Ray–Plane Intersection

Floors, walls, ceilings, the still surface of water — the flat world of a game is made of planes. Firing a ray at one is the cheapest intersection in the whole tracer: where the sphere gave us a quadratic, the plane gives us a single linear equation, solved with one division. No discriminant, no \pm — just one t.

From plane equation to one division, line by line

A plane is fixed by a normal n and a constant d: a point P lies on it exactly when its projection onto the normal equals d.

n\cdot P = d.

Step 1 — substitute the ray. A point is on the plane when it satisfies this, and on the ray when it equals O + tD. Demanding both, put the ray in for P:

n\cdot(O + tD) = d.

Step 2 — distribute the dot product. The dot product is linear, so it splits the sum and pulls the scalar t out:

n\cdot O + t\,(n\cdot D) = d.

Step 3 — isolate the only unknown. Everything except t is a known number. Move n\cdot O across and divide by n\cdot D:

t = \frac{d - n\cdot O}{n\cdot D}.

That is the entire intersection — one dot product on top, one on the bottom, one division.

Step 4 — watch for n\cdot D = 0: the parallel case. If the ray's direction is perpendicular to the normal, the ray runs flat along the plane and the denominator vanishes. There is no single crossing point: either the ray never meets the plane (it floats above or below it forever) or it lies entirely within it. Geometrically n\cdot D = 0 means the ray is parallel to the surface — the one case the formula refuses to answer, so we test for it first and report "no hit".

n\cdot D = 0 \quad\Longrightarrow\quad \text{ray parallel to the plane — no intersection.}

Step 5 — reject hits behind the origin. A solved t < 0 sits behind the camera — the plane is in the rear-view mirror, not in front of the ray. Keep it only when t \ge 0.

Step 6 — recover the hit point. As always, feed the surviving t back into the ray:

P_{\text{hit}} = O + t\,D.

For a ray P(t) = O + tD and a plane n\cdot P = d:

One linear solve — the hit parameter is t = \dfrac{d - n\cdot O}{n\cdot D}.
Parallel ⇒ no hit — when n\cdot D = 0 the ray is parallel to the plane; there is no single crossing (test this first).
Forward only — keep the hit when t \ge 0; t < 0 is behind the origin.
Hit point — the surface point is P_{\text{hit}} = O + t\,D.

The formula intersects an infinite plane: a sheet of glass stretching forever in every direction. Solving for t always finds the crossing (unless the ray is parallel), even if that crossing is a mile off the edge of the actual floor tile.

Real geometry is bounded — a floor, a quad, a window. So a ray–plane hit is usually only step one: compute P_{\text{hit}} = O + tD on the infinite plane, then test whether that point lies inside the bounded patch. For an axis-aligned rectangle that is a pair of range checks on two coordinates; for an arbitrary quad or triangle it's a containment test in the plane. The plane gives you the where on the infinite sheet cheaply; the boundary test decides whether you actually landed on the surface or shot past its edge into the void.