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

Equation of a Plane in Cartesian Form

The Cartesian form ax + by + cz = d is the cleanest plane equation — no parameters, just three coefficients and a constant. The coefficients (a, b, c) are the components of a normal vector perpendicular to the plane.

📘 What you need to know

Cartesian form: ax + by + cz = d (in the formula booklet).
Normal-form equation: r · n = a · n (in the formula booklet).
Normal vector n = (a, b, c) is perpendicular to the plane.
Find n from vector form: take n = b × c (vector product of the two direction vectors).
Find d: d = n · a for any anchor a on the plane (or substitute any known point’s coordinates).
Same plane if both LHS and RHS scale by the same factor: 2x + 3y + z = 5 and 4x + 6y + 2z = 10 are the same plane.
Different plane if only the LHS scales (parallel but not identical): 4x + 6y + 2z = 9 differs from 2x + 3y + z = 5.
Point-on-plane check: substitute coordinates into LHS — equals d? On plane.

The Cartesian form

Cartesian equation of a plane ax + by + cz = d

The coefficients (a, b, c) are the components of n, a normal vector to the plane. d is whatever n · a works out to — same value for every point on the plane.

Coefficients

(a, b, c) = n

a normal vector perpendicular to the plane

Constant

d = n · a

found by plugging any known point on the plane into the LHS

Converting from vector form

From r = a + λb + μc to Cartesian n = b × c, d = n · a

The cross product of the two direction vectors gives a vector perpendicular to both — exactly the normal direction. Then any point on the plane (e.g., the anchor a) gives the constant d.

When two equations give the same plane

Two Cartesian equations describe the same plane if and only if one is a scalar multiple of the other — meaning both sides scale by the same factor. If only the LHS scales but the RHS doesn’t match, the planes are parallel but distinct.

Equation 1	Equation 2	Relationship
2x + 3y + z = 5	4x + 6y + 2z = 10	Same plane (×2 throughout)
2x + 3y + z = 5	4x + 6y + 2z = 9	Parallel but different planes
2x + 3y + z = 5	4x + 5y + 2z = 10	Not parallel (LHS not proportional)

🧭 Recipe — convert vector form to Cartesian

Identify a, b, c from r = a + λb + μc.
Compute n = b × c — the normal vector.
Simplify n by removing common factors (optional but cleaner).
Compute d = n · a.
Write: ax + by + cz = d with the components of n as coefficients.

Worked examples

WE 1

Cartesian form given a point and a normal vector

The plane Π contains the point P(1, 4, −2) and has normal vector n = 2i − 3j + k. Find the Cartesian equation of Π.

Step 1: Read off coefficients a = 2, b = −3, c = 1 Step 2: Find d using d = n · P d = (2)(1) + (−3)(4) + (1)(−2) = 2 − 12 − 2 = −12 2x − 3y + z = −12 the normal’s components ARE the LHS coefficients — direct read-off

WE 2

Convert vector form to Cartesian form

Find the Cartesian equation of the plane r = (1, 0, 2) + λ(2, 1, −1) + μ(0, 3, 1).

Step 1: Find normal n = b × c i: (1)(1) − (−1)(3) = 4 j: −[(2)(1) − (−1)(0)] = −2 k: (2)(3) − (1)(0) = 6 n = (4, −2, 6) → simplify by 2: n = (2, −1, 3) Step 2: d = n · a = (2)(1) + (−1)(0) + (3)(2) d = 2 + 0 + 6 = 8 2x − y + 3z = 8 simplifying the normal by a common factor gives a cleaner equation — same plane

WE 3

Cartesian equation through three points

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

Step 1: Direction vectors AB = (2, 1, 2); AC = (−1, 3, −2) Step 2: n = AB × AC i: (1)(−2) − (2)(3) = −8 j: −[(2)(−2) − (2)(−1)] = 2 k: (2)(3) − (1)(−1) = 7 n = (−8, 2, 7) Step 3: d = n · A d = (−8)(2) + (2)(−1) + (7)(1) = −16 − 2 + 7 = −11 −8x + 2y + 7z = −11 (or equivalently 8x − 2y − 7z = 11) verify by checking n · B = n · C = d

WE 4

Determine which points lie on a plane

Determine whether the points P(2, −1, 0) and Q(1, 0, −1) lie on the plane with Cartesian equation 4x + 3y − 2z = 5.

Test P: substitute (2, −1, 0) 4(2) + 3(−1) − 2(0) = 8 − 3 − 0 = 5 ✓ Test Q: substitute (1, 0, −1) 4(1) + 3(0) − 2(−1) = 4 + 0 + 2 = 6 ≠ 5 ✗ P lies on the plane; Q does not Cartesian form makes point checks much faster than vector form — no system to solve

WE 5

Find a missing coordinate so a point lies on the plane

The point (1, k, 2) lies on the plane 2x − y + 3z = 5. Find the value of k.

Substitute the point’s coordinates 2(1) − k + 3(2) = 5 2 − k + 6 = 5 8 − k = 5 k = 3 verify: 2(1) − 3 + 3(2) = 2 − 3 + 6 = 5 ✓

WE 6

Find a value so two equations represent the same plane

The Cartesian equations 2x + 3y − z = 6 and 4x + 6y − 2z = k represent the same plane. Find the value of k.

Step 1: Compare LHS coefficients (4, 6, −2) = 2 × (2, 3, −1) ✓ — scaled by factor 2 Step 2: For the SAME plane, RHS must scale by the same factor k = 2 × 6 k = 12 if k were anything other than 12, the planes would be parallel but not identical

💡 Top tips

Coefficients = normal vector. From ax + by + cz = d, read off n = (a, b, c) directly.
Simplify the normal by a common factor before computing d — cleaner numbers.
Cartesian is fastest for point-on-plane checks — just one substitution.
For “same plane”, check that both the coefficient vector and the constant scale by the same factor.
Multiply through by −1 to make leading coefficients positive if you prefer.

⚠ Common mistakes

Treating the anchor a as the normal — the normal comes from b × c, not from a.
Sign errors in the cross product — j-component flips sign in the determinant expansion.
Forgetting to update d when scaling the equation — must scale both sides.
Confusing parallel with identical planes — same direction (LHS scales) doesn’t mean same plane unless RHS scales too.
Missing the negative sign when the answer comes out negative — e.g., −12 ≠ 12.

Next: Intersections of a Line & a Plane. Three possibilities: the line crosses the plane at one point, runs parallel and never touches, or lies entirely in the plane. The Cartesian form makes this fast — substitute the line’s parametric equations into ax + by + cz = d and solve for the parameter.

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