# Distance between two parallel lines

## 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!

## Answers and Replies

Dick
Science Advisor
Homework Helper
No, you are doing it right. You just made a booboo. Dividing 8x+8y–2z=12 by -2 gives –4x–4y+1z=–6, doesn't it?

Aa! Can't believe I made such a stupid mistake. Thanks!!

So, the answer would just be (5/33)(sqrt33), correct?...

As I would keep P as (0,0,-1), but make A (0,0,-6); and get the vector AP = (0,0,5)

So, projecting AP onto the normal, and then getting the distance of the projection would result in: (5/33)(sqrt33)

Dick
Science Advisor
Homework Helper
That seems right.

HallsofIvy
Science Advisor
Homework Helper
But why bother with projecting? You know that a point on plane 1 is (0, 0, -1) and that the normal vector is given by <-4, -4 , 1> so a line through that point, normal to the plane is x= -4t, y= -4t, z= -1+ t. Where does that line intersect plane 2? The distance between those two points is the distance between the planes.