Let the normal vector of plane π and straight line L be x+y = 1...①; The direction vector of 2x+y = 3 ... ② is parallel, and the plane π passes through this point.

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.