Linear Algebra Help, Area and Volume using Vectors

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Homework Help Overview

The discussion revolves around finding the volume of a parallelepiped defined by four-dimensional vectors. The original poster expresses confusion regarding the methods applicable for calculating volume in higher dimensions, particularly R^n, as most resources focus on three-dimensional cases.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the use of Gramian matrices for higher dimensions and question whether traditional three-dimensional techniques can be adapted. There is a discussion about ignoring constant dimensions and how that might relate to calculating volume.

Discussion Status

Participants are actively sharing their thoughts on the problem, with some suggesting methods involving dot products and cross products. There is acknowledgment of the complexity of the problem, and while some guidance has been provided, there is no clear consensus on the correct approach or solution.

Contextual Notes

The original poster notes a lack of resources for understanding volume in R^n, which contributes to the complexity of the discussion. Additionally, there is mention of discrepancies with online tools like Webwork, which adds to the uncertainty regarding the correctness of the proposed solutions.

Axoren
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Homework Statement


Find the volume of the parallelepiped defined by the vectors.

x = {2, 0, 1, 1}
y = {-1, -1, 1, 1}
z = {-2, 3, 1, -2}

Homework Equations


Don't know.

The Attempt at a Solution


Don't know where to start.The main problem is that most people on the internet only discuss three dimensional volumes, leaving me searching for information regarding this and finding nothing but cross-product and the like. I need to understand the method of finding volume for R^n, not just R^3
 
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Well, if you're interested in the volumes of parallelotopes in dimensions higher than 3, you can look into Gramian matrices and the like.

However, I don't think you need that. Note that the third element of your four-dimensional vectors is constant, which seems to indicate that maybe the usual three-dimensional techniques will work.
 
hgfalling said:
Well, if you're interested in the volumes of parallelotopes in dimensions higher than 3, you can look into Gramian matrices and the like.

However, I don't think you need that. Note that the third element of your four-dimensional vectors is constant, which seems to indicate that maybe the usual three-dimensional techniques will work.

I tried that just now and it doesn't work. Is there really no way to answer this question?
 
What did you try? I got an answer by dotting one vector with the cross product of the other two, ignoring the dimension where all three vectors are constant.

I mean, if I gave you these two vectors:

a = {1,1,2}
b = {3,0,2}

and said, "give me the area of the parallelogram formed by these two vectors," you would just pretend like the z-dimension didn't exist, because parallelograms are in two dimensions, and everything is happening in a plane here. The same thing is true of your vectors: parallelepipeds are in three dimensions, and everything's happening in three dimensions here.
 
hgfalling said:
What did you try? I got an answer by dotting one vector with the cross product of the other two, ignoring the dimension where all three vectors are constant.

I mean, if I gave you these two vectors:

a = {1,1,2}
b = {3,0,2}

and said, "give me the area of the parallelogram formed by these two vectors," you would just pretend like the z-dimension didn't exist, because parallelograms are in two dimensions, and everything is happening in a plane here. The same thing is true of your vectors: parallelepipeds are in three dimensions, and everything's happening in three dimensions here.

I find the answer to be 7, however, Webwork says it's incorrect. I've tried plugging in -7, still nothing.
 
Yeah, my visualization was off here. The dimension where they are all the same does matter. Sorry about that.

Anyway, you can do this, which works for any dimension of vectors:

Let:

A = \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}

Now if the dimension of the vectors equals 3 (ie, A is square), then the absolute value of the determinant of that is the volume.

If it doesn't, then:

V = \sqrt{\left| det(A^T A) \right|}

I get \sqrt{308}.
 
Last edited:
Thank you very much. That works wonderfully.
 

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