

▶

Step 1 — Find magnitude: ‖v‖ = √(9+16) = √25 = 5

Step 2 — Divide each component by magnitude: v̂ = (1/5)·[3, 4] = [3/5, 4/5] = [0.6, 0.8]

Check: ‖v̂‖ = √(0.36+0.64) = √1 = 1 ✓

Unit vector: [0.6, 0.8]

Medium

Test for collinearity

Are u = [4, 6] and v = [6, 9] collinear (parallel)?

▶

Method — check if v = k·u: If [6, 9] = k·[4, 6], then k = 6/4 = 3/2 and k = 9/6 = 3/2

Same k value? Yes, k = 3/2 for both components → v = (3/2)·u

Yes, u and v are collinear (parallel). v = (3/2)·u

Checkpoint 4 — Vector Operations

Given u = [1, 5] and v = [−3, 2], find 2u + v and its magnitude
Are p = [6, −4] and q = [−9, 6] collinear?
Find the unit vector in the direction of w = [−8, 6]

2u+v = [2+(-3), 10+2] = [−1, 12]; magnitude = √(1+144) = √145 ≈ 12.04
p = [6,−4]; q = [−9,6]. Check: k = −9/6 = −3/2 and −4·(−3/2)=6 ✓ → Yes, collinear (q = −(3/2)p)
‖w‖ = √(64+36) = 10; ŵ = [−0.8, 0.6]

The Scalar (Dot) Product

Multiplying two vectors to get a number

The dot product takes two vectors and returns a scalar (a plain number). It measures how much one vector points in the direction of the other — essential for finding angles between vectors.

💭

What does it measure? How much of one vector points in the direction of the other. Two perpendicular vectors share no common direction → dot product = 0. Parallel vectors share maximum direction → maximum dot product.

Two Ways to Compute u · v

Component form: u · v = x₁x₂ + y₁y₂
Geometric form: u · v = ‖u‖ · ‖v‖ · cos(θ)
These two always give the same answer — use whichever is convenient
To find angle: cos(θ) = (u · v) / (‖u‖ · ‖v‖), then θ = arccos(...)
Perpendicular test: u ⊥ v if and only if u · v = 0

Dot product value	Meaning	Angle θ
= 0	Perpendicular (orthogonal)	90°
> 0	Acute angle between vectors	0° < θ < 90°
< 0	Obtuse angle between vectors	90° < θ < 180°
= ‖u‖·‖v‖	Parallel, same direction	0°
= −‖u‖·‖v‖	Parallel, opposite directions	180°

Medium

Finding the angle between two vectors

Find the angle between u = [3, 5] and v = [7, 2]

▶

Step 1 — Dot product: u · v = 3×7 + 5×2 = 21 + 10 = 31

Step 2 — Magnitudes: ‖u‖ = √(9+25) = √34 ≈ 5.831; ‖v‖ = √(49+4) = √53 ≈ 7.280

Step 3 — Find angle: cos θ = 31/(5.831×7.280) ≈ 31/42.45 ≈ 0.730 → θ = arccos(0.730) ≈ 43.1°

θ ≈ 43.1°

Basic

Test for perpendicularity

Are u = [4, −3] and v = [6, 8] perpendicular?

▶

Compute dot product: u · v = 4×6 + (−3)×8 = 24 − 24 = 0

Conclusion: Dot product = 0 → vectors are perpendicular ✓

Yes, u and v are perpendicular (orthogonal)

Medium

Find an unknown component to make vectors perpendicular

Find k such that u = [3, k] and v = [4, −2] are perpendicular

▶

Step 1 — Set dot product = 0: 3·4 + k·(−2) = 0

Step 2 — Solve: 12 − 2k = 0 → k = 6

Check: u·v = 3·4 + 6·(−2) = 12 − 12 = 0 ✓

k = 6

Hard

Angle between vectors given magnitudes

‖u‖ = 6, ‖v‖ = 10, and u · v = −18. Find the angle between u and v.

▶

Formula: cos θ = (u · v)/(‖u‖·‖v‖) = −18/(6·10) = −18/60 = −0.3

Solve: θ = arccos(−0.3) ≈ 107.5°

Check: Negative dot product → obtuse angle. 107.5° is obtuse ✓

θ ≈ 107.5° (obtuse angle)

Checkpoint 5 — Dot Product

Compute u · v where u = [−2, 5] and v = [3, 4]
Find the angle between u = [1, 0] and v = [0, 1] (should be obvious!)
Find k so that [k, 3] ⊥ [5, −k]
‖u‖ = 4, ‖v‖ = 7, angle between them = 60°. Find u · v.

