Part 3: More with Equations of Planes

Using the points in a different order may result in a different normal, and (2) may also appear to be different. However, the functional form will be the same regardless of how the points are labeled.      

EXAMPLE 5    Redo example 3 with P₂( 1,2,2) , P₃( 4,6,1) and P₁( 0,5,4) .

Solution: To do so, we form the vectors

u  =   P1P2   = á1-0,2-5,2-4 ñ = á 1,-3,-2 ñ v  =   P1P3   = á4-0,6-5,1-4 ñ = á 4,1,-3 ñ

Once again, 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-0) -5( y-5) +13( z-4) = 0

Reducing to functional form then yields the equation

z = - 11 13 x +    5 13 y  +   27 13

           

A plane can also be determined by a line and a point not on that line, or by two intersecting lines. In particular, if given the vector equation of a line L( t) , points in the plane can be obtained by choosing different values of the parameter t.

       

EXAMPLE 6    Find the equation of the plane containing the point P₁( 1,2,2) and the line L( t) =  ( 4t+8,t+7,-3t-2)

Solution: If we let t = 0, then L( 0) = (8,7,-2) . If we let t = 1, then L( 1) = (12,8,-5) . Thus, we need to find the equation of the plane through P₁( 1,2,2) , P₂( 8,7,-2) , and P₃(12,8,-5) . We first form two vectors

u =  P1P2   = á 7,5,-4 ñ,        v =   P1P3   = á11,6,-7 ñ

Thus, a normal to the plane is

n = u×v = á -11,5,-13 ñ

so that the equation of the plane is

-11( x-1) +5( y-2) -13( z-2) = 0

Solving for z then yields

z = - 11 13 x +    5 13 y  +   27 13