The Plane Equation

A flat surface — a wall, a floor, the edge of the camera's view — is a plane, and the question a game asks about it constantly is "which side am I on?". Is the player above the floor or through it? Inside the view, or off-screen? Both answers fall out of one tidy equation, built from a single point on the plane and the normal sticking out of it.

From a point and a normal to "which side", line by line

Pin the plane down with one known point on it, \vec{p}_0, and its normal \vec{n} (the direction it faces). Every other point \vec{p} on the plane is reachable from \vec{p}_0 by a step that lies in the plane.

Step 1 — the in-plane step is perpendicular to the normal. If \vec{p} is on the plane, the vector \vec{p} - \vec{p}_0 lies flat in the plane, so it is perpendicular to \vec{n} — and a dot product of perpendicular vectors is zero:

\vec{n}\cdot(\vec{p} - \vec{p}_0) = 0.

That is the plane: the set of all points whose displacement from \vec{p}_0 has no component along the normal.

Step 2 — expand to the standard form. Distribute the dot product and move the constant to the right. Writing d = \vec{n}\cdot\vec{p}_0:

\vec{n}\cdot\vec{p} - \vec{n}\cdot\vec{p}_0 = 0 \quad\Longrightarrow\quad \vec{n}\cdot\vec{p} = d, \qquad d = \vec{n}\cdot\vec{p}_0.

(In coordinates with \vec{n} = (a,b,c) this is the familiar ax + by + cz = d.) One vector \vec{n} and one number d pin the whole plane.

Step 3 — with a unit normal, the leftover is a signed distance. Take \vec{n} to be a unit vector, \lVert\vec{n}\rVert = 1. For any point \vec{p} in space — on the plane or not — the quantity

s = \vec{n}\cdot\vec{p} - d

is the signed distance from \vec{p} to the plane: its actual perpendicular distance, carrying a sign.

Step 4 — read the sign as the side. Because the normal has unit length, s is positive on the side the normal points toward, negative on the far side, and exactly zero on the plane itself:

s > 0\ (\text{normal's side}), \qquad s = 0\ (\text{on the plane}), \qquad s < 0\ (\text{the other side}).

"Which side am I on?" is the sign of one dot product minus a constant — no square roots, no division.

A plane is fixed by a point \vec{p}_0 on it and a normal \vec{n}:

Point–normal form — a point \vec{p} is on the plane exactly when \vec{n}\cdot(\vec{p} - \vec{p}_0) = 0.
Standard form — equivalently \vec{n}\cdot\vec{p} = d with d = \vec{n}\cdot\vec{p}_0.
Signed distance — with a unit normal, the signed distance from any point \vec{p} to the plane is s = \vec{n}\cdot\vec{p} - d.
Sign ⇒ side — s > 0 on the normal's side, s < 0 on the other, s = 0 on the plane.

The view a camera can see is a box — the frustum — bounded by six planes (left, right, top, bottom, near, far), each with its normal pointing inward. An object is visible only if it sits on the positive side of all six. So frustum culling is just six signed-distance tests: compute s = \vec{n}\cdot\vec{p} - d against each plane, and the moment any comes out negative, the object is outside the view and can be skipped. Whole armies are discarded this way before they ever touch the GPU.

Collision against a flat surface is the same test wearing different clothes. The signed distance of a ball's centre to the floor plane is how far the ball is from touching: when s drops below the radius the ball has penetrated, and the depth r - s tells you exactly how far to push it back out — along \vec{n}, of course. One equation, \vec{n}\cdot\vec{p} - d, quietly running both the renderer and the physics.