1. The problem statement, all variables and given/known data Determine the distance between the parallel planes –4x–4y+1z=–1 and 8x+8y–2z=12 2. Relevant equations Proj_n_v = ((vn)/(nn))n 3. The attempt at a solution I thought I understood how to do this, but I am not getting a correct answer for it. What I did was: I made the equations to be: –4x–4y+1z=–1 and –4x–4y+1z=–12 and therefore got the normal to these two planes to be: (-4,-4,1) Then, I took a point P (0,0,-1) from plane 1. Then, took a point A on plane 2 to be (0,0,-12). From these two points, I got vector AP = (0,0,11). I projected vector AP onto the normal... so proj_n_AP = (((AP)n)/(nn))n And got the result: (11/33)[(-4,-4,1)] I then got the distance by ll(11/33)[(-4,-4,1)]ll to be (11/33)sqrt33 Am I doing this totally wrong? Thanks!