How this instrument works
A plane written as Ax+By+Cz+D=0 hides a second object inside it for free: the vector (A, B, C) is normal to the plane — it points straight out, perpendicular to every line drawn on the plane's surface. That normal vector is the key to the distance formula. Plug the point (x₀, y₀, z₀) into the plane's own left-hand side and you get a number, Ax₀+By₀+Cz₀+D, that measures how far off the plane the point sits, but scaled by the length of that normal vector. Dividing by √(A²+B²+C²) undoes the scaling and leaves an honest length.
The geometry behind that division is a projection. Picture the short segment running from any point on the plane to (x₀, y₀, z₀); only its component along the normal direction survives once you divide by the normal's own length, and that surviving component is exactly the perpendicular drop from the point to the plane — the shortest route there is, since any other path to the plane is the hypotenuse of a right triangle with this one as a leg. Drop the absolute value bars and the sign tells you which side of the plane the point is on, a detail 3D graphics engines use constantly to decide what to clip.
One property worth sitting with: the formula is scale-invariant in its inputs. Multiply A, B, C, and D all by the same nonzero number — say, doubling 1x+2y−2z+3=0 into 2x+4y−4z+6=0 — and you get the identical plane, just written differently, so the distance it reports for any point does not move at all. The numerator and denominator both pick up that same factor and it cancels. The one case the formula cannot handle is A=B=C=0, since there is no plane, and no normal vector, left to divide by.
- Enter your plane's coefficients into A, B, C, and D, matching the standard form Ax+By+Cz+D=0 shown beside the A field.
- Enter the point you are measuring from into Point: x₀, Point: y₀, and Point: z₀.
- Read Distance — the shortest, perpendicular length from that point to the plane, reported in whatever length unit your coordinates already use.
- Adjust any single coefficient or coordinate and Distance recalculates immediately, which is a quick way to see how sensitive the result is to each number.
Worked example — a point one third of a unit off the plane
Take the plane x+2y−2z+3=0, so A=1, B=2, C=−2, D=3, and check the point (4, −1, 3). Plugging the point straight into the plane's left-hand side gives 1(4)+2(−1)+(−2)(3)+3 = 4−2−6+3 = −1, so the numerator's absolute value is |−1| = 1.
The normal vector for this plane is (1, 2, −2), whose length is √(1²+2²+(−2)²) = √9 = 3. Dividing gives d = 1 ⁄ 3 = 0.3333333333333333 — exactly what Distance reports, confirming the point sits precisely a third of a unit off the plane, the same shortest-distance-to-a-line idea from flat geometry lifted one dimension up.
Questions
How is the point-to-plane distance formula actually derived?
It comes from projecting the vector between the point and any point on the plane onto the plane's unit normal. Ax₀+By₀+Cz₀+D already measures that offset scaled by the normal's length √(A²+B²+C²), so dividing by that same length converts the scaled number back into a true, unscaled distance.
Does it matter how I write the plane's equation?
No. Multiplying every coefficient by the same nonzero constant describes the identical plane and returns the identical distance, since the factor appears in both the numerator and the denominator and cancels out. 1x+2y−2z+3=0 and 2x+4y−4z+6=0 are the same plane and give the same answer for any point.
What does the sign inside the absolute value bars actually mean?
Without the absolute value, Ax₀+By₀+Cz₀+D divided by the normal's length is a signed distance: positive on the side the normal vector points toward, negative on the other side, zero exactly on the plane. Rendering engines use that sign to decide which side of a cutting plane an object falls on.
How does this relate to the distance from a point to a line in 2D?
It's the same idea one dimension up. A line Ax+By+C=0 has distance |Ax₀+By₀+C| ⁄ √(A²+B²) from a point, using the line's own two-number normal instead of a plane's three-number one. Both formulas plug the point into the boundary's equation, then normalize by the length of whatever direction is perpendicular to that boundary.
What happens if the point already lies on the plane?
The distance comes out to exactly zero, because plugging a point that satisfies Ax+By+Cz+D=0 into the numerator makes it vanish. For the plane x=0 and the point (0, 3, 3), for instance, this sheet returns 0 — the point isn't off the plane at all, it's sitting on it.
Why does the formula reduce to the point's height for the plane z=0?
For A=0, B=0, C=1, D=0, the formula simplifies to |z₀| ⁄ 1 = |z₀|, plain and simple. A point at (5, 7, 4) is a distance of exactly 4 from the xy-plane, matching the plain intuition that height above a flat floor is already the perpendicular distance to it.