We establish that a plane is determined by a point and a normal vector, and use this information to derive a general equation for planes in .
RRN-0030: Planes in
You are probably familiar with the expression “two points determine a line.” A line is also determined by one point and a vector parallel to the line. You are probably also familiar with the fact that three non-collinear points determine a plane (This is why photographers use tripods for stability, while four-legged chairs often wobble!) Is there another way to determine a plane?
It is evident geometrically that among all planes that are perpendicular to a given straight line there is exactly one containing any given point. The diagram below shows several planes perpendicular to vector . There are infinitely many such planes, but only one contains point .
Normal Vector A nonzero vector is called a normal for a plane if it is orthogonal to
every vector in the plane.
The standard unit vector is orthogonal to every vector in the -plane, therefore it is a
normal for the -plane.
Given a point and a nonzero vector , there is a unique plane through with normal .
This fact can be used to give a very simple description of a plane. Observe that a point lies on this plane if and only if the vector is orthogonal to —that is, if and only if .
Let . By “head-tail” formula (Formula form:headminustailr3 of VEC-0010), we have: . So, lies in the plane if and only if We summarize this result as a theorem.
As demonstrated in Example ex:planewithnormalvector, we can distribute coefficients , and of equation (eq:plane) as follows: Setting , shows that every plane with a normal vector has a linear equation of the form
for some constant . Conversely, the graph of this equation is a plane with as a normal vector (assuming that , , and are not all zero). Find an equation of the plane through that is parallel to the plane with equation
.
The plane with equation has normal . Because the two planes are parallel,
serves as a normal for the plane we seek, so the equation is for some by (eq:eqofline).
Insisting that lies on the plane determines ; that is, . Hence, the equation is
.
Practice Problems
Text Source
The text in this module is an adaptation of Section 4.2 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)
W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p 233-234.