borovecm
- 13
- 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: