The document provides an example of finding the equation of a plane using the point-normal form of a plane equation. It gives the steps to: 1) find two vectors parallel to the plane using three points on the plane, 2) take the cross product of the vectors to find the normal vector, and 3) use the normal vector and a point to find the equation in the form a(x-x0)+b(y-y0)+c(z-z0)=0. Applying this process to the three points P1(-2,1,1), P2(0,2,3), and P3(1,0,-1) results in the plane equation 2y - z - 1

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Finding the Equation of a Plane
The equation of a plane in 3-D space can be found by using the Point-Normal Form
equation for a plane.
a(x-x0)+b(y-y0)+c(z-z0)=0
In order to use this formula, a point in the plane and a vector perpendicular to the plane
are needed. A vector perpendicular to the plane is called a normal to the plane.
Example:
Find an equation for the plane that passes through the given points P1(-2,1,1), P2(0,2,3),
and P3(1,0,-1).
Step 1:
Since the points P1, P2, and P3 lie on the plane, the vectors P1P2 and P1P3 are parallel to
the plane. To find the vectors P1P2 and P1P3 the do the following
P1(-2,1,1) = P1(x1,y1,z1)
P2(0,2,3) = P2(x2,y2,z2)
P3(1,0,-1) = P3(x3,y3,z3)
P1P2 = u = <x2-x1, y2-y1, z2-z1> = <0-(-2), 2-1, 3-1> = <2, 1, 2>
P1P3 = v = <x3-x1, y3-y1, z3-z1> = <1-(-2), 0-1, -1-1> = <3, -1, -2>
Therefore,
u = <2, 1, 2> and v = <3, -1, -2> are the parallel vectors to the plane

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Step 2:
Find the normal vector to the plane using the Cross Product of vectors v and u
u x v =
213
212
−−
kji
= i
21
21
−−
- j
23
22
−
+ k
13
12
−
= i(-2-(-2)) - j(-4-6) + k(-2-3)
= i(-2+2) - j(-10) + k(-5)
= 0i + 10j - 5k
Hence the normal vector is 0i + 10j – 5k or <0, 10, -5>
Step 3:
Use the Point-Normal Form, the normal vector, and the point on the plane to find the
equation
a(x-x0)+b(y-y0)+c(z-z0)=0
Normal vector = <a, b, c> = <0, 10, -5>
Point on the plane = P(x0,y0,z0) = P2(0,2,3)
Simplify
0(x-0) + 10(y-2) - 5(z-3) = 0
10y - 20 - 5z + 15 = 0
10y - 5z - 5 = 0
2y - z - 1 = 0
Therefore, the points P1(-2,1,1), P2(0,2,3) and P3(1,0,-1) are all contained by the plane
2y – z – 1 = 0.
Note: you may have also used either of the other two points in the Point-Normal Form
and still acquired the same equation. You can double check your work by trying the other
points.