IB Maths AI HL Vector Properties Paper 1 & 2 Quadrilateral proofs ~10 min read

Geometric Proof with Vectors

To prove a geometric property with vectors, match the property to a vector condition: parallel ⇔ scalar multiple (or zero cross product), perpendicular ⇔ zero dot product, equal length ⇔ equal magnitude, same line ⇔ collinear (one shared parallel direction). Stack these checks to identify shapes (parallelogram, rectangle, square, rhombus, kite, trapezium) and to find midpoints or split a segment in a given ratio.

📘 What you need to know

Parallel: u = kv for some scalar k ⇔ u × v = 0.
Perpendicular: u · v = 0 (with both non-zero).
Equal length: |u| = |v|.
Equal AND parallel: u = v (same direction) or u = −v (opposite direction).
Midpoint M of AB: position vector m = ½(a + b), and AM = ½AB.
Ratio split: if X divides AB in ratio p:q, then AX = pp+qAB.
Collinear A, B, C: any two of AB, AC, BC are parallel (i.e. scalar multiples).
Quadrilateral toolkit: identify shape from side/diagonal conditions (table below).

Vector conditions for geometric properties

Most proofs reduce to checking one or two of these vector facts. Pick the matching condition for what you need to prove, compute it, conclude.

Core vector conditions parallel: u = kv | perpendicular: u · v = 0 equal length: |u| = |v| | equal & parallel: u = v (or u = −v) midpoint of AB: m = ½(a + b) · collinear A, B, C: AB ∥ AC

Each quadrilateral has a unique fingerprint in vector terms. Stack the right conditions in the right order to identify the shape.

Quadrilateral identification — what to check

Label the four vertices in order (e.g. A, B, C, D) and form the four side vectors AB, BC, CD, DA. Then run the checklist:

Shape	Vector conditions to verify
Parallelogram	AB = −CD and BC = −DA (both pairs of opposite sides equal & parallel)
Rectangle	Parallelogram conditions plus AB · BC = 0 (adjacent sides perpendicular)
Rhombus	Parallelogram conditions plus \|AB\| = \|BC\| (adjacent sides equal length)
Square	Parallelogram conditions plus \|AB\| = \|BC\| plus AB · BC = 0
Trapezium	Exactly one pair of opposite sides parallel (e.g. AB = kCD for some k, but BC not parallel to DA)
Kite	AC · BD = 0 (diagonals perpendicular) plus no opposite sides parallel

Order of work: identify the most restrictive shape needed and check its conditions in order. For a square: parallelogram first → then equal sides → then perpendicular sides. Skipping the parallelogram step is the most common mistake.

🧭 Recipe — geometric proof with vectors

Sketch a diagram and label the points and known vectors clearly.
Form the relevant vectors from given position vectors (e.g. AB = b − a).
Match the property to a vector condition (parallel, perpendicular, equal length, collinear).
Compute the dot product, cross product, scalar multiple check, or magnitude as needed.
State the conclusion using “since … therefore …” linking the vector fact to the geometric claim.
For quadrilaterals: identify the most specific shape needed and run the conditions in order.

Worked examples

WE 1

Prove a quadrilateral is a parallelogram (from the PDF)

Points A, B, C, D have position vectors a = 3i − 5j − 4k, b = 8i − 7j − 5k, c = 3i − 2j + 4k, d = −2i + 5k. Prove ABCD is a parallelogram.

find the four side vectors AB = b − a = (5−2−1) BC = c − b = (−559) CD = d − c = (−521) DA = a − d = (5−5−9) check opposite-side equality AB = −CD ✓ BC = −DA ✓ ABCD is a parallelogram ✓ opposite sides equal & opposite directions → parallelogram (PDF’s exact answer).

