What is the key to teaching determinants with the volume of a parallelepiped?

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The discussion focuses on calculating the volume of a parallelepiped using the position vectors of its vertices. The volume is determined using the triple scalar product, specifically through the vectors formed by the edges connecting to one vertex, A. Participants emphasize the importance of understanding neighboring vertices, clarifying that these are directly connected by edges rather than diagonally across the shape. A final calculation shows the volume to be 16 cubic units, highlighting the method's effectiveness in teaching determinants intuitively. The conversation also touches on the broader implications of using volume to teach concepts in linear algebra.
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Homework Statement



A corner,A , of a parallelepiped ABCDEFGH, has position vector <3,7,4> and the points B,G,D that are neighboring vertices A of have position vectors< 2,9,7>, < 5,10,10> ,< 4,11,9 > , respectively. Find the volume of the parallelepiped in cubic units

Homework Equations



Triple scalar product in general a cross b dot c

The Attempt at a Solution


OK, so I know what to do but I don't know if I did the right things. First I tried to draw it this was kind of hard because I'm not so good at orientating my axes so I can see what I'm looking at.
I want..I think
(g-d) cross ( a -b) dot (b-d)

OK I did

(g-d) = < 5,10,10> -< 4,11,9 > = < 1 ,-1, 1>
(a - b) = <3,7,4> -< 2,9,7> = < 1,-2,-3>
(g-d) cross (a - b) = < 5, 4, -1>
Then, dot this with vector ( b - d) = < 2,9,7> - <4,11,9> = <-2,-2,-2>
< 5, 4, -1> dot <-2,-2,-2> = 39 cubic units?


I feel like if I did this wrong it is because of my drawing maybe. Thanks.
 
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Jbreezy said:
(g-d) cross ( a -b) dot (b-d)
That does seem an odd choice. Try thinking in terms of translating all the vectors to make one of the vertices the origin (A, say).
 
There is a reason why you are given the corner A first, then the other three points!

The three edges, terminating at A, are the vectors B- A, D- A, and G- A.
 
The three edges, terminating at A, are the vectors B- A, D- A, and G- A.

How do you know this ?

Since I can't draw if you pause this video about 39 seconds in. If you look at the diagram if you did G-A you would be subtracting across diagonally? I see how you are saying A was given first but how do I know that G-A isn't across the parallelepiped diagonally? thx
 
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I guess to rephrase my question is G-A is a vector diagonally across the parallelepiped you can't use this as the length because it will actually be longer then needed?
 
Jbreezy said:
I guess to rephrase my question is G-A is a vector diagonally across the parallelepiped you can't use this as the length because it will actually be longer then needed?
No. G-A is the displacement vector between A and G.

The key is that you stated that A and G are neighboring vertices.
 
Look, This is my drawing. I know G-A is the displacement vector. I do not know what is meant by neighboring vertices. Look at my drawing and forget about how crappy it is it serves its purpose :)
I'm given A,B,G,D
If I want the volume I A-B I need some the length of say A-E or B-F. A to G is a diagonal it is not the same length as A to E which I need. I have the height which is B to D. The diagonal is throwing me off.
 

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Jbreezy said:
I do not know what is meant by neighboring vertices.
Then that's your problem. It means there is an edge of the object connecting A to each. I.e. AB., AD and AG are edges of the shape.
 
Did you look at my drawing is it correct?
 
  • #10
'Neighboring vertices' means that each vertex which is a neighbor to vertex A has no other vertex between A and it. You know, your neighbor lives next door, not across town.
 
  • #11
Jbreezy said:
Did you look at my drawing is it correct?

It came out blank for me.
 
  • #12
OK, So I worked something out I think.
a = < 3,7,4>
b = <2,9,7>
g = <5,10,10>
d = < 4,11,9>

So ,
a' = b-a = <2,9,7>-< 3,7,4> = < -1, 2 ,3>
b' = g-a = <5,10,10> - < 3,7,4> = < 2,3,6>
c' = d -a = < 4,11,9> - < 3,7,4> = <1,4,5>

Then,

( b' cross c') dot a' = volume

( b' cross c') = <-9,-4,5>
Now dot with a'
<-9,-4,5> dot < -1, 2 ,3> = 9 - 8 + 15 = 16

So volume is 16 cubic units?
Thanks.
 
  • #13
Looks right.
 
  • #14
Jbreezy said:
How do you know this ?

Since I can't draw if you pause this video about 39 seconds in. If you look at the diagram if you did G-A you would be subtracting across diagonally? I see how you are saying A was given first but how do I know that G-A isn't across the parallelepiped diagonally? thx

YOU said that "points B,G,D that are neighboring vertices A". Diagionally opposite points are not "neighboring"..
 
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  • #15
This is a fine site and there are so much riches that if you don't grab something when it appears you may never find it again. I have searched and never found again this one:

- roughly early last year there was a post from someone saying the volume of a paralellipiped was the key! I am not sure now what it was the key to - whether just determinants or to linear algebra. That this volume was the way to teach determinants - to make them intuitive and easy, which frankly they are not (is it just me?) and could very well do with being made. I don't know whether this was just an once off opinion and maybe the way the poster teaches them, or whether he had some supporting material developed.

OK, as a journeyman math user I can work out the volume of the ppped with conrners (x1, y1, z1) , 2, 3 without too much difficulty. Given that, I can see how it illustrates the idea of linear dependence. But the formula, if not difficult, is to me anything but self-evident, like most of deterninant theory or linear algebra.

If that poster is here or anyone remembers the post, and if this idea was ever elaborated it would be useful to at least one person to hear again.
 
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