Practice Maths

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Planes and Intersections

Key Terms

Equation of a plane
ax + by + cz = d, where n = (a, b, c) is the normal vector to the plane.
Point-normal form
n · (r − a) = 0, i.e. a(x−x₀) + b(y−y₀) + c(z−z₀) = 0, where (x₀, y₀, z₀) is a known point on the plane.
Normal from two vectors in the plane
if u and v lie in the plane, then n = u × v.
Plane through 3 points
Use two edge vectors, find their cross product for n, then apply point-normal form.
Distance from point P₀=(x₀,y₀,z₀) to plane ax+by+cz=d
  dist = |ax₀+by₀+cz₀−d| / √(a²+b²+c²)
Line-plane intersection
substitute the parametric equations of the line into the plane equation and solve for λ.
Plane-plane intersection
Two non-parallel planes intersect in a line. Find direction d = n₁ × n₂, then find a point on both planes.
Angle between planes
= angle between their normal vectors: cosθ = |n₁ · n₂| / (|n₁||n₂|).
Key Formulas:
  • Plane equation: n · r = d (where d = n · a for any point a on the plane)
  • Normal: n = u × v for vectors u, v in the plane
  • Point-to-plane distance: |ax₀+by₀+cz₀−d| / √(a²+b²+c²)
  • Angle between planes: cosθ = |n₁·n₂| / (|n₁||n₂|)
Hot Tip To find the equation of a plane through three points A, B, C: compute AB = B−A and AC = C−A (two vectors in the plane), then n = AB × AC. Substitute any of the three points into n · r = d to find d. Check by verifying all three points satisfy the equation.

Worked Example 1 — Plane through a point with given normal

Find the equation of the plane through (1, 2, 3) with normal n = (2, −1, 4).

d = n · a = (2)(1) + (−1)(2) + (4)(3) = 2 − 2 + 12 = 12

Plane equation: 2x − y + 4z = 12

Worked Example 2 — Distance from a point to a plane

Find the distance from P = (3, 1, −2) to the plane 2x − y + 4z = 12.

dist = |2(3) − 1(1) + 4(−2) − 12| / √(4+1+16)

= |6 − 1 − 8 − 12| / √21 = |−15| / √21 = 15/√21 = 15√21/21 = 5√21/7 ≈ 3.27 units

Planes in 3D: The Point-Normal Characterisation

A plane in three dimensions is uniquely determined by a point on it and a normal vector (a vector perpendicular to every vector in the plane). If a is a position vector of a point on the plane and n = (a, b, c) is the normal, then any point P = (x, y, z) on the plane satisfies n · (r − a) = 0, which expands to ax + by + cz = d, where d = n · a.

This is the Cartesian (scalar) equation of the plane. Note that the coefficients (a, b, c) of x, y, z are exactly the components of the normal vector. This is the fastest way to read off the normal from a plane equation.

Finding the Normal from Two Vectors in the Plane

If two non-parallel vectors u and v lie in a plane, their cross product n = u × v is perpendicular to the plane. This is the key method for finding the plane through three points A, B, C: take u = AB = B−A and v = AC = C−A.

Line-Plane Intersection

Given a line r = a + λd and a plane n · r = d: substitute the parametric form of the line into the plane equation. If the result has a unique solution for λ, the line intersects the plane at that point. If the equation reduces to 0=0, the line lies entirely in the plane. If it reduces to 0 = nonzero, the line is parallel to the plane (no intersection).

The angle between a line with direction d and a plane with normal n is the complement of the angle between d and n: sinφ = |d·n|/(|d||n|).

Plane-Plane Intersection

Two non-parallel planes intersect in a line. The direction of this line is d = n₁ × n₂ (perpendicular to both normals, hence parallel to both planes). To find the full line equation, find one point common to both planes by setting one variable to a convenient value and solving the two plane equations simultaneously.

Angle Between Two Planes

The dihedral angle θ between two planes is the angle between their normal vectors (or its supplement if obtuse). Since we typically want the acute angle: cosθ = |n₁ · n₂| / (|n₁||n₂|).

Exam Tip: After finding a plane equation ax+by+cz=d, always verify by substituting all given points. For three-point problems, all three must satisfy the equation.
Exam Tip: For the distance formula, be careful with the sign: the numerator is |ax₀+by₀+cz₀ − d|. If your plane equation is written as ax+by+cz+e=0 (with e = −d), the formula becomes |ax₀+by₀+cz₀+e|/√(a²+b²+c²). Either form works — just be consistent.

Mastery Practice

See Answers ➔

  1. Find the Cartesian equation of the plane through the given point with the given normal. Fluency

    1. (a) Point (2, 1, −3), normal n = (1, 2, −1)
    2. (b) Point (0, 0, 4), normal n = (3, −2, 1)

  2. Find the equation of the plane through the three points A = (1, 0, 0), B = (0, 2, 0), C = (0, 0, 3). Fluency

  3. Find the distance from the given point to the given plane. Fluency

    1. (a) P = (1, 2, 3), plane 2x − y + 2z = 5
    2. (b) P = (0, 0, 0), plane 3x + 4y − 12z = 26

  4. Find the point where the line r = (1, 0, 2) + λ(2, 1, −1) intersects the plane x + 2y − z = 3. Fluency

  5. Find the line of intersection of the planes 2x + y − z = 4 and x − y + 2z = 1. Understanding

  6. Find the angle between the planes x + y + z = 6 and 2x − y + z = 3. Understanding

  7. Find the equation of the plane containing the point P = (2, 1, 3) and the line r = (0, 1, 0) + λ(1, 0, 2). Understanding

  8. Find the value of k such that the point (1, k, 2) lies on the plane 3x − 2y + 4z = 7. Understanding

  9. Find the equation of the plane through A = (2, 1, 0), B = (3, −1, 2), C = (1, 2, −1). Hence find the distance from the origin to this plane. Problem Solving

  10. Two planes Π₁: x + 2y − z = 3 and Π₂: 2x − y + kz = 5 are perpendicular. Find k. Then find the line of intersection. Problem Solving