(−2)(3) + (5)(4) = −6 + 20 = 14
u·v = 0 → 90° (East and North are perpendicular)
5k + 3·(−k) = 0 → 5k − 3k = 0 → 2k = 0 → k = 0. Check: [0,3]·[5,0]=0 ✓
u·v = 4·7·cos(60°) = 28·0.5 = 14

Applications

Navigation, forces, and real-world problems

Most application problems involve multiple vectors being combined. The strategy is always the same: decompose, add components, then reassemble the result.

Four-Step Strategy

Step 1: Draw a clear diagram labelling all known quantities
Step 2: Break each vector into components: x = ‖v‖cos β, y = ‖v‖sin β
Step 3: Add all x-components; add all y-components
Step 4: Find magnitude and direction of the resultant: ‖r‖ = √(Σx² + Σy²), β = arctan(Σy/Σx) (check quadrant!)

Hard

Navigation with two legs

A ship travels 80 km East, then 60 km at 40° North of East. Find total displacement.

▶

Step 1 — Components:
v₁ = [80, 0] (East = 0°)
v₂: β = 40° → x₂ = 60·cos(40°) ≈ 45.96; y₂ = 60·sin(40°) ≈ 38.57

Step 2 — Add: Total = [80+45.96, 0+38.57] = [125.96, 38.57]

Step 3 — Magnitude: ‖r‖ = √(125.96² + 38.57²) ≈ √17 354 ≈ 131.7 km

Step 4 — Direction: β = arctan(38.57/125.96) ≈ arctan(0.306) ≈ 17° North of East

≈ 131.7 km at 17° North of East

Medium

Equilibrium of forces

Two forces act on an object: F₁ = 30 N at 0° and F₂ = 40 N at 90°. Find the resultant.

▶

Step 1 — Components:
F₁ = [30, 0]; F₂ = [0, 40]

Step 2 — Add: R = [30+0, 0+40] = [30, 40]

Step 3 — Magnitude: ‖R‖ = √(900+1600) = √2500 = 50 N

Step 4 — Direction: β = arctan(40/30) = arctan(4/3) ≈ 53.1° (from East, toward North)

Resultant: 50 N at 53.1° (North of East)

Hard

Three forces in equilibrium

Three forces act on a point: F₁ = [5, 8], F₂ = [−3, 2], F₃ = [a, b]. If the system is in equilibrium, find F₃.

▶

Equilibrium condition: F₁ + F₂ + F₃ = 0 (zero vector)

Find x-component: 5 + (−3) + a = 0 → 2 + a = 0 → a = −2

Find y-component: 8 + 2 + b = 0 → 10 + b = 0 → b = −10

F₃ = [−2, −10]

Hard

Boat crossing a river with current

A boat heads due North at 12 km/h. A river current flows East at 5 km/h. Find the actual velocity (magnitude and direction).

▶

Step 1 — Identify vectors:
Boat: v_boat = [0, 12] (North = 90°)
Current: v_current = [5, 0] (East = 0°)

Step 2 — Add: v_actual = [0+5, 12+0] = [5, 12]

Step 3 — Magnitude: ‖v‖ = √(25+144) = √169 = 13 km/h

Step 4 — Direction: β = arctan(12/5) = arctan(2.4) ≈ 67.4° from East = N 22.6° E

13 km/h in direction N 22.6° E (the current pushes the boat off course)

Checkpoint 6 — Applications

Two forces: F₁ = 50 N at 30°, F₂ = 30 N at 120°. Find the resultant magnitude and direction.
A plane flies N 60° E at 400 km/h. Wind blows due West at 80 km/h. Find actual velocity.

F₁=[50cos30°, 50sin30°]=[43.3, 25]; F₂=[30cos120°, 30sin120°]=[−15, 26.0]; Sum=[28.3, 51.0]; ‖R‖=√(800+2601)=√3401≈58.3 N; β=arctan(51/28.3)≈61°
N 60° E = standard 30°: plane=[400cos30°, 400sin30°]=[346.4, 200]; wind=[−80,0]; actual=[266.4, 200]; ‖v‖=√(70969+40000)≈333 km/h; β=arctan(200/266.4)≈36.9° (≈N 53° E in compass notation)

Checkpoint 7 — Mixed Review

Vector v = [a, 6] has magnitude 10. Find a (there may be two answers).
u = [2, 5] and v = [−1, 3]. Find the angle between them.
A hiker walks 5 km N 30° E, then 8 km due East. Find total displacement.

√(a²+36) = 10 → a²+36 = 100 → a² = 64 → a = ±8
u·v = (2)(−1)+(5)(3) = −2+15 = 13; ‖u‖=√29≈5.385; ‖v‖=√10≈3.162; cosθ=13/17.03≈0.763; θ≈40.2°
N 30° E = 60° standard: v₁=[5cos60°, 5sin60°]=[2.5, 4.33]; v₂=[8,0]; total=[10.5, 4.33]; ‖r‖=√(110.25+18.75)=√129≈11.36 km; β=arctan(4.33/10.5)≈22.4° → N 67.6° E