IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read HL only

Position & Displacement Vectors

A position vector OA = a places point A relative to the origin — its components are just the coordinates. A displacement vector AB takes you from A to B, and the master formula is AB = b − a.

📘 What you need to know

Position vector OA = a: components = coordinates of A. Point (3, −2, −1) → 3i − 2j − k.
Displacement vector AB: shortest route from A to B.
Key formula: AB = b − a (end position minus start position).
Going backwards: BA = −AB = a − b.
Route via a third point: AB = AC + CB. Useful when A→B isn’t direct but you have the via-points.
Round trip: AB + BC + CA = 0 (you end where you started).
A position vector is just a displacement from the origin: OA is “how to get from O to A”.

Two related ideas

Position vector

OA = a

where the point sits relative to the origin; components = coordinates

Displacement vector

AB = b − a

how to get from A to B; depends on both points

The master formula

Displacement vector between two points AB = b − a = OB − OA

This comes from going A → O → B: AB = AO + OB = −a + b = b − a. End minus start.

Mnemonic: AB = “to B from A” → second letter minus first letter as position vectors. b − a, not a − b.

Routes via other points

AB = AC + CB

Walking A → C → B totals the same as walking A → B directly. Use this when you’re given displacements but not all positions.

A “round trip” closes back to zero: AB + BC + CA = 0. If a vector chain returns to its start, the displacements cancel.

🧭 Recipe — find a displacement vector between two points

Write the position vectors of both points (just take coordinates).
Use AB = b − a — end minus start.
Subtract componentwise: x − x, y − y, z − z.
Watch the order — AB ≠ BA (they’re opposite vectors).
Convert form if the question asks for column or base vector.

Worked examples

WE 1

Position vector from coordinates

Determine the position vector of the point with coordinates (−2, 5, 6).

Position vector = OA — components ARE the coordinates −2i + 5j + 6k or as a column: (−2, 5, 6)

WE 2

Coordinates from a position vector

The position vector of point P is p = 7i − 3j + k. Write down the coordinates of P.

Read off the coefficients of i, j, k as x, y, z x = 7, y = −3, z = 1 P (7, −3, 1) remember the missing j-term means y = 0; here +k means z = 1

WE 3

Displacement vector between two points

The points A and B have coordinates (2, −1, 4) and (5, 3, −2) respectively. Find the displacement vector AB.

Step 1: Write position vectors a = (2, −1, 4); b = (5, 3, −2) Step 2: Apply AB = b − a x: 5 − 2 = 3 y: 3 − (−1) = 4 z: −2 − 4 = −6 AB = (3, 4, −6) = 3i + 4j − 6k

WE 4

Find a point given another and a displacement

The point A has coordinates (1, −3, 2). Given that AB = 4i + 5j − k, find the coordinates of B.

Step 1: Rearrange AB = b − a → b = a + AB Step 2: Add componentwise b = (1, −3, 2) + (4, 5, −1) = (1+4, −3+5, 2−1) = (5, 2, 1) B (5, 2, 1) B’s coordinates ARE the components of its position vector

WE 5

Find AB using a third point

Given that AC = 3i − 2j + 5k and CB = i + 4j − 2k, find AB.

Step 1: Use the route formula AB = AC + CB Step 2: Add componentwise i: 3 + 1 = 4 j: −2 + 4 = 2 k: 5 + (−2) = 3 AB = 4i + 2j + 3k A → C → B totals the same as A → B directly. The “via” point cancels in the algebra.

WE 6

Closed-loop displacements sum to zero

The points A, B, and C have coordinates (2, 0, 1), (−1, 3, 4), and (5, −2, 0). Show that AB + BC + CA = 0.

Step 1: Compute each displacement AB = b − a = (−1−2, 3−0, 4−1) = (−3, 3, 3) BC = c − b = (5−(−1), −2−3, 0−4) = (6, −5, −4) CA = a − c = (2−5, 0−(−2), 1−0) = (−3, 2, 1) Step 2: Add the three x: −3 + 6 + (−3) = 0 y: 3 + (−5) + 2 = 0 z: 3 + (−4) + 1 = 0 AB + BC + CA = 0 a closed walk through any number of points always returns the zero vector — every displacement cancels out

💡 Top tips

Coordinates = position vector components. Conversion is read-and-write.
Always use end minus start for displacement: AB = b − a.
If you have a “via” point, chain the displacements: AB = AC + CB.
BA = −AB — same magnitude, opposite direction.
Closed loops give zero — useful as a sanity check on your displacements.

⚠ Common mistakes

Subtracting in the wrong order. AB = b − a, NOT a − b. The arrow over AB tells you the direction.
Sign errors with negatives: 3 − (−1) = 4, not 2.
Confusing position with displacement. Position depends on origin; displacement only depends on start and end points.
Misreading coordinates from base vectors. 7i − 3j + k = (7, −3, 1) — the +k alone means coefficient 1.
Going A → B by adding “AB to a” when it should be “a + AB”. (Same thing — but it’s easy to write down “B + AB” by mistake.)

Next note: Magnitude of a Vector & Unit Vectors. The size (length) of a vector via Pythagoras in 3D, the distance between two points, and how to scale any vector down to length 1.

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