P (2 2,6,-1), find the equation of plane π.
Solution: the normal vector n of plane ①? ={ 1, 1,0}; Normal vector n of plane ②? ={2, 1,0};
The straight line L is the intersection of planes ① and ②. What about L⊥n? What about L⊥n? ; Let the direction vector s of l = {m, n, p };;
So s=n? ×n? ; Namely:
②-① x=2,y=- 1,z = 0; That is, l intersection (2,-1, 0);
So the equation of the straight line L is: (x-2)/0 = (y+1)/0 = z/(-1);
Plane π ∨ l; The direction vector of the straight line L is the normal vector of the plane π, and π crosses the point (2,6,-1).
So the equation of plane π is 0(x-2)+0(y-6)-(z+ 1)=0, that is, z=- 1.