# tetrahedron

• Jun 29th 2008, 09:31 AM
Apprentice123
tetrahedron
Calculate the volume of the tetrahedron limited by the plan $\displaystyle n1: 3x+2y-4z-12=0$ and plans for coordinated.

12
• Jun 29th 2008, 10:36 AM
Reckoner
Quote:

Originally Posted by Apprentice123
Calculate the volume of the tetrahedron limited by the plan $\displaystyle n1: 3x+2y-4z-12=0$ and plans for coordinated.

12

I think you are asking for the volume of the tetrahedron bounded by $\displaystyle n_1$ and the coordinate planes?

If so, one vertex will be at the origin, and the other three will be located at the points where the plane cuts the $\displaystyle x, y,\text{ and }z$ axes, namely at $\displaystyle (4,0,0),\;(0,6,0),\text{ and }(0,0,-3)$.

Thus, the volume is

$\displaystyle V = \pm\frac16\left\lvert\begin{matrix} 0 & 0 & 0 & 1\\ 4 & 0 & 0 & 1\\ 0 & 6 & 0 & 1\\ 0 & 0 & -3 & 1 \end{matrix}\right\rvert = \frac16\cdot72 = 12$
• Jun 29th 2008, 01:43 PM
Soroban
Hello, Apprentice123!

The volume of a pyramid with base area $\displaystyle B$ and height $\displaystyle h$ is: .$\displaystyle V \;=\;\frac{1}{3}Bh$

Quote:

Calculate the volume of the tetrahedron limited by the plane $\displaystyle n_1:\;\;3x+2y-4z-12\:=\:0$
and the coordinate planes.

The vertices are: .$\displaystyle (0,0,0),\:(4,0,0),\;(0,6,0,),\:(0,0,-3)$

The base is a right triangle with legs 4 and 6.
. . Its area is: .$\displaystyle B \:=\:\frac{1}{2}(4)(6) \:=\:12\text{ units}^2$

The height is: .$\displaystyle h \,= \,3\text{ units}$

Therefore: .$\displaystyle V \;=\;\frac{1}{3}(12)(3) \;=\;12\text{ units}^3$

• Jun 29th 2008, 02:26 PM
Apprentice123
Thank you for the help of all