Question

Find equation of a plane which is equidistant to parallel planes 2x-2y-z+7=0 and 8x-8y-4z+1=0.

Answer 1

express 2º equation like this:
2x-2y-1z=-1/4
and 2º
2x-2y-z=-7
so, notice that bouth planes are at the "same side" of the origin (bouth have negative distance) so just rest 1/4 from 7, to get ?

Answer 2

got it @Valdas ?

Answer 3

What do you mean by "rest 1/4 from 7, to get ?"?

Answer 4

\[7-\frac{1}{4}=\frac{27}{4}\]

Answer 5

ok?

Answer 6

And what does that number represent? The distance between those planes?

Answer 7

it is the distance that you need in the plane formula.

Answer 8

2x-2y-1z=-27/4

Answer 9

since the planes are parallel, i would find a point in one plane, and anchor the normal to it to form a line; i would then determine where that line punctures the other plane at. this gives two points to deduce a midpoint with; that midpoint would be used to define the plane between them

Answer 10

to complicated

Answer 11

im a complicated person :)

Answer 12

:)

Answer 13

little mistake:
27/4 is the distance between planes, what you need is take half of it and add it to 1/4

Answer 14

so 27/8+1/4=29/8

Answer 15

and plane formula:
2x-2y-1z=-29/8

Answer 16

Distance between original planes is 9/4 (right?), so the distance between any those planes and the new one should be 9/8. But The distance between 2x-2y-z+7=0 and 2x-2y-1z=-29/8 is 3/8 if I did all my calculations correctly

Answer 17

distance between original planes is 27/4

Answer 18

This is how I calculated the distance. I took point (0,0,7) from plane 2x-2y-z+7=0 and found distance between this point and 8x-8y-4z+1=0 using distance formula.
\[\frac{ |8*0-8*0-4*7+1| }{ \sqrt{64+64+16} }=\frac{ 27 }{ 12 }=\frac{ 9 }{ 4 }\]

Answer 19

as myko points out; we can determine the needed point by taking the average of the points where they cross the z axis

Answer 20

x,y = 0

2x-2y-z+7=0
0 0
--------------
-z+7=0 ; z=7


8x-8y-4z+1=0
0 0
---------------
-4z+1=0 ; z=1/4

(7+1/4)/2 = 29/8

Answer 21

use the point (0,0,29/8) and the normal <2,-2,-1> as your midplane

Answer 22

Is the way I calculated the distance is somehow wrong?

Answer 23

the more complicated we try to do things, the more room for error creeps in :) if you found the perp distance, that would be fine as well

Answer 24

the mathing looks fine; so, 9/4 has to be applied to the line setup

Answer 25

or rather 9/4 divided by 2 is 9/8 for the half distance

Answer 26

So, as I understand, the distance between 2x-2y-z+7=0 and 2x-2y-1z+29/8=0 should be equal to 9/8? But I get, that it's 3/8
\[\frac{ |2*0-2*0-7+29/8| }{ \sqrt{4+4+1} }=\frac{ 27/8 }{ 9 }=\frac{ 3 }{ 8 }\]

Answer 27

2x-2y-z+7=0

the line is:

x = 0 + 2t
y = 0 - 2t
z = 7 - t

using the other plane and these defs

8x-8y-4z +1=0
2t -2t (7-t)
---------------
-28+4t+1=0
-27+4t =0 ; t = 27/4 is the distance between the planes in order
for the line to pierce the other plane; half the
distance defines it midpoint.

t=27/8

x = 0 + 2t; x=27/4
y = 0 - 2t; y=-27/4
z = 7 - t; z= 29/8

Answer 28

I did a mistake in my calculation. The distance is correct. Thanks for help

Answer 29

;) good luck