Equation of a Plane
Lesson · Intermediate
Geometry: Coordinate Geometry
Equation of a Plane
A plane in three-dimensional space can be written as
where
is a normal vector to the plane.
Point-Normal Form
If a plane passes through
and has normal vector
then its equation is
Equivalently,
where
Finding a Plane Through Three Points
Suppose the plane passes through three noncollinear points
Two vectors lying in the plane are
and
Therefore, a normal vector is
Then use one of the points in the point-normal form.
For any point P=(x,y,z) on the plane,
lies in the plane and is therefore perpendicular to n. Hence
which gives the equation.
Two planes
and
are parallel when their normal vectors are parallel:
They are perpendicular when their normal vectors are perpendicular:
The angle between two planes is the acute angle between their normal vectors.
If their normal vectors are n₁ and n₂, then
The absolute value gives the acute angle between the planes.
If
then the x-intercept is found by setting
and similarly for the other axes.
If all three intercepts are nonzero, the plane can also be written in intercept form
where p, q, r are its x-, y-, and z-intercepts.
Find the equation of the plane through
We have
Their cross product is
so we may use
Using A=(1,2,0),
Thus the plane is
Find the equation of the plane with intercepts
Using intercept form,
Multiplying by 12,
Find the plane through P=(2,-1,4) parallel to
Parallel planes have parallel normal vectors, so the required plane has the form
Substituting P,
Therefore,
Find a plane through P=(1,2,3) that is perpendicular to both
and
Their normal vectors are
The normal vector of the required plane must be perpendicular to both, so take
which gives
so we can use
Using P=(1,2,3),
Therefore,
Find k so that the planes
and
are perpendicular.
Their normal vectors are
and
For perpendicular planes,
Thus
so