WE 2

Prove it’s a rectangle (not just a parallelogram)

Show that the quadrilateral with vertices P(0, 0, 0), Q(4, 0, 0), R(4, 3, 0), S(0, 3, 0) is a rectangle.

side vectors PQ = (4, 0, 0) QR = (0, 3, 0) RS = (−4, 0, 0) SP = (0, −3, 0) opposite sides equal & parallel PQ = −RS ✓, QR = −SP ✓ adjacent sides perpendicular? PQ · QR = 4(0) + 0(3) + 0(0) = 0 ✓ parallelogram + adjacent ⊥ → rectangle ✓

WE 3

Prove three points are collinear

Points A(1, 2, 3), B(3, 5, 4), C(7, 11, 6). Prove A, B, C are collinear.

form AB and AC AB = b − a = (2, 3, 1) AC = c − a = (6, 9, 3) check scalar multiple AC = 3 · AB AB ∥ AC & share point A A, B, C are collinear ✓ “share a point” is automatic since AB and AC both start at A.

WE 4

Find the midpoint & verify it’s equidistant

A(1, −2, 4), B(5, 6, 0). Find the midpoint M of AB, and verify |AM| = |MB|.

midpoint formula m = ½(a + b) = ½(6, 4, 4) = (3, 2, 2) AM and MB AM = m − a = (2, 4, −2) MB = b − m = (2, 4, −2) magnitudes |AM| = √(4 + 16 + 4) = √24 |MB| = √24 ✓ M = (3, 2, 2); |AM| = |MB| = √24 AM = MB as vectors too — M sits exactly halfway.

WE 5

Ratio split of a line segment

Point P divides AB in the ratio 2 : 3, where A(0, 1, −2) and B(10, 6, 8). Find the position vector of P.

AP = (2/5) AB AB = b − a = (10, 5, 10) AP = (2/5)(10, 5, 10) = (4, 2, 4) OP = OA + AP p = (0, 1, −2) + (4, 2, 4) = (4, 3, 2) P = (4, 3, 2) ratio p:q means AP = (p / (p+q)) AB. Here 2 : 3 → 2/5.

WE 6

Identify the quadrilateral — rhombus or square?

The quadrilateral EFGH has E(0, 0, 0), F(2, 1, 2), G(3, 3, 0), H(1, 2, −2). Determine the most specific name for EFGH.

side vectors EF = (2, 1, 2), FG = (1, 2, −2) GH = (−2, −1, −2), HE = (−1, −2, 2) parallelogram? EF = −GH ✓, FG = −HE ✓ side lengths equal? (rhombus check) |EF| = √(4+1+4) = 3 |FG| = √(1+4+4) = 3 ✓ adjacent sides perpendicular? (square check) EF · FG = 2(1) + 1(2) + 2(−2) = 0 ✓ EFGH is a square ✓ parallelogram + equal sides + ⊥ adjacent = square. All three checks pass.

💡 Top tips

Always sketch the diagram — vector proofs are far easier when you can see the shape and label vectors directly on it.
Match property to vector condition: parallel → scalar multiple, perpendicular → dot product 0, equal length → equal magnitudes, collinear → AB ∥ AC.
Stack conditions for special shapes: rectangle = parallelogram + adjacent ⊥. Square = parallelogram + equal sides + adjacent ⊥.
For “is it a rectangle?” — verify parallelogram first, then check one pair of adjacent sides is perpendicular.
Label vertices in order (A → B → C → D going around the shape) so opposite sides are AB & CD, and BC & DA.
Use the dot product = 0 test for perpendicularity instead of cross product — fewer computations.

⚠ Common mistakes

Checking AB = CD instead of AB = −CD — going around the quadrilateral, opposite sides point in opposite directions, so the equality involves a minus sign.
Concluding “parallelogram” from one pair of opposite sides being equal — you need both pairs (in 3D, one pair alone doesn’t force the shape to close up).
Stopping at parallelogram when asked for rectangle/rhombus/square — always run the extra perpendicular or equal-length check.
Using positions instead of displacements — vector AB is b − a, not a + b or a.
Forgetting that “collinear” needs at least 2 of the 3 vectors checked — one parallel pair is enough since they share endpoints.
Mixing up the ratio formula — for ratio p:q, use AX = pp+qAB, not p/q.

Chapter complete — you now have all eleven Vector Properties sub-topics: introduction to vectors, parallel vectors, adding & subtracting, position & displacement, magnitude & unit vectors, the scalar product, angle between vectors & perpendicularity, the vector product, areas using the vector product, components of vectors, and geometric proof with vectors. Together they cover every Paper 1 & 2 vector-properties question on the AI HL syllabus.

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