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

Equation of a Plane in Vector Form

A plane in 3D needs three pieces of information: an anchor point, plus two non-parallel direction vectors that lie in the plane. The vector equation r = a + λb + μc uses two parameters because a plane is two-dimensional — every point is reached by sliding along b by λ and c by μ.

📘 What you need to know

The vector equation

Vector equation of a plane r  =  a  +  λb  +  μc

Two parameters because a plane is 2D: λ = 0, μ = 0 lands on the anchor; sliding λ moves you along b, sliding μ moves you along c. Every point on the plane corresponds to one pair (λ, μ).

Anchor point
a
any known point on the plane
Two directions
b, c
non-parallel; both lie in the plane

Plane through three points

Plane through A, B, C r = a + λ AB + μ AC

Pick any of the three points as the anchor; the other two give the directions via subtraction. Three points define a plane only if they’re not all on the same line — i.e., AB and AC must not be parallel.

Quick non-collinearity check: compute the ratios of AB to AC componentwise. If they’re not all equal, the vectors aren’t parallel — three points define a plane.

Does a point lie on the plane?

To check if Q with position vector q lies on r = a + λb + μc: substitute q for r, write the three component equations in λ and μ, solve any two simultaneously, and verify the third. If all three are consistent — Q is on the plane.

🧭 Recipe — vector equation of a plane through three points

  1. Pick an anchor: take a as one of the three points.
  2. Find two direction vectors: AB = ba and AC = ca.
  3. Verify non-collinearity: AB and AC must not be parallel.
  4. Write: r = a + λAB + μAC.
  5. Sanity check: at λ = 1, μ = 0 you should land on B; at λ = 0, μ = 1 you land on C.

Worked examples

WE 1

Vector equation given a point and two direction vectors

Find a vector equation of the plane Π containing the point P(2, −1, 3) and parallel to the vectors b = i + 2jk and c = 3ij + 2k.

Apply r = a + λb + μc directly a = (2, −1, 3);   b = (1, 2, −1);   c = (3, −1, 2) Check b and c not parallel Ratios: 1/3, 2/(−1), −1/2 — not equal ✓ r = (2, −1, 3) + λ(1, 2, −1) + μ(3, −1, 2) two non-parallel directions are essential — parallel ones would only give a line
WE 2

Vector equation through three points

Find a vector equation of the plane passing through A(1, 0, 2), B(3, 4, −1), and C(−2, 5, 1).

Step 1: Two direction vectors AB and AC AB = B − A = (2, 4, −3) AC = C − A = (−3, 5, −1) Step 2: Check AB and AC not parallel Ratios 2/(−3), 4/5, −3/(−1) — not all equal ✓ Step 3: Use A as anchor r = (1, 0, 2) + λ(2, 4, −3) + μ(−3, 5, −1) choosing B or C as anchor gives different but equivalent equations
WE 3

Show a point lies on a plane

Determine whether the point Q(4, 8, 7) lies on the plane r = (1, 2, −1) + λ(2, 1, 3) + μ(−1, 4, 2).

Step 1: Set components equal x: 4 = 1 + 2λ − μ → 2λ − μ = 3 … (1) y: 8 = 2 + λ + 4μ → λ + 4μ = 6 … (2) z: 7 = −1 + 3λ + 2μ → 3λ + 2μ = 8 … (3) Step 2: Solve (1) and (2) From (1): μ = 2λ − 3; sub (2): 9λ = 18 → λ = 2, μ = 1 Step 3: Check (3) 3(2) + 2(1) = 8 ✓ Q lies on the plane (λ = 2, μ = 1) all three component equations must agree on the same λ, μ pair
WE 4

Show a point does NOT lie on a plane

Determine whether the point R(0, 0, −1) lies on the plane r = (1, −1, 2) + λ(1, 2, −1) + μ(2, 1, 3).

Step 1: Three component equations x: 0 = 1 + λ + 2μ → λ + 2μ = −1 … (1) y: 0 = −1 + 2λ + μ → 2λ + μ = 1 … (2) z: −1 = 2 − λ + 3μ → −λ + 3μ = −3 … (3) Step 2: Solve (1) and (2) 2×(1): 2λ + 4μ = −2; subtract (2): 3μ = −3 → μ = −1, λ = 1 Step 3: Check (3) −(1) + 3(−1) = −4 ≠ −3 ✗ R does NOT lie on the plane contradiction in the third equation — one mismatch is enough
WE 5

Plane through three points; check a fourth point

(a) Find a vector equation of the plane through A(2, 1, 0), B(4, 5, 2), and C(0, 3, 4). (b) Show that the point D(2, 7, 6) lies on this plane.

Part (a): Direction vectors AB = (2, 4, 2);   AC = (−2, 2, 4) r = (2, 1, 0) + λ(2, 4, 2) + μ(−2, 2, 4) Part (b): Set r = D and form equations x: 2 = 2 + 2λ − 2μ → λ − μ = 0 … (1) y: 7 = 1 + 4λ + 2μ → 2λ + μ = 3 … (2) z: 6 = 0 + 2λ + 4μ → λ + 2μ = 3 … (3) Solve (1) and (2): from (1) λ = μ, sub (2) → 3μ = 3 → μ = 1, λ = 1 Check (3): 1 + 2(1) = 3 ✓ D lies on the plane (λ = 1, μ = 1) multi-step questions are common — write the equation first, then test additional points
WE 6

Show three points define a plane, then find its equation

The points A, B, and C have coordinates (1, 2, 3), (3, 5, 1), and (2, 4, 6). (a) Show that A, B, and C do not lie on the same line. (b) Find a vector equation of the plane through A, B, and C.

Part (a): Compute AB and AC, check not parallel AB = (2, 3, −2) AC = (1, 2, 3) Ratios 2/1 = 2, 3/2 = 1.5, −2/3 ≈ −0.67 — not equal → AB and AC not parallel → A, B, C not collinear Part (b): Vector equation with A as anchor r = (1, 2, 3) + λ(2, 3, −2) + μ(1, 2, 3) the non-collinearity check is essential — without it, three points might just lie on a single line

💡 Top tips

⚠ Common mistakes

Next: Equation of a Plane in Cartesian Form. The Cartesian form ax + by + cz = d is much cleaner — the coefficients are the components of a normal vector to the plane (perpendicular to it). You’ll get there from the vector form by taking n = b × c.

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