Calculate the volume of the prism ABD.A'B'D'?

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Violette

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A square-based right prism ABCD.A'B'C'D' has its lateral edges with a length of 2a. Let M and O be the midpoints of A'B' and A'C' respectively. Given that the distance between AM and CO is 4a/9, what is the volume of the prism ABD.A'B'D'?
I drew the square-based right prism with all the information given. But I am stuck at this because I have no idea how to find the shared perpendicular line between AM and CO and I also don't know which plane should i put one of the two lines into it and simplify the math problem from distance between two lines to distance from a point of this line to the plane that is parallel to the line contains the other line.
 

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I'd look at the projection of the points and lines along the direction of AM', i.e. project the whole picture on a plane orthogonal to AM'.
 
Consider plane P which is orthogonal to AM' and contains M'O. What are the
projections of lines AM' and CO onto P?
 
blamocur said:
Consider plane P which is orthogonal to AM' and contains M'O. What are the
projections of lines AM' and CO onto P?
Click to expand...
I'm slightly confused by your use (introduction) of M'. Is that a different point from M (on the OP's diagram)?
 
Violette said:
A square-based right prism ABCD.A'B'C'D' has its lateral edges with a length of 2a. Let M and O be the midpoints of A'B' and A'C' respectively. Given that the distance between AM and CO is 4a/9, what is the volume of the prism ABD.A'B'D'?
I drew the square-based right prism with all the information given. But I am stuck at this because I have no idea how to find the shared perpendicular line between AM and CO and I also don't know which plane should i put one of the two lines into it and simplify the math problem from distance between two lines to distance from a point of this line to the plane that is parallel to the line contains the other line.
Click to expand...
Hmmm, Am I missing something? ?
(Or have I just completely misunderstood what's going on here? ?‍♂️)

Surely the volume of the prism ABD.A'B'D' is just half the volume of the right cubic prism ( ABCD.A'B'C'D')?

1686809304218.png​

If we add point N to the diagram, will MN not be parallel to A'D' & B'C' thus making the base area of the right cubic prism: [imath]\small\left(\frac{8a}{9}\right)^2~[/imath]? And we are given its height as: "2a".

Or is the "distance between AM and CO" different from that between M & O?
(3D Geometry not a strong point. ?)
 
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The Highlander said:
Hmmm, Am I missing something? ?
(Or have I just completely misunderstood what's going on here? ?‍♂️)

Surely the volume of the prism ABD.A'B'D' is just half the volume of the right cubic prism ( ABCD.A'B'C'D')?


If we add point N to the diagram, will MN not be parallel to A'D' & B'C' thus making the base area of the right cubic prism: [imath]\small\left(\frac{8a}{9}\right)^2~[/imath]? And we are given its height as: "2a".

Or is the "distance between AM and CO" different from that between M & O?
(3D Geometry not a strong point. ?)
Click to expand...
yeah MO is not the distance between AM and CO you have to find that
 
I don't really understand this to apply it on my own with different math problem.This is how my teacher find the distance
 

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Violette said:
I don't really understand this to apply it on my own with different math problem.This is how my teacher find the distance
Click to expand...
If our task is to calculate the volume of the prism cut from the cube (cut along the plane BB'DD'), then all those dotted lines are "red herrings".

Volume of the original cube is [(2*a)3 =] 8*a3

The plane BB'DD' is plane of symmetry - it cuts the cube into two equal halves.

Thus the volume of the prism is [8*a3/2 =] 4*a3
 
khansaheb said:
If our task is to calculate the volume of the prism cut from the cube (cut along the plane BB'DD'), then all those dotted lines are "red herrings".

Volume of the original cube is [(2*a)3 =] 8*a3

The plane BB'DD' is plane of symmetry - it cuts the cube into two equal halves.

Thus the volume of the prism is [8*a3/2 =] 4*a3
Click to expand...
Your solution is actually wrong because it's a not a cube it's a prism with the base is a square we only know the height of the prism we doesn't know the side length of the square base, so we can't calculate the area of the base;therefore we can't use the formula Volume=Area of the base x height (noted that the face of the prism is not a square but a rectangle)
 
Violette said:
Your solution is actually wrong because it's a not a cube it's a prism with the base is a square we only know the height of the prism we doesn't know the side length of the square base, so we can't calculate the area of the base;therefore we can't use the formula Volume=Area of the base x height (noted that the face of the prism is not a square but a rectangle)
Click to expand...
Your problem statement is wrong (confusing), You state:
Violette said:
Given that the distance between AM and CO is 4a/9
Click to expand...
AM and CO are two skewed lines in 3-D space. You can only define minimum distance between these two lines.
1686913355831.pngwhich line (AC or MO) is being defined by you as the distance? Or the distance is something else?
Is this statement a translation created by you?

Please post a photo-copy of the problem as it was presented to you.
 
khansaheb said:
Your problem statement is wrong (confusing), You state:

AM and CO are two skewed lines in 3-D space. You can only define minimum distance between these two lines.
View attachment 36003which line (AC or MO) is being defined by you as the distance? Or the distance is something else?
Is this statement a translation created by you?

Please post a photo-copy of the problem as it was presented to you.
Click to expand...
yes you're right about AM and CO are two skewed lines in 3-D space...
I just connected MO together ( sorry for confusing you) MO is not the distance between two lines as I replied above the distance is indeed C'H(solved by my teacher).And yeah I translated the math problem into english.Here is the original context.
 

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Violette said:
I don't really understand this to apply it on my own with different math problem.This is how my teacher find the distance
Click to expand...
I believe your teacher uses the fact that the distance between two lines in space is equal to the distance between the first line and a plane which contains the second line but is parallel to the first line.

In your teacher's diagram CP is parallel to AM, so the plane COP is parallel to AM and contains CO. Thus the distance between CO and AM is twice the difference between C' and the plane COP. The latter is equal to the distance from C' to CK, i.e. one of the heigts of the CC'K triangle.

Does this make sense now?
 
blamocur said:
I believe your teacher uses the fact that the distance between two lines in space is equal to the distance between the first line and a plane which contains the second line but is parallel to the first line.

In your teacher's diagram CP is parallel to AM, so the plane COP is parallel to AM and contains CO. Thus the distance between CO and AM is twice the difference between C' and the plane COP. The latter is equal to the distance from C' to CK, i.e. one of the heigts of the CC'K triangle.

Does this make sense now?
Click to expand...
Yeah I guess but apply this on my own is somehow hard for me to search for those additional lines and surface
 
Violette said:
Yeah I guess but apply this on my own is somehow hard for me to search for those additional lines and surface
Click to expand...
One thing at a time:
  1. In the OC'P triangle can you figure out the height C'K (in terms of the uknown size PC' or C'D') ?
  2. Once you get an expression for C'K you can can take a look at the CC'K triangle and get an expression for the height coming from the C' vertex. This expression is in terms of the horizontal side (e.g. C'D'), but it must be equal to one half of 4a/9 (because 4a/9 = d(N,COP) = 2d(C', COP)), which you can solve to express C'D' in terms of 'a'.
Does this help?
 
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