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

Areas using the Vector Product

The magnitude |v × w| is the area of the parallelogram with v and w as adjacent sides. Halve it to get the area of the matching triangle. Two formulas, one fast tool for finding areas in 3D.

📘 What you need to know

Parallelogram area: A = |v × w| — in the formula booklet.
Triangle area: A = ½|v × w| — NOT in the booklet (a parallelogram is two triangles, so halve it).
v and w must be adjacent sides — both starting from the same vertex.
For triangle from 3 vertices A, B, C: form AB and AC, take their cross product, halve the magnitude.
Either form works: |v × w| = |v||w|sin θ — use whichever the question gives you.
Order of v and w doesn’t matter for area. (v × w and w × v have opposite signs, but same magnitude.)
Simplify surds — areas often come out as expressions like 5√6 or 8√2.

Two shapes, one cross product

Parallelogram

A = |v × w|

v and w are adjacent sides from the same vertex

Triangle

A = ½ |v × w|

half the parallelogram with the same two side vectors

Why the half? Two identical triangles sharing a side make a parallelogram. So a triangle’s area is half the parallelogram with the same base vectors.

Starting from coordinates

Given three points A, B, C in 3D space:

Triangle area from three vertices Area of △ABC = ½ |AB × AC|

Form two displacement vectors out of the same vertex (here A), cross-multiply, and halve the magnitude. Pick whichever vertex you want — same answer.

🧭 Recipe — find the area of a triangle from three points

Pick a vertex (call it A) and form two vectors leaving it: AB = b − a and AC = c − a.
Compute the cross product AB × AC.
Take the magnitude — square components, sum, take √.
Halve it for triangle (skip this step for parallelogram).
Simplify the surd if possible (√150 = 5√6, etc.).

Worked examples

WE 1

Parallelogram area from two adjacent vectors

Find the exact area of the parallelogram with adjacent sides v = (3, −1, 2) and w = (1, 2, −1).

Step 1: Compute v × w i: (−1)(−1) − (2)(2) = 1 − 4 = −3 j: (2)(1) − (3)(−1) = 2 + 3 = 5 k: (3)(2) − (−1)(1) = 6 + 1 = 7 v × w = (−3, 5, 7) Step 2: Magnitude |v × w|² = 9 + 25 + 49 = 83 Area = √83

WE 2

Triangle area from three vertices

The points A, B, and C have coordinates (2, 1, −1), (4, 0, 3), and (1, 2, 2). Find the exact area of triangle ABC.

Step 1: Form AB and AC AB = B − A = (2, −1, 4) AC = C − A = (−1, 1, 3) Step 2: Cross product i: (−1)(3) − (4)(1) = −7 j: (4)(−1) − (2)(3) = −10 k: (2)(1) − (−1)(−1) = 1 AB × AC = (−7, −10, 1) Step 3: Magnitude |AB × AC|² = 49 + 100 + 1 = 150 = 25 × 6 |AB × AC| = 5√6 Step 4: Halve for triangle Area = 5√62

WE 3

Triangle area from two side vectors

Two adjacent sides of a triangle are u = (4, 0, −3) and v = (2, 5, 1). Find the exact area of the triangle.

Step 1: Cross product i: (0)(1) − (−3)(5) = 0 + 15 = 15 j: (−3)(2) − (4)(1) = −6 − 4 = −10 k: (4)(5) − (0)(2) = 20 u × v = (15, −10, 20) Step 2: Magnitude |u × v|² = 225 + 100 + 400 = 725 = 25 × 29 |u × v| = 5√29 Step 3: Halve for triangle Area = 5√292

WE 4

Parallelogram area using the sin formula

A parallelogram has adjacent sides of magnitudes |a| = 8 and |b| = 6, with the angle between them 60°. Find the exact area.

Use Area = |a × b| = |a||b| sin θ A = 8 × 6 × sin 60° = 48 × √32 Area = 24√3 use this when you don’t have components — just lengths & angle

WE 5

Parallelogram area from four vertices

The points A(1, 1, 0), B(4, 2, 1), C(5, 5, 4), and D(2, 4, 3) form a parallelogram ABCD. Find the exact area.

Step 1: Form two adjacent sides from A AB = B − A = (3, 1, 1) AD = D − A = (1, 3, 3) Step 2: Cross product i: (1)(3) − (1)(3) = 0 j: (1)(1) − (3)(3) = −8 k: (3)(3) − (1)(1) = 8 AB × AD = (0, −8, 8) Step 3: Magnitude |AB × AD|² = 0 + 64 + 64 = 128 = 64 × 2 Area = 8√2 no halving here — parallelogram, not triangle

WE 6

Find an unknown given the area

The triangle with vertices A(0, 0, 0), B(2, 3, 0), and C(0, k, 2) has area √14. Find the positive value of k.

Step 1: Form vectors AB and AC AB = (2, 3, 0); AC = (0, k, 2) Step 2: Cross product i: (3)(2) − (0)(k) = 6 j: (0)(0) − (2)(2) = −4 k: (2)(k) − (3)(0) = 2k AB × AC = (6, −4, 2k) Step 3: Magnitude squared |AB × AC|² = 36 + 16 + 4k² = 52 + 4k² Step 4: Set Area = ½|AB × AC| = √14, square both sides ¼(52 + 4k²) = 14 13 + k² = 14 → k² = 1 k = 1 (positive)

💡 Top tips

Always start two vectors from the SAME vertex — they must be adjacent sides for the formula to work.
Halve only for triangles. Parallelograms get the full magnitude.
Simplify surds at the end — √150 = 5√6, √128 = 8√2.
Squaring kills negatives — the magnitude depends only on the squared components.
For “find unknown given area” questions, set up |v × w|² and equate to (Area × 2)² for triangles.

⚠ Common mistakes

Forgetting the half for triangle area.
Using non-adjacent vectors like AB and BC. Both must come out of the same vertex.
Halving the parallelogram area by mistake when the question asks for parallelogram, not triangle.
Using the dot product by accident — area uses cross product, not dot product.
Not simplifying the final surd. The answer 5√6 is cleaner than √150.

Next note: Geometric Proof with Vectors. Use parallel, perpendicular, equal-length, and midpoint conditions to prove shapes are parallelograms, rectangles, rhombi, and to show that points are collinear or are at midpoints.

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