Volume of tetrahedron when you are given four planes

AI Thread Summary
To find the volume of a tetrahedron defined by four planes, the user identified the intersection points and calculated the volume of the parallelepiped formed by vectors derived from these points, resulting in a volume of 4. Consequently, the volume of the tetrahedron was determined to be 2/3. The discussion also highlighted an alternative formula for volume, V = 1/3 * B * h, where B is the area of the base and h is the height. The area of the triangular base can be calculated using vector cross products, while the height is the distance from the fourth point to the plane formed by the other three points. The conversation emphasizes the importance of understanding both methods for volume calculation.
borovecm
Messages
13
Reaction score
0

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


\vec{a}=\vec{AB}=(X2-X1)\vec{i}+(y2-y1)\vec{j}+(z2-z1)\vec{k}
\vec{b}=\vec{AC}=(X2-X1)\vec{i}+(y2-y1)\vec{j}+(z2-z1)\vec{k}
\vec{c}=\vec{AD}=(X2-X1)\vec{i}+(y2-y1)\vec{j}+(z2-z1)\vec{k}
V(parallelepiped)=\vec{a}\ast(\vec{b}\times\vec{c})
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 \vec{a}\ast(\vec{b}\times\vec{c}) 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:
Physics news on Phys.org
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?
 
Last edited:
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 \vec{u} and \vec{v} from one of the vertices to the other two. Then the area of the base, the triangle, is B= (1/2)|\vec{u}\times\vec{v}|. 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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top