Find the distance between two planes: C1:x+y+2z=4 and C2:3x+3y+6z=18 And find the other plane C3≠C1...

Michael Maggard

Michael Maggard



Find the distance between two planes:
C1:x+y+2z=4 and C2:3x+3y+6z=18
And find the other plane C3C1 that has the distance d po the plain C2

Answer & Explanation

Karen Robbins

Karen Robbins


2021-12-28Added 49 answers

Well, we can rewrite C2:3x+3y+6z=18
Two plain have normal n=[3,3,6]
Chose P,Q from C1,C2 respectively.
Say, P(5,0,0) and Q(12,0,0)
Then, QP=[112,0,0]
Thus, the distance is ||Proj{n}QP||=(1112)2+(1112)2+(116)2=12124
Bernard Lacey

Bernard Lacey


2021-12-29Added 30 answers

For a plane defined by ax+by+cz=d the normal is said to be (a,b,c). This is a direction, so we can normalise it (1,1,2)1+1+4=(3,3,6)9+9+36, which means these two planes are parallel and we can write the normal as 16(1,1,2).
Now we should find two points on the planes.
Let y=0 and z=0, and find the corresponding x values. For C1, x=4 and for C2, x=6. So we know C1 contains the point (4,0,0) and C2 contains the point (6,0,0).
The distance between these points is 2 and the direction is (1,0,0). We now that this is not the shortest distance between these two points as (1,0,0)16(1,1,2) so the direction is not perpendicular to these planes. But this is ok as we can use the dot product between (1,0,0) and 16(1,1,2) to work out the proportion of the distance that is perpendicular to the planes.
So the distance between the two planes is 26.
The last part is to find the plane which is the same distance away from C2 as C1 but in the opposite direction. We know the normal must be the same, 16(1,1,2). Using this we can write C3:x+y+2z=a and determine a. When y=0,z=0 we moved from (4,0,0)(6,0,0), so if we move the same distance again we go (6,0,0)(8,0,0) and (8,0,0) is on C3.
Therefore, a=8. So final equation of the plane is: C3:x+y+2z=8



2022-01-08Added 573 answers

Thanks a lot!

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