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## Homework Statement

Determine the distance between the parallel planes –4x–4y+1z=–1 and 8x+8y–2z=12

## Homework Equations

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:

–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!