Hesse Normal Form (Distance Point-Plane)

In three steps with the Hesse normal form. First: bring the plane into coordinate form ax + by + cz = d. Second: insert the point and subtract d; the absolute value of this is the numerator. Third: divide by the length of the normal vector √(a² + b² + c²). Example: E: 2x + y + 2z = 9 and P(3|1|5): inserting gives 6 + 1 + 10 = 17, minus 9 leaves 8; the denominator is √9 = 3, so d(P;E) = 8/3 ≈ 2.67 LU. Result 0 means the point lies in the plane. Forget neither subtracting d nor the division, those are the two classics.

Because the same plane has infinitely many coordinate equations: 2x + y + 2z = 9 and 4x + 2y + 4z = 18 describe identical point sets. If you only insert a point, you get different numbers depending on the equation, here 8 or 16; that cannot be a distance yet. Only the division by |n⃗| normalizes the equation and makes the result unique: 8/3 = 16/6. Geometrically a projection is behind it: the distance is the length of the projection of the connection vector onto the normal direction, and for that you need the unit normal vector n⃗/|n⃗| of length 1. The division is thus not a convention but the actual core of the formula.

Without the absolute value the Hesse normal form yields the signed distance: the sign tells on which side of the plane the point lies. Positive means the point is on the side the normal vector points to; negative means the opposite side. With this you can check whether two points lie on the same side of a plane (equal signs) or whether a segment pierces the plane (different signs) without computing the intersection. Example: for E: 2x + y + 2z = 9, P(3|1|5) gives +8/3, while the origin gives −9/3 = −3; so they lie on different sides. For the geometric distance you take the absolute value at the end.

Bring both planes to the same normal vector; then the equations differ only in the right-hand side, and d = |d₁ − d₂|/|n⃗| holds. Example: 2x + y + 2z = 9 and 2x + y + 2z = 3 have |n⃗| = 3, so the distance is |9 − 3|/3 = 2 LU. If the second plane comes with a multiple of the normal vector (say 4x + 2y + 4z = 6), divide the equation by the factor first. Alternatively pick any point of one plane and compute its distance to the other with the ordinary HNF formula; both lead to the same result. Non-parallel planes intersect, their distance is 0.

For relative positions of sphere and plane the distance of the centre M from the plane compared with the radius r decides. If d(M;E) > r, the plane misses the sphere; if d(M;E) = r, it touches the sphere (tangent plane); if d(M;E) < r, it cuts out a circle. Example: sphere with M(3|1|5), r = 3 and E: 2x + y + 2z = 9: d = 8/3 ≈ 2.67 < 3, so the plane cuts the sphere; by Pythagoras the intersection circle has radius √(r² − d²) = √(9 − 64/9) ≈ 1.37. Tangent planes are constructed the same way: all planes at distance r from the centre.