Equation of a Plane and Converting Plane Representations
A plane in space has the general equation Ax + By + Cz + D = 0, where (A, B, C) is a normal vector perpendicular to the plane; the same plane can be given parametrically as r = P + s·u + t·v or by three non-collinear points, and each form can be converted into the others. In brief: the plane through P(1, 2, 3) with the normal n = (2, −1, 4) is 2(x − 1) − (y − 2) + 4(z − 3) = 0, that is 2x − y + 4z − 12 = 0. The distance from the origin to it is 12/√21 ≈ 2.62.
A plane is the three-dimensional counterpart of a line in the plane: one linear equation, one normal direction. This guide shows how to build a plane from a point and a normal, from three points, and from a parametric description, how to convert between forms, and how to find distances and angles, with worked examples and a table of special planes.
What is the general equation of a plane?
Let P = (x0, y0, z0) lie in the plane and n = (A, B, C) be perpendicular to it. A point X = (x, y, z) is in the plane exactly when PX · n = 0, that is A(x − x0) + B(y − y0) + C(z − z0) = 0, or Ax + By + Cz + D = 0 with D = −(Ax0 + By0 + Cz0). Multiplying the equation by a nonzero number gives the same plane; the normal is determined up to a multiple.
The other representations are the parametric form r = P + s u + t v (with independent vectors u, v lying in the plane) and the intercept form x/p + y/q + z/r = 1 for a plane meeting the axes at (p, 0, 0), (0, q, 0), (0, 0, r).
How do you build a plane from given data?
Example 1: a point and a normal
P(1, 2, 3), n = (2, −1, 4): 2(x − 1) − (y − 2) + 4(z − 3) = 0, so 2x − y + 4z − 12 = 0. Check P: 2 − 2 + 12 − 12 = 0 ✓.
Example 2: three points
A(1, 0, 0), B(0, 2, 0), C(0, 0, 3). The normal is AB × AC = (6, 3, 2). The equation is 6(x − 1) + 3y + 2z = 0, that is 6x + 3y + 2z = 6. In the intercept form, x/1 + y/2 + z/3 = 1, the same plane.
Example 3: a point and a line
The plane through P(1, 1, 1) and the line r = t(1, 2, 3). The direction d = (1, 2, 3) and the vector from a point of the line to P, (1, 1, 1), lie in the plane, so n = d × (1, 1, 1) = (2 · 1 − 3 · 1, 3 · 1 − 1 · 1, 1 · 1 − 2 · 1) = (−1, 2, −1). Equation: −x + 2y − z = 0 (the line passes through the origin). Check P: −1 + 2 − 1 = 0 ✓ and d: −1 + 4 − 3 = 0 ✓.
Example 4: a plane parallel to another
The plane through (2, 1, −1) parallel to x + 2y − 2z = 5 has the same normal: x + 2y − 2z = 2 + 2 + 2 = 6.
How do you convert between representations?
Example 5: from parametric to general
r = (1, 0, 2) + s(1, 1, 0) + t(0, 1, 1). The normal is u × v = (1 · 1 − 0 · 1, 0 · 0 − 1 · 1, 1 · 1 − 1 · 0) = (1, −1, 1). So (x − 1) − y + (z − 2) = 0, that is x − y + z − 3 = 0. Check the point with s = t = 1: (2, 2, 3): 2 − 2 + 3 − 3 = 0 ✓.
Example 6: from general to parametric
x + 2y − z = 4. Choose a point, for instance (4, 0, 0), and two independent directions orthogonal to the normal (1, 2, −1): u = (2, −1, 0) (check: 2 − 2 = 0) and v = (1, 0, 1) (check: 1 − 1 = 0). So r = (4, 0, 0) + s(2, −1, 0) + t(1, 0, 1). Alternatively, take x = s, y = t as free parameters and solve z = s + 2t − 4.
Example 7: to the intercept form
3x − 2y + 6z = 12. Divide by 12: x/4 + y/(−6) + z/2 = 1. The axis intercepts are (4, 0, 0), (0, −6, 0), (0, 0, 2).
How do you find distances and angles?
The distance from P0 = (x0, y0, z0) to Ax + By + Cz + D = 0 is |Ax0 + By0 + Cz0 + D|/√(A2 + B2 + C2). For parallel planes with the same normal, the distance is |D1 − D2|/|n|. The angle between planes equals the angle between their normals (acute): cos θ = |n1 · n2|/(|n1||n2|).
Example 8: a point and a plane
The distance from (1, 2, 3) to 2x − 2y + z − 9 = 0: |2 − 4 + 3 − 9|/3 = 8/3 ≈ 2.67.
Example 9: parallel planes
x + 2y + 2z − 3 = 0 and x + 2y + 2z + 6 = 0: the distance is |−3 − 6|/3 = 3.
Example 10: the angle between planes
x + y = 1 and y + z = 1: n1 = (1, 1, 0), n2 = (0, 1, 1), cos θ = 1/2, so θ = 60°.
Table of special planes
| Plane | Equation | Normal |
|---|---|---|
| the xy-plane | z = 0 | (0, 0, 1) |
| parallel to the yz-plane | x = 3 | (1, 0, 0) |
| parallel to the z-axis | x + y = 1 | (1, 1, 0) |
| through the origin | x − 2y + z = 0 | (1, −2, 1) |
| with intercepts 1, 2, 3 | 6x + 3y + 2z = 6 | (6, 3, 2) |
| through (1, 2, 3), normal (2, −1, 4) | 2x − y + 4z = 12 | (2, −1, 4) |
Where does the plane equation come from?
The linear equation of a plane appeared with the extension of Descartes’ coordinates to space in the seventeenth and eighteenth centuries; Euler’s works from the 1740s contain a systematic treatment of surfaces of the first and second degree. The normal form with a unit normal is named after Ludwig Otto Hesse.
Which mistakes are common?
Using a point instead of a normal
The coefficients (A, B, C) form the normal vector, not a point. A point is found by substituting values into the equation.
Forgetting the constant D
Computing D from a point of the plane is a required step; planes with the same normal but different D are parallel, not equal.
Choosing dependent directions
In the parametric form, u and v must be independent (u × v ≠ 0), otherwise the description gives only a line.
Mixing the angle between normals with the angle to a line
The angle between a line and a plane is 90° minus the angle between the direction and the normal.