# tetrahedron

• Jun 29th 2008, 10:31 AM
Apprentice123
tetrahedron
Calculate the volume of the tetrahedron limited by the plan $n1: 3x+2y-4z-12=0$ and plans for coordinated.

12
• Jun 29th 2008, 11:36 AM
Reckoner
Quote:

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

12

I think you are asking for the volume of the tetrahedron bounded by $n_1$ 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 $x, y,\text{ and }z$ axes, namely at $(4,0,0),\;(0,6,0),\text{ and }(0,0,-3)$.

Thus, the volume is

$V = \pm\frac16\left\lvert\begin{matrix}
0 & 0 & 0 & 1\\
4 & 0 & 0 & 1\\
0 & 6 & 0 & 1\\
0 & 0 & -3 & 1
\end{matrix}\right\rvert = \frac16\cdot72 = 12$
• Jun 29th 2008, 02:43 PM
Soroban
Hello, Apprentice123!

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

Quote:

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

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

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

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

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

• Jun 29th 2008, 03:26 PM
Apprentice123
Thank you for the help of all