Ok, the vector equation of a plane is:n∗(r−r0)=0where n is a vector normal to the plane, r is the position vector of an arbitrary point in the plane and r0 is the position vector of a given point in the plane. We need a normal vector to solve this. To get a normal vector, we use the given information that our plane is perpendicular to the y-axis. Therefore, our normal vector must be parallel to the y-axis. Any vector parallel to the y-axis will do, so I choose the unit vector <0,1,0> for simplicity. We have:<0,1,0>∗<x−2,y−3,z−4>=0giving,y=3This makes sense, since the y-coordinate must be constant, and there is no restriction on the values x,z that lie on the plane in 3-space.
