IB Maths AA HL
Topic 3 — Geometry & Trigonometry
Paper 1 & 2
~7 min read
HL only
Intersections of Two Planes
Two non-parallel planes meet along a line (not a point) — like two pages of an open book joining at the spine. Two methods to find it: algebra (set one variable as λ, solve for the others) or the cross product (n1 × n2 gives the line’s direction).
📘 What you need to know
- Three cases: planes intersect at a line, are parallel and distinct (no intersection), or are coincident (same plane, infinite intersection).
- Parallel test: normal vectors are scalar multiples of each other.
- Same plane if both sides of the Cartesian equations scale by the same factor.
- Direction of intersection line: n1 × n2 (perpendicular to both normals).
- Algebra method: set one variable = λ, solve the two equations for the other two variables in terms of λ, write parametric form.
- Cross product method: take direction = n1 × n2; find a single shared point by setting one variable to 0 and solving the resulting 2×2 system.
- Multiple valid forms: the line equation can be written with different anchors and direction scalings — all describe the same line.
The three cases
| Case | Normals | RHS scaling | Intersection |
|---|
| Intersect along a line | not scalar multiples | — | a line |
| Parallel, distinct | scalar multiples | RHS doesn’t match | none |
| Same plane (coincident) | scalar multiples | RHS scales by the same factor | entire plane |
Two methods for the intersection line
Algebra method
set one variable = λ
solve the 2×2 system for the other two variables in terms of λ
Cross product method
direction = n1 × n2
find one shared point by setting a variable to 0; combine into r = a + λb
Which to use: algebra is more reliable on the no-calculator paper. Cross product is faster if the resulting 2×2 system is easy to solve. Either method is correct.
🧭 Recipe — find line of intersection (algebra method)
- Confirm not parallel: check the normals aren’t scalar multiples.
- Set one variable = λ (any of x, y, z — pick whichever simplifies).
- Solve the two equations simultaneously for the other two variables in terms of λ.
- Write parametric form: x = …, y = …, z = … (each in terms of λ).
- Stack into r = a + λb: read off the constants as the anchor and the λ-coefficients as the direction.
Worked examples
WE 1Find the line of intersection (algebra method)
Find a vector equation of the line of intersection of the planes Π1: 3x − y + z = 7 and Π2: x + y + 3z = 5.
Step 1: Normals (3, −1, 1) and (1, 1, 3) — not scalar multiples → not parallel ✓
Step 2: Let z = λ
3x − y = 7 − λ … (1)
x + y = 5 − 3λ … (2)
Step 3: Add (1) + (2)
4x = 12 − 4λ → x = 3 − λ
Step 4: Sub into (2)
y = 5 − 3λ − x = 5 − 3λ − (3 − λ) = 2 − 2λ
Step 5: Stack parametric → vector form
r = (3, 2, 0) + λ(−1, −2, 1)
verify at λ = 1: (2, 0, 1) → Π₁: 6 − 0 + 1 = 7 ✓; Π₂: 2 + 0 + 3 = 5 ✓
WE 2Find the line of intersection (cross product method)
Find a vector equation of the line of intersection of the planes Π1: 2x + y − z = 4 and Π2: x − y + 2z = 5.
Step 1: Direction = n₁ × n₂
i: (1)(2) − (−1)(−1) = 1
j: −[(2)(2) − (−1)(1)] = −5
k: (2)(−1) − (1)(1) = −3
Direction b = (1, −5, −3)
Step 2: Find a shared point — set z = 0
2x + y = 4; x − y = 5
Add: 3x = 9 → x = 3, y = −2
Step 3: Combine into vector form
r = (3, −2, 0) + λ(1, −5, −3)
verify: at point, 2(3) + (−2) − 0 = 4 ✓ and 3 − (−2) + 0 = 5 ✓
WE 3Show two planes are parallel and do not intersect
Show that the planes Π1: 2x − y + 3z = 4 and Π2: 4x − 2y + 6z = 5 are parallel and do not intersect.
Step 1: Compare normal vectors
n₁ = (2, −1, 3); n₂ = (4, −2, 6)
n₂ = 2 × n₁ ✓ → normals scalar multiples → parallel
Step 2: Compare RHS
For same plane, RHS would need to be 2 × 4 = 8, but Π₂ has 5
8 ≠ 5 → planes are different
Parallel and distinct — no intersection
two parallel non-coincident planes never meet, so the line of intersection doesn’t exist
WE 4Show two equations represent the same plane
Show that the equations 3x + 2y − z = 5 and 6x + 4y − 2z = 10 represent the same plane.
Step 1: Compare normals
(6, 4, −2) = 2 × (3, 2, −1) ✓ → parallel
Step 2: Compare RHS
2 × 5 = 10 ✓ → matches
Both equations represent the same plane
infinitely many points of “intersection” — every point of the plane satisfies both equations
WE 5Find a value to make two planes parallel
Find the value of k for which the planes Π1: 2x − y + 3z = 5 and Π2: 4x + ky + 6z = 7 are parallel. State whether they represent the same plane.
Step 1: For parallel, normals are scalar multiples
(4, k, 6) = c × (2, −1, 3)
Step 2: Use known components to find c
From x: 4 = 2c → c = 2
From z: 6 = 3c → c = 2 ✓
Step 3: Apply c to y-component
k = −1 × c = −2
Step 4: Same plane check
RHS would need to be 2 × 5 = 10, but Π₂ has 7
k = −2; planes parallel but NOT the same plane
parallel and distinct — no line of intersection
WE 6Find the line of intersection, then verify a point lies on it
(a) Find a vector equation of the line of intersection of Π1: 2x + y + z = 5 and Π2: x − y + 2z = 4. (b) Show that the point P(2, 0, 1) lies on this line.
Part (a): Algebra method, let x = λ
y + z = 5 − 2λ … (1)
−y + 2z = 4 − λ … (2)
Add: 3z = 9 − 3λ → z = 3 − λ
Sub (1): y = 5 − 2λ − (3 − λ) = 2 − λ
r = (0, 2, 3) + λ(1, −1, −1)
Part (b): Test P(2, 0, 1)
x: 0 + λ = 2 → λ = 2
y: 2 − λ = 0 → λ = 2 ✓
z: 3 − λ = 1 → λ = 2 ✓
P lies on the line (λ = 2)
P also satisfies Π₁: 2(2)+0+1 = 5 ✓ and Π₂: 2−0+2 = 4 ✓
💡 Top tips
- Check parallel first by comparing normals — saves time if there’s no line to find.
- Choose your λ-variable wisely — pick the one that gives the simplest 2×2 system.
- Cross product gives a quick check: the direction of the line should be perpendicular to both normals.
- Multiple valid answers — different anchors and scaled directions all describe the same line.
- Always sanity check by plugging a point on your line back into both plane equations.
⚠ Common mistakes
- Forgetting to check parallel first — you’ll waste time deriving “the line” before realising none exists.
- Treating “normals are scalar multiples” as “same plane” — must also check the RHS scales by the same factor.
- Sign errors when adding or subtracting the two plane equations.
- Computing the cross product backwards (j-component sign).
- Accepting “no solution” when expecting a line — re-pick the parameter variable; sometimes one choice leads to inconsistency.
Next: Angles Between a Line & a Plane. The angle between a line and a plane is the angle between the line and its projection onto the plane — found via cos−1 on the direction and the normal, then subtracted from 90° (or π/2). Same dot-product idea, with one extra step.
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