< User talk:Paul WormerRevision as of 06:07, 30 March 2010 by imported>Paul Wormer
PD Image Equation for plane. X is arbitary point in plane;
and
are collinear.
In analytic geometry several closely related algebraic equations are known for a plane in three-dimensional Euclidean space. One such equation is illustrated in the figure. Point X is an arbitrary point in the plane and O (the origin) is outside the plane. The point A in the plane is chosen such that vector

is orthogonal to the plane. The collinear vector

is a unit (length 1) vector normal (perpendicular) to the plane. Evidently a is the distance of O to the plane. The following relation holds for an arbitrary point X in the plane

This equation for the plane can be rewritten in terms of coordinates with respect to a Cartesian frame with origin in O. Dropping arrows for component vectors (real triplets) that are written bold, we find

with

Conversely, given the following equation for a plane

it is easy to derive the same equation.
Write

It follows that

Hence we find the same equation,

where f , a, and n are collinear. The equation may also be written in the following mnemonically convenient form

which is the equation for a plane through a point A perpendicular to
.