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 1Parallelogram 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 2Triangle 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 3Triangle 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 4Parallelogram 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 5Parallelogram 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 6Find 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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