IB Maths AI HLVector PropertiesPaper 1 & 2AB = b − a~7 min read
Position & Displacement Vectors
A position vector is the vector from the origin O to a point — its components are simply the point’s coordinates. A displacement vector goes from one point to another: AB means “from A to B”. The single relationship at the heart of this topic is AB = b − a, the difference of the two position vectors. Master that and almost every “find the vector from P to Q” question reduces to a one-line subtraction.
📘 What you need to know
Position vector: OA = a = (xA, yA, zA)T — identical to A’s coordinates, written as a vector from the origin.
Displacement from A to B: AB = b − a = (xB−xA, yB−yA, zB−zA)T.
Reverse direction: BA = −AB = a − b. The two are equal in magnitude, opposite in direction.
Triangle relation: AB + BC = AC for any three points — a “via a third point” identity that’s often the cleanest way to combine displacements.
Recover a point from a displacement: OB = OA + AB, so to find B from A and AB, add the displacement to A’s position vector.
Notation: an arrow over two letters (e.g. above AB) indicates a displacement vector; a single letter in bold (e.g. a) usually means a position vector unless context says otherwise.
The triangle O–A–B
Every displacement question reduces to the same picture: the origin O, two points A and B, and the triangle formed by the three displacement arrows OA, OB, and AB. Reading the triangle nose-to-tail gives OA + AB = OB, which rearranges to AB = OB − OA = b − a. So to find the displacement from one point to another, subtract their position vectors — end minus start, always in that order. Reversing the order reverses the vector: BA = a − b = −AB.
The triangle O–A–B: position vectors OA = a (teal) and OB = b (orange) from the origin; displacement AB = b − a (blue) connects A to B directly, with components (5−3, 4−1) = (2, 3).
Position & displacement formulasOA = a = (xA, yA, zA)T · AB = b − a · BA = −AB · AB + BC = AC“end minus start” — always subtract the start point’s coordinates from the end point’s
Recovering a point from a displacement
The same triangle works in reverse. If you know point A and the displacement AB, then point B sits at OB = OA + AB: add the displacement components to A’s coordinates. This is the standard pattern for “find the coordinates of B such that…” questions, and it extends naturally to scaled displacements like AB = 2AC — compute the displacement on the right, multiply, then add to A.
Direction matters: AB ≠ BA. They have the same length but opposite signs in every component. Always check the order: the second letter is the destination.
🧭 Recipe — displacement between two points
Write each point as a position vector using its coordinates: OA = (xA, yA, zA)T, and similarly for B.
Subtract end minus start: AB = b − a, computing each component as (end − start).
If reversed direction needed: BA = −AB — flip every sign.
To find a point from a displacement: add the displacement vector to the start point’s position vector.
For chains via a third point: use the triangle relation AB = AC + CB (or rearrange to get any side from the other two).
Worked examples
WE 1
Position vector from coordinates
Find the position vector of the point P with coordinates (5, −3, 2). Write your answer in both column form and base-vector form.
position vector = coordinates as a vectorOP = (5, −3, 2)TOP = (5, −3, 2)T = 5i − 3j + 2kno calculation needed — the coordinates *are* the components of the position vector.
WE 2
Displacement from two points
Points A and B have coordinates (1, 4, −2) and (6, −1, 3) respectively. Find the displacement vector AB.
write each position vectorOA = (1, 4, −2)T; OB = (6, −1, 3)TAB = b − a (end minus start)AB = (6−1, −1−4, 3−(−2)) = (5, −5, 5)TAB = (5, −5, 5)T = 5i − 5j + 5kthird component: 3 − (−2) = 3 + 2 = 5; watch the double negative.
WE 3
Compare AB and BA
Given A = (4, 0, 7) and B = (−3, 2, 5), find BA and compare with AB.
BA = a − b (start of BA is B)BA = (4 − (−3), 0 − 2, 7 − 5) = (7, −2, 2)TAB = b − aAB = (−3 − 4, 2 − 0, 5 − 7) = (−7, 2, −2)TBA = (7, −2, 2)T; AB = (−7, 2, −2)T = −BAcomponent by component, BA flips every sign of AB — opposite direction, same length.
WE 4
Find B given A and AB
The point A has position vector 2i − j + 5k. The displacement vector AB = 3i + 4j − 2k. Find the coordinates of point B.
OB = OA + ABOA = (2, −1, 5)TAB = (3, 4, −2)Tadd component by componentOB = (2+3, −1+4, 5+(−2)) = (5, 3, 3)TB = (5, 3, 3)B is found by walking from A in the direction of AB — its components become A’s coordinates.
WE 5
Combine two displacements
Three points satisfy AB = (2, −3, 4)T and BC = (−1, 5, −2)T. Find the displacement vector AC.
triangle relation: AC = AB + BCAC = (2 + (−1), −3 + 5, 4 + (−2)) = (1, 2, 2)TAC = (1, 2, 2)T = i + 2j + 2k“A to C via B” = A to B + B to C — adding nose-to-tail works regardless of where the points are.
WE 6
Find B from a scaled displacement
The points A and C have position vectors (3, 1, −2)T and (−1, 5, 4)T respectively. The point B is such that AB = 2AC. Find the coordinates of B.
find AC = c − aAC = (−1−3, 5−1, 4−(−2)) = (−4, 4, 6)Tscale: AB = 2 ACAB = (−8, 8, 12)Tfind OB = OA + ABOB = (3 + (−8), 1 + 8, −2 + 12) = (−5, 9, 10)TB = (−5, 9, 10)geometrically, B lies on the line from A through C, twice as far from A as C is — extension by factor 2.
💡 Top tips
Always “end minus start”: for AB the start is A so subtract A’s coordinates; for BA subtract B’s coordinates. Order matters.
Draw a quick sketch — even rough — to spot which way each arrow points before you compute.
Use the triangle relation for “via” questions: AC = AB + BC avoids having to know B’s actual coordinates if both displacements are given.
To find a point from a displacement: OB = OA + AB. Adding displacements to position vectors keeps the geometry intact.
Double-check signs: subtracting a negative coordinate flips the sign, e.g. 3 − (−2) = 5.
⚠ Common mistakes
Reversing the subtraction: writing AB = a − b instead of b − a. That gives BA, not AB.
Treating coordinates as a displacement — (3, −1, 2) is a point, not a vector from one named point to another unless context makes that clear.
Forgetting the triangle direction: AB + BC = AC only works if the middle letters match (B at the end of the first matches B at the start of the second).
Subtracting a negative wrong: 4 − (−3) = 7, not 1. Two negatives = plus.
Confusing position with displacement: OB is a position vector (from O), AB is a displacement (from A) — they’re computed differently.
Next up — Magnitude of a Vector & Unit Vectors. The magnitude |v| is the vector’s length: a Pythagoras-style square root of the squared components. Combined with the displacement rule AB = b − a you just learned, the magnitude of AB gives you the distance between any two points — the most commonly tested application of vectors on the AI HL syllabus.
Need help with Vectors?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
