IB Maths AA HL
Topic 3 — Geometry & Trigonometry
Paper 1 & 2
~6 min read
HL only
Applications to Kinematics
A line equation is also a motion equation: r = r0 + vt describes an object moving with constant velocity. The starting point becomes r0, the direction becomes the velocity v, and the parameter λ becomes the time t.
📘 What you need to know
- Position vector at time t: r = r0 + vt (not in the formula booklet — memorise).
- r0 = position vector at t = 0 (the start).
- v = the (constant) velocity vector — direction × magnitude.
- Displacement: position relative to a fixed reference point.
- Velocity: rate of change of displacement (vector).
- Speed: |v|, the magnitude of velocity (scalar).
- Distance travelled (constant velocity): speed × time = |v| × t.
- Two objects collide if the same t gives equal position vectors in all components.
The motion equation
Position at time t (constant velocity)
r = r0 + vt
It’s the same equation as r = a + λb with new labels: a = r0 (starting point), b = v (velocity), λ = t (time). At t = 0 you’re at the start; each unit of time moves you by v.
Velocity (vector)
v
tells you how fast AND in what direction
Speed (scalar)
|v|
just how fast — magnitude only
Common question types
| Asked for… | Use… |
|---|
| position at given time | plug t into r = r₀ + vt |
| velocity from two points + time | v = (r₂ − r₁) / Δt |
| speed | |v| = √(v₁² + v₂² + v₃²) |
| distance travelled | distance = speed × time |
| time to reach a point | solve target = r₀ + vt for t componentwise |
| do two objects collide? | set r₁(t) = r₂(t); same t in all components → yes |
🧭 Recipe — set up a kinematics problem with vectors
- Identify r0: the position at time zero (often given as a coordinate point).
- Identify v: the velocity vector (given directly, or compute as displacement ÷ time).
- Write: r = r0 + vt.
- For position at a time — plug in t; for time at a position — solve componentwise and check all components agree.
- For speed/distance — find |v| first, then multiply by time if needed.
Worked examples
WE 1Position of a particle after a given time
A particle starts at the point (3, −1, 2) and moves with constant velocity v = 2i + 5j − k m/s. Find its position vector after 4 seconds.
Apply r = r₀ + vt
r = (3, −1, 2) + 4(2, 5, −1)
= (3, −1, 2) + (8, 20, −4)
= (3+8, −1+20, 2−4)
r = (11, 19, −2)
just multiply velocity by time and add to the start position
WE 2Find velocity vector and speed
A boat travels in a straight line from P(2, 8) to Q(14, −1) in 3 hours at constant velocity. Find the velocity vector and the speed of the boat.
Step 1: Find displacement PQ
PQ = Q − P = (14−2, −1−8) = (12, −9)
Step 2: Velocity = displacement ÷ time
v = (12, −9) / 3 = (4, −3) km/h
Step 3: Speed = |v|
|v| = √(4² + (−3)²) = √(16 + 9) = √25
v = (4, −3) km/h, speed = 5 km/h
velocity per unit time is total displacement divided by time taken
WE 3Find the time to reach a given point
A particle moves with position vector r = (1, −3, 5) + t(2, 4, −3). At what time t does it pass through the point (7, 9, −4)?
Set r = (7, 9, −4) and solve each component for t
x: 7 = 1 + 2t → t = 3
y: 9 = −3 + 4t → t = 3 ✓
z: −4 = 5 − 3t → t = 3 ✓
t = 3 (seconds)
all three components must agree on the same t — otherwise the particle never passes through that point
WE 4Distance travelled in a given time
A particle has initial position vector (5, −2) and velocity v = −3i + 4j m/s. Find the distance travelled by the particle in 6 seconds.
Step 1: Find the speed |v|
|v| = √((−3)² + 4²) = √(9 + 16) = √25 = 5 m/s
Step 2: Distance = speed × time
distance = 5 × 6
distance = 30 m
starting position doesn’t matter for distance — only speed and time
WE 5Determine whether two objects collide
Two boats move with position vectors rA = (2, −1) + t(3, 4) and rB = (10, 19) + t(−1, −6), where t is in hours. Determine whether the boats collide; if so, when and where.
Set r_A = r_B and solve componentwise
x: 2 + 3t = 10 − t → 4t = 8 → t = 2
y: −1 + 4t = 19 − 6t → 10t = 20 → t = 2 ✓
Same t in both components → collision
Find collision point: substitute t = 2
r_A = (2, −1) + 2(3, 4) = (8, 7)
Collide at (8, 7) at t = 2 hours
“collide” means same place at the SAME time — both components must give equal t
WE 6Find the initial position from a later observation
A car moves in a straight line with constant velocity v = (4, −3) m/s. After 5 seconds, the car is at the point (15, −8). Find the initial position of the car.
Step 1: Rearrange r = r₀ + vt → r₀ = r − vt
Step 2: Substitute t = 5, r = (15, −8), v = (4, −3)
r₀ = (15, −8) − 5(4, −3)
= (15, −8) − (20, −15)
= (15 − 20, −8 + 15)
Initial position: (−5, 7)
working “backwards in time” — subtract velocity × time from the known later position
💡 Top tips
- Velocity is a vector; speed is its magnitude — don’t confuse them in word problems.
- Distance travelled = speed × time only when velocity is constant.
- For collisions, the same t must satisfy all components — different t‘s mean the paths cross but the objects miss each other.
- Watch the units: if velocity is m/s and time is in minutes, convert.
- Visualise r0 as the launchpad and v as the arrow showing where one second of motion takes you.
⚠ Common mistakes
- Computing displacement as v × t and calling it distance — distance is |v| × t, not the vector itself.
- Mixing up “paths cross” with “objects collide”. Paths can intersect at different times — not a collision.
- Forgetting to verify all components when solving for time — one component agreeing is not enough.
- Using r instead of r − vt when working backwards to find r0.
- Treating speed as a vector (writing it with components) — speed is a single non-negative number.
Next: Coincident, Parallel, Intersecting & Skew Lines. In 3D, two lines can do four very different things — sit on top of each other, run in parallel, cross at a point, or pass without ever meeting. The dot/scalar product and parameter-matching tell you which.
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