Plane Equation Visualization | Normal, Distance, Intercepts, Traces

The idea: a quadric with no squares

Take the general quadric Ax² + By² + Cz² + Dxy + Eyz + Fzx + Gx + Hy + Iz + J = 0 and set every second-degree coefficient to 0. What is left is linear, and it is written

Ax + By + Cz + D = 0, with A, B, C not all 0.

Its points form a plane, the flat, degenerate member of the quadric family. Four numbers describe it, but they do two different jobs: (A, B, C) sets which way the plane faces, and D sets how far it is from the origin.

The plane x + y + 2z − 2 = 0 drawn as a tilted parallelogram in 3D axes. From its foot point F = (1/3, 1/3, 2/3) the normal vector n = (1, 1, 2) points away, at a right angle to the plane. An orange segment from the origin O to F is the distance 2/√6 ≈ 0.82 — (A, B, C) is the normal n; the plane's closest point to O is F, at distance |D|/|n|.

(A, B, C) is the normal vector

Write the left side as a dot product: Ax + By + Cz = n · p with n = (A, B, C) and p = (x, y, z). Take two points p and q of the plane. Both give n · p = −D, so n · (q − p) = 0: n is at a right angle to every direction that lies in the plane. That makes n the normal vector. Change A, B or C and the normal turns, so the plane turns with it. (Demo 2: A, B, C turn the normal) The canvas draws n as an arrow starting on the plane.

D slides the plane: distance and foot point

Keep A, B, C and change D: the normal does not change, so the new plane is parallel to the old one. It has slid along n. (Demo 1: D slides the plane along the normal) The point of the plane nearest the origin is on the line through O along n, p = t·n. Putting it in the equation gives t|n|² + D = 0, so

what	formula	for x + y + 2z − 2 = 0
foot point F	−D·n / \|n\|²	(2/6)(1, 1, 2) = (1/3, 1/3, 2/3)
distance from O	\|D\| / √(A² + B² + C²)	2/√6 ≈ 0.82
D = 0	distance 0	the plane goes through the origin

The same idea gives the distance from any point p₀: |n · p₀ + D| / |n|.

Intercepts and the intercept form

On the x-axis y = z = 0, so Ax + D = 0 and the plane crosses it at x = −D/A. In the same way it crosses the y-axis at −D/B and the z-axis at −D/C. If A = 0 there is no x-intercept: Ax + D = 0 has no solution (or every x, when D = 0 too), and the plane is parallel to the x-axis. (Demo 3: intercepts and x/p + y/q + z/r = 1)

When D ≠ 0 and the intercepts are p, q, r, divide the equation by −D:

x/p + y/q + z/r = 1

For x + y + 2z − 2 = 0 that is x/2 + y/2 + z/1 = 1. Drawing a plane by hand starts here: mark the three intercepts and join them.

Left: the plane x + y + 2z − 2 = 0 in the first octant as a triangle with corners at the intercepts (2, 0, 0), (0, 2, 0) and (0, 0, 1); its sides are the traces x + y = 2 in green, x + 2z = 2 in purple and y + 2z = 2 in teal. Right: with A = 0 the plane y + 2z − 2 = 0 has no x-intercept and runs parallel to the x-axis like a ramp — The intercepts are the corners of the triangle; the traces are its sides. A = 0 leaves the plane parallel to the x-axis.

Traces and the cut z = h

A trace is where the plane meets a coordinate plane. Put z = 0, y = 0 or x = 0 into the equation:

trace	equation	for x + y + 2z − 2 = 0
with z = 0 (green)	Ax + By + D = 0	x + y = 2
with y = 0 (purple)	Ax + Cz + D = 0	x + 2z = 2
with x = 0 (teal)	By + Cz + D = 0	y + 2z = 2

Each one is a line through two intercepts. Cutting at any height z = h gives Ax + By = −D − Ch: the same left side as the trace z = 0, only another right side. So all cuts z = h are parallel lines, the plane's contour lines, and they move evenly as h grows. (Demo 6: the cut z = h slides a parallel line) If A = B = 0 the left side is empty: the cut is the whole plane when h = −D/C and nothing otherwise.

Special planes: C = 0, A = B = 0, D = 0

case	equation	the plane
C = 0	Ax + By + D = 0	vertical, parallel to the z-axis: z is free, so the trace line is pushed straight up (a "cylinder" over a line)
A = B = 0	z = −D/C	horizontal; its normal points straight up
B = C = 0	x = −D/A	parallel to the yz-plane
D = 0	Ax + By + Cz = 0	through the origin; all intercepts are 0

A missing variable means the plane is parallel to that axis, just as a missing z makes the elliptic cylinder a tube along the z-axis. (Demo 4: special planes: C = 0, A = B = 0, D = 0)

Three small 3D sketches. C = 0: the vertical plane x + y = 2 standing on its green trace with a horizontal normal. A = B = 0: the horizontal plane z = 1 with its normal pointing up. D = 0: the plane x + y + 2z = 0 passing through the origin O — A zero in the equation shows up as a parallel axis or a plane through O.

Same plane, scaled equation

Multiply all four numbers by k ≠ 0: kAx + kBy + kCz + kD = k · 0 = 0, so exactly the same points satisfy it. 2x + 2y + 2z − 2 = 0 is the plane x + y + z − 1 = 0. The normal gets k times longer, and for k < 0 it points the other way, but the distance |kD| / |kn| does not change. (Demo 5: 2·(A, B, C, D) is the same plane) That is why a plane is often written with a unit normal, |n| = 1: then D itself is the signed distance.

See also Quadric Surfaces, Elliptic Cylinder, Hyperbolic Paraboloid (a curved surface that still contains straight lines), Vectors and Dot Product and Changing Coordinates in 3D.

What the page leaves out

The plane through three points (take n as the cross product of two edges), the point-normal form n · (p − p₀) = 0, the angle between two planes (the angle between their normals), the line where two planes meet, the two half-spaces Ax + By + Cz + D > 0 and < 0, and parametric planes p₀ + s·u + t·v.