Points, lines and planes: the basic rules
A plane is a flat surface that goes on for ever in every direction. Some simple rules (axioms) are true in space:
- Through two different points there is exactly one line.
- Through three points that are not on one line there is exactly one plane.
- If two points of a line are in a plane, the whole line is in that plane.
- If two different planes share a point, they share a whole line.
A plane is also fixed by a line and a point not on it, by two crossing lines, or by two parallel lines.
Relative position of lines and planes (parallelism)
Two lines
They intersect (one common point), are parallel (same direction, in one plane, no common point), or are skew (not in any one plane, so they never meet and are not parallel).
A line and a plane
The line lies in the plane, is parallel to it (no common point), or cuts it at exactly one point. Test: a line is parallel to a plane if it is parallel to some line inside the plane.
Two planes
They are parallel or they meet in a line. Test: two planes are parallel if two crossing lines of one are parallel to the other.
Perpendicular lines and planes; the normal vector
A line is perpendicular to a plane if it is at 90° to every line in the plane. To check, it is enough to be at 90° to two crossing lines of the plane.
The direction of such a line is the normal vector n = (a, b, c). If P₀ = (x₀, y₀, z₀) is on the plane, every point (x, y, z) of the plane satisfies a(x − x₀) + b(y − y₀) + c(z − z₀) = 0, that is ax + by + cz = d.
Three perpendiculars theorem: if a slanted line meets a plane and its shadow (projection) on the plane is perpendicular to a line in the plane, then the slanted line is also perpendicular to that line.
Two planes are perpendicular when their normals are perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0.
Describing lines and planes with equations
A line through point A with direction vector u: r = A + t·u (vector form). In coordinates (parametric form): x = x₁ + t·u₁, y = y₁ + t·u₂, z = z₁ + t·u₃. Removing t gives the symmetric form (x − x₁)/u₁ = (y − y₁)/u₂ = (z − z₁)/u₃.
A plane through A with two direction vectors u and v: r = A + s·u + t·v. Its normal is n = u × v, which gives ax + by + cz = d.
Where a line meets a plane: put the parametric x, y, z into the plane equation and solve for t.
Angles and distances in space
- Angle between two lines (even skew ones): cos θ = |u·v| ÷ (|u||v|).
- Angle between a line and a plane: the angle φ between the line and its projection on the plane; sin φ = |u·n| ÷ (|u||n|).
- Angle between two planes (dihedral angle): the angle between their normals, cos θ = |n₁·n₂| ÷ (|n₁||n₂|).
- Distance from a point to a plane: D = |ax₀ + by₀ + cz₀ − d| ÷ √(a² + b² + c²).
- Distance between parallel planes ax + by + cz = d₁ and = d₂: |d₁ − d₂| ÷ √(a² + b² + c²).
Key formulas and definitions
- Line: r = A + t·u; (x − x₁)/u₁ = (y − y₁)/u₂ = (z − z₁)/u₃
- Plane: ax + by + cz = d, normal n = (a, b, c); r = A + s·u + t·v with n = u × v
- Angle between lines: cos θ = |u·v| / (|u||v|)
- Angle between line and plane: sin φ = |u·n| / (|u||n|)
- Angle between planes: cos θ = |n₁·n₂| / (|n₁||n₂|)
- Point to plane: D = |ax₀ + by₀ + cz₀ − d| / √(a² + b² + c²)
Worked examples
1. Find the equation of the plane through (1, 2, 3) with normal (2, −1, 4).
2(x − 1) − 1(y − 2) + 4(z − 3) = 0 → 2x − y + 4z − 2 + 2 − 12 = 0 → 2x − y + 4z = 12.
2. Write parametric equations of the line through A(1, 0, 2) and B(3, 4, 1).
Direction u = B − A = (2, 4, −1). So x = 1 + 2t, y = 4t, z = 2 − t.
3. Find the distance from P(1, 2, 3) to the plane 2x + 2y + z = 3.
D = |2·1 + 2·2 + 1·3 − 3| ÷ √(4 + 4 + 1) = |6| ÷ 3 = 2 units.
4. Where does the line x = 1 + t, y = 2t, z = 3 − t meet the plane x + y + z = 10?
(1 + t) + 2t + (3 − t) = 10 → 4 + 2t = 10 → t = 3. Point: (4, 6, 0).
5. Are the planes x + 2y − z = 4 and 2x − y = 7 perpendicular?
n₁·n₂ = 1·2 + 2·(−1) + (−1)·0 = 0. Yes, the normals are at 90°, so the planes are perpendicular.
6. Find the angle between the line with direction (1, 1, 0) and the plane z = 0.
Normal n = (0, 0, 1). u·n = 0, so sin φ = 0 and φ = 0°: the line is parallel to the plane (or lies in it).
7. Show that the lines L₁: (t, 0, 0) and L₂: (0, s, 1) are skew.
Directions (1, 0, 0) and (0, 1, 0) are not parallel. A common point needs z = 0 and z = 1 at once: impossible. Not parallel and no common point, so they are skew.
Common mistakes
- Thinking two lines that do not meet must be parallel. In space they may be skew.
- Checking perpendicularity to only ONE line of the plane. You need two crossing lines.
- Forgetting the absolute value or the √(a² + b² + c²) in the distance formula.
- Using cos instead of sin for the angle between a line and a plane (the normal is at 90° to the plane).