Part 2: Equation of a Plane

Given any plane, there must be at least one nonzero vector n = á a,b,c ñ that is perpendicular to every vector v parallel to the plane.

In particular, the plane through the point P1(x1,y1,z1) with normal vector n = áa,b,c ñ is the set of points P(x,y,z) such that the vectors

v =  

P1P

  = áx-x1,y-y1,z-z1 ñ

are perpendicular to n. However, n·v = 0 is equivalent to
á a,b,c ñ · á x-x1, y-y1, z-z1 ñ = 0
Computing the inner product then leads us to the following definition:      

Definition 4.3: The equation of the plane with normal n = á a,b,c ñ through the point áx1,y1,z1 ñ is
a( x-x1) +b( y-y1) +c( z-z1) = 0
(2)

       

If c ¹ 0, then we can transform (2) into functional form, which is
z = mx+ny+d
where m = -a/c, n = -b/c and d = ( ax1+by1+cz1) /c.       

EXAMPLE 3    Find the equation of the plane with normal n = á 1,2,7 ñ which contains the point P1(5,3,4).       

Solution: To do so, we substitute into the equation (2). The result is that
1( x-5) +2( y-3) +7( z-4) = 0
We then solve for z to obtain the functional form:
z = -
1
7
x -   
2
7
y  +  
39
7

      

To find the equation of the plane through three non-collinear points P1, P2, and P3, we first form the two vectors
u =   P1P2        and        v =   P1P3
Since u and v are both parallel to the plane, the cross product u×v must be perpendicular to the plane.

We use Definition 4.3 with n = u×v to write the equation of the plane.       

EXAMPLE 4    Find the equation of the plane passing through P1( 1,2,2) , P2( 4,6,1) and P3(0,5,4) .

Solution: We form the vectors

u = 

P1P2

  = á4-1,6-2,1-2 ñ = á 3,4,-1 ñ
v = 

P1P3

  = á0-1,5-2,4-2 ñ = á -1,3,2 ñ
The cross product of u and v is
n = u×v = á 11,-5,13 ñ
Thus, the equation of the plane through P1, P2, and P3 is
11( x-1) -5( y-2) +13( z-2) = 0
which in functional form is
z = -
11
13
x +   
5
13
y  +  
27
13
(3)
               

Check your Reading: Explain why P2(1,2,2) satisfies (3).