1. The problem statement, all variables and given/known data Find the equation of the plane which contains the line r(t) = (2-t, 1+t, t) and the point (1,0,1). 3. The attempt at a solution Since r(t) = (2-t, 1+t, t), we know that vector a, which is parallel to the line, is (-1, 1, 1). I'm assuming that since the point is (1,0,1), that means if we look to that point from the origin (0,0,0) we will have vector b = (1,0,1). Okay so with two vectors a and b, I can find a vector n such that n is orthogonal to vectors a and b. So a x b = (1-0,1+1,0-1) = (1,2-1) = vector n. The scalar equation of a plane is a(x-x0) + b(y-y0 + c(z-z0). So x + 2y -z = 0... But somehow I have a feeling that my result is wrong. Is there supposed to be a "y" at all in the equation for this plane? Please advise. Thanks in advance!