Planes with common unit normal have distance . Every joining displacement has this normal component, giving a lower bound by its absolute value; a displacement parallel to the normal attains the bound.
A line intersects a plane exactly when the scalar equation can be solved. If , there is one point. If that product vanishes, the line is either contained in the plane or disjoint from it. Direction and normal vectors must be nonzero.
For three non-collinear points with position vectors , a nonzero normal vector isThe plane is the set of satisfying . Expanding the cross product and using the scalar triple product givesNon-collinearity is essential: otherwise and the equation fails to select a unique plane.
The Euclidean plane is with its standard Euclidean norm and the resulting distance, angles, and Riemannian metric.
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