latentspace.

Plane Equation


A plane in 3D coordinate space is determined by a point and a vector that is perpendicular to the plane.

Let \(P_0 = (x_0, y_0, z_0) \) be the point given, and \(n\) the orthogonal vector. Also let \(P = (x, y, z)\) be any point in the plane, and \(r\) and \(r_0\) the position vectors of points \(P\) and \(P_0\), respectively. Now, if we let \(n = (a,b,c) \), then since \(\overrightarrow{P_0 P} \) is perpendicular to \(n\), we have \[ \,\\ \begin{align} \overrightarrow{P_0 P} \cdot n &= (P - P_0) \cdot n \\ &= (x - x_0, y - y_0, z - z_0) \cdot (a, b, c) \\ &= a(x - x_0) + b(y - y_0) + c(z - z_0) \\ &= ax + by + cz - (ax_0 + by_0 + cz_0) \\ \end{align} \,\\ \] This expression must be zero for all points \( P \) on the plane, because the vector \( \overrightarrow{P_0P} \) is orthogonal to \( n \). We define the constant term \[ \,\\ d = -(ax_0 + by_0 + cz_0) \,\\ \] and thus obtain the plane equation in its scalar form: \[ \,\\ ax + by + cz + d = 0 \,\\ \] In vector form, the plane can be written as below if we let \(p = P - P_0\) \[ \,\\ p \cdot n = 0 \,\\ \] That is, for every point \( P \) on the plane, the vector \( P - P_0 \) is orthogonal to \( n \), meaning the dot product between \( n \) and \( P - P_0 \) is \(0\). Conversely, any point that satisfies this condition must lie on the plane defined by \( n \) and \( P_0 \).