How do I find the equation of the plane A (1,1,0), B (1,2,1) and C (-2,2,-1) and hence what is the distance between the plane and the line x-6/3=y-3/-1= z+2/1?

Answer 1

To find the equation of a plane from three points A, B, and C, form two vectors A-B and C-B (or any two vectors joining two pairs of the given points A,B,C). The cross product of these vectors will form a normal to the plane, and the equation of the plane will be N_x*(x-x_0)+N_y*(y-y_0)+N_z*(z-z_0)=0 where the point (x_0,y_0,z_0) is any point in the plane.

To find the distance between the plane and a line, note that the plane and line must be parallel. (Otherwise they will intersect and the distance is 0). Therefore form a vector that joins any point in the plane to any point on the line. The projection of that vector onto a unit normal to the plane will be the distance between them.