IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read HL only

Intersections of a Line & a Plane

A line and a plane in 3D can do exactly three things: meet at a single point, run parallel without ever meeting, or have the line lying entirely in the plane. The dot product of the line’s direction with the plane’s normal tells you which case you’re in.

📘 What you need to know

Three possible cases: intersect at one point, parallel (no intersection), or line lies in the plane.
Parallel test: b · n = 0, where b is the line’s direction and n is the plane’s normal.
If parallel: substitute the line’s anchor into the plane equation. Satisfies it → line lies in plane; doesn’t → parallel but distinct (no intersection).
If not parallel: a unique intersection point exists.
Method (Cartesian plane): write parametric equations of the line, substitute into ax + by + cz = d, solve for λ, plug back into the line.
Method (vector plane): equate r-line = r-plane, solve the system in λ, μ, ν.
Always check the parallel condition first — saves time if it turns out to be parallel.

The three cases

Case	Direction · Normal	Anchor on plane?
Single intersection	≠ 0 (not parallel)	not relevant
Line in plane	= 0 (parallel)	yes
Parallel, no intersection	= 0 (parallel)	no

Decision flow: first compute b · n. If non-zero → unique intersection (find λ). If zero → check the anchor: on the plane means line in plane; off the plane means parallel and disjoint.

Finding the intersection point

Substitution method Sub x = x₀ + λl, y = y₀ + λm, z = z₀ + λn into ax + by + cz = d

This collapses three unknowns into one equation in λ. Solve, then substitute that λ back into the line’s parametric equations to find the (x, y, z) coordinates.

🧭 Recipe — find the intersection of a line and a plane

Check parallel: compute b · n. If 0, you’re in case 2 or 3 — check the anchor next.
If not parallel: write parametric equations of the line.
Substitute into the Cartesian plane equation.
Solve for the parameter λ.
Substitute λ back into the line’s parametric equations to get the intersection point. Verify it satisfies the plane equation.

Worked examples

WE 1

Find the point of intersection (basic)

Find the point of intersection of the line r = (1, 0, 2) + λ(2, −1, 1) with the plane x + y + 2z = 8.

Step 1: Check not parallel: b · n = (2)(1) + (−1)(1) + (1)(2) = 3 ≠ 0 ✓ Step 2: Parametric form x = 1 + 2λ; y = −λ; z = 2 + λ Step 3: Substitute into plane equation (1 + 2λ) + (−λ) + 2(2 + λ) = 8 1 + 2λ − λ + 4 + 2λ = 8 → 5 + 3λ = 8 → λ = 1 Step 4: Sub λ = 1 into line (1+2, 0−1, 2+1) = (3, −1, 3) Intersection at (3, −1, 3) verify: 3 + (−1) + 2(3) = 8 ✓

WE 2

Find the intersection (negative parameter value)

Find the point of intersection of the line r = (3, 4, 5) + λ(1, −1, 2) with the plane 2x + y − 3z = 5.

Step 1: Check not parallel b · n = (1)(2) + (−1)(1) + (2)(−3) = 2 − 1 − 6 = −5 ≠ 0 ✓ Step 2: Substitute parametric into plane 2(3+λ) + (4−λ) − 3(5+2λ) = 5 6 + 2λ + 4 − λ − 15 − 6λ = 5 −5 − 5λ = 5 → λ = −2 Step 3: Sub λ = −2 into line (3−2, 4+2, 5−4) = (1, 6, 1) Intersection at (1, 6, 1) negative λ just means the intersection is on the “backward” side of the anchor

WE 3

Show line is parallel and does not intersect

Show that the line r = (1, 2, 3) + λ(1, −2, 1) is parallel to the plane x + y + z = 10 but does not lie in it.

Step 1: Check b · n = 0 b · n = (1)(1) + (−2)(1) + (1)(1) = 0 ✓ → direction is perpendicular to the normal → parallel Step 2: Test anchor (1, 2, 3) on plane 1 + 2 + 3 = 6 ≠ 10 ✗ Line is parallel to the plane but does NOT lie in it no point on the line satisfies x + y + z = 10 — they all give 6

WE 4

Show a line lies in a plane

Show that the line r = (2, −1, 5) + λ(1, −2, 1) lies entirely in the plane x + y + z = 6.

Step 1: Check parallel: b · n (1)(1) + (−2)(1) + (1)(1) = 0 ✓ → parallel Step 2: Test anchor on plane 2 + (−1) + 5 = 6 ✓ Both parallel AND anchor on plane → line lies in plane Optional check at λ = 1: point (3, −3, 6) 3 + (−3) + 6 = 6 ✓ The line lies entirely in the plane infinitely many intersection points — every point of the line is on the plane

WE 5

Intersection of a line through two points with a plane

The line l passes through A(1, 0, 2) and B(3, 4, −2). Find the point at which l meets the plane x + y − z = 4.

Step 1: Direction AB = (2, 4, −4), simplify by 2: b = (1, 2, −2) Step 2: Line: r = (1, 0, 2) + λ(1, 2, −2) x = 1 + λ; y = 2λ; z = 2 − 2λ Step 3: Substitute into plane (1+λ) + 2λ − (2−2λ) = 4 1 + λ + 2λ − 2 + 2λ = 4 −1 + 5λ = 4 → λ = 1 Step 4: Sub back (1+1, 2, 2−2) = (2, 2, 0) Intersection at (2, 2, 0) verify: 2 + 2 − 0 = 4 ✓

WE 6

Find an unknown for the line to be parallel to the plane

Find the value of a for which the line r = (1, 2, 3) + λ(2, a, −1) is parallel to the plane 3x + y − 2z = 5. With this value of a, determine whether the line lies in the plane.

Step 1: For parallel, b · n = 0 (2)(3) + (a)(1) + (−1)(−2) = 0 6 + a + 2 = 0 → a = −8 Step 2: Test anchor (1, 2, 3) on plane 3(1) + 2 − 2(3) = 3 + 2 − 6 = −1 −1 ≠ 5 → anchor not on plane a = −8; line is parallel but does NOT lie in the plane parallel and disjoint — there is no point of intersection

💡 Top tips

Check b · n first — if 0, save time and skip to the anchor test.
Use the Cartesian form of the plane for quickest substitution — avoid full vector form unless given.
Verify the answer: substitute the intersection point into the plane equation; it should satisfy it.
Simplify the direction by a common factor before substituting — fewer arithmetic steps.
“Line lies in plane” ≠ intersect at a point — they intersect at every point of the line.

⚠ Common mistakes

Skipping the parallel check — leading to algebra that gives 0 = (non-zero) (parallel, disjoint) or 0 = 0 (line in plane).
Sign errors when substituting the parametric equations into the plane.
Forgetting to plug λ back into the line — the answer is the (x, y, z) point, not just λ.
Confusing the anchor with the normal — the line’s anchor is a point; the plane’s normal is a direction.
Calling it “parallel” when b · n ≠ 0 — that’s the case of unique intersection.

Next: Intersections of Two Planes. Two non-parallel planes intersect along a line, not a single point. Two ways to find it: solve the two Cartesian equations simultaneously by parameterising one variable, or use n₁ × n₂ as the direction and find a single shared point.

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