Volume of tetrahedron when you are given four planes

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Homework Help Overview

The discussion revolves around finding the volume of a tetrahedron defined by the intersection of four planes. The planes are given by specific equations, and the original poster has identified the vertices where these planes intersect.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the volume using vectors derived from the vertices of the tetrahedron and the formula for the volume of a parallelepiped. They question whether their approach is correct and the shortest way to find the solution. Other participants raise questions about the notation used for volume and base area, and suggest methods for calculating the area of the base and the height of the tetrahedron.

Discussion Status

Participants are exploring different methods to calculate the volume of the tetrahedron, including vector approaches and formulas involving the area of the base and height. There is no explicit consensus on the best method, and the discussion includes various interpretations of the formulas and calculations involved.

Contextual Notes

There is some confusion regarding the notation used for volume and base area, as well as the difficulty of calculating determinants of matrices. The original poster has provided specific points of intersection, but the area of the base and the height remain under discussion.

borovecm
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Homework Statement


I have to find volume of tetrahedron that is bounded between 4 planes.
Planes are
x+y+z-1=0
x-y-1=0
x-z-1=0
z-2=0

Homework Equations


[tex]\vec{a}[/tex]=[tex]\vec{AB}[/tex]=(X2-X1)[tex]\vec{i}[/tex]+(y2-y1)[tex]\vec{j}[/tex]+(z2-z1)[tex]\vec{k}[/tex]
[tex]\vec{b}[/tex]=[tex]\vec{AC}[/tex]=(X2-X1)[tex]\vec{i}[/tex]+(y2-y1)[tex]\vec{j}[/tex]+(z2-z1)[tex]\vec{k}[/tex]
[tex]\vec{c}[/tex]=[tex]\vec{AD}[/tex]=(X2-X1)[tex]\vec{i}[/tex]+(y2-y1)[tex]\vec{j}[/tex]+(z2-z1)[tex]\vec{k}[/tex]
V(parallelepiped)=[tex]\vec{a}[/tex][tex]\ast[/tex]([tex]\vec{b}[/tex][tex]\times[/tex][tex]\vec{c}[/tex])
V(tetrahedron)=1/6*V(parallelepiped)

The Attempt at a Solution



I found four points where planes meet. These are:
A(1,0,0)
B(0,-1,2)
C(3,2,2)
D(3,-4,2)

From that I made vectors AB, AC, AD and then I put that into [tex]\vec{a}[/tex][tex]\ast[/tex]([tex]\vec{b}[/tex][tex]\times[/tex][tex]\vec{c}[/tex]) and got that volume of parallelepiped is 4. From there I got that volume of this tetrahedron is 2/3. Is this the correct and shortest way to get a solution? My teacher said that I can use formula V=B*v/3 but I don't know where to use it.
 
Last edited:
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I can't tell you where you would use B= B*v/2 since you haven't said what B or v mean in that formula!
 
Sorry. That's Croatian notation. I think american would be Volume=1/3*B*h where B is area of the base and h is height of tetrahedron. I can calculate h from formula for distance between point where first three planes intersect and the fourth plane. I don't know how to calculate area of the base. Is it correct that volume of this tetrahedron is 2/3?
 
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You can choose any 3 of the 4 vertices to be a triangular base. A quick way of finding the area is to construct vectors [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] from one of the vertices to the other two. Then the area of the base, the triangle, is [itex]B= (1/2)|\vec{u}\times\vec{v}|[/itex]. The height of the distance from the fourth point to the plane defined by the first three points.
 
Last edited by a moderator:
NumberedEquation1.gif



You can find out the volume by this formula but it is difficult to calculate the determinant of a 4*4 matrix
 
@borovcm Don't put tex tags around every expression. Type whatever equations would nicely fit on one line and put the tags around that.
 

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