Determine the distance between the parallel planes –4x–4y+1z=–1 and 8x+8y–2z=12
Proj_n_v = ((vn)/(nn))n
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:
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!