1. The problem statement, all variables and given/known data Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Show algebraically. 2. Relevant equations 3. The attempt at a solution I really have no idea where to start so I put the equations into parametric form to try to jog my brain but still got me nowhere! (x,y,z)=(5,-4,6) + u(1,4,-1) x=5+u y=-4+4u z=6-u (x,y,z)=(3,0,2) + s(1,1,-1) + t(2,-1,1) x=3+s+2t y=s-t z=2-s+t EDIT: I was looking back in some of my notes and although there was nothing on how to determine if a line lies within a plane, I found how to determine if a point does. So I discovered that the point from the first vector equation (5,-4,6) does indeed lie within the plane (x,y,z)= (3, 0, 2) + s(1,1,-1) + t(2, -1, 1). After discovering this however, i still don't see how i can use this information to see if the line (x,y,z)=(5,-4,6) + u(1,4,-1) lies within the plane Not sure but maybe i have to sub in a value for s or t and solve the equations? i really have no leads so any help is very much appreciated thank you! EDIT 2: I think i may have figured it out, here is what i've done: substitute a value for u, ex. let u = 0 solve (x,y,z)=(5,-4,6)+0(1,4,-1) therefore (5,-4,6) is a point when u=0 sub the values of x,y, and z into the parametric equations of the plane: x=3+s+2t, y=s-t, z=2-s+t 6=3+s+2t, 0=s-t, 5=2-s+t ................s=t sub s=t in 6=3+s+2t 6=3+t+2t t=1 Sub t=1 in s=t s=t s=1 Now sub s=1 and t=1 into 5=2-s+t LS=5 RS=2-s+t RS=2-1+1 RS=2 Therefore the line (x,y,z)=(5,-4,6) + u(1,4,-1) does not lie in the plane???????