AcademyVectors In Rn
Academy
Planes In R3
Level 1 - Linear Algebra topic page in Vectors In Rn.
Principle
A plane in \(\mathbb R^3\) is a flat two-dimensional set of points. It can be described by a point on the plane and a normal vector perpendicular to the plane, or by one point and two non-parallel direction vectors lying in the plane.
Notation
\(\mathbf r\)
a variable position vector for a point on the plane
\(\mathbf r_0\)
a fixed position vector for one known point on the plane
\(\mathbf n\)
a normal vector perpendicular to the plane
\(\mathbf a,\mathbf b\)
direction vectors lying in the plane
\(s,t\)
real parameters
The Core Method
Point-normal form says that the displacement from the fixed point to any point on the plane is perpendicular to the normal vector:
Point-normal plane
\[(\mathbf r-\mathbf r_0)\cdot\mathbf n=0\]
If \(\mathbf r=(x,y,z)\), \(\mathbf r_0=(x_0,y_0,z_0)\), and \(\mathbf n=(A,B,C)\), this becomes
Cartesian plane
\[A(x-x_0)+B(y-y_0)+C(z-z_0)=0\]
Parametric form uses two non-parallel direction vectors in the plane:
Parametric plane
\[\mathbf r=\mathbf r_0+s\mathbf a+t\mathbf b\]
The normal vector can be found from the direction vectors by \(\mathbf n=\mathbf a\times\mathbf b\), provided \(\mathbf a\) and \(\mathbf b\) are not parallel.
Worked Cases
Question
Find the equation of the plane through \((1,-2,3)\) with normal vector \((2,1,-4)\).
Answer
Use \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\). Here \((x_0,y_0,z_0)=(1,-2,3)\) and \((A,B,C)=(2,1,-4)\). Substitute: \(2(x-1)+1(y+2)-4(z-3)=0\). Expand: \(2x-2+y+2-4z+12=0\). Combine constants: \(2x+y-4z+12=0\).
Examples
Question
Does \((2,0,4)\) lie on the plane \(2x+y-4z+12=0\)?
Answer
Substitute \(x=2\), \(y=0\), and \(z=4\): \(2(2)+0-4(4)+12=4-16+12=0\). Since the equation is satisfied, the point lies on the plane.