How Do You Find the Minimal Volume of a Tetrahedron Passing Through a Point?

  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Tetrahedron Volume
MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
i have the point P(x0,y0,z0) i need to find the minimal volume of a tetrahedron which is constructed by a plane which crosses over point P, and by the axis planes.

i got that the side of the tetrahedron is sqrt[(x-x0)^2+(y-y0)^2+(z-z0)^2], but I am not sure it's correct because then the answer is that the volume of the tetrahedron is 0.

your help is appreciated.
 
Physics news on Phys.org
The plane that crosses the axes at (a, 0, 0), (0, b, 0), and (0, 0, c) has equation x/a+ y/b+ z/c= 1 (do you see why that's obvious?). What is the volume of that tetrahedron? Of course, to pass through the point (x0,y0,z0), it must also satisfy
x0/a+ y0/b+ z0/c= 1.

So, minimize that formula for volume of a tetrahedron (in terms of a, b, c) subject to that constraint.
 
HallsofIvy said:
The plane that crosses the axes at (a, 0, 0), (0, b, 0), and (0, 0, c) has equation x/a+ y/b+ z/c= 1 (do you see why that's obvious?). What is the volume of that tetrahedron? Of course, to pass through the point (x0,y0,z0), it must also satisfy
x0/a+ y0/b+ z0/c= 1.

So, minimize that formula for volume of a tetrahedron (in terms of a, b, c) subject to that constraint.
can you tell me how did you arrive at the equation?
cause from what i can remember, you start by constructing vectors from the three points:
(a,0,-c),(a,-b,0),(0,-b,c) and then you substract them and you get the next parameter equation:
(0,-b,c)+s(-a,0,c)+t(-a,-b,2c)
and then multiply by coeffiecient vector (A,B,C), and then plug in (0,-b,c)
which i get the next equation:
x/a+y/b+z/c=0 without the 1, where did i get it wrong?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top