Calculate the volume of the tetrahedron limited by the plan $\displaystyle n1: 3x+2y-4z-12=0$ and plans for coordinated.
Answer:
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
$\displaystyle V = \pm\frac16\left\lvert\begin{matrix}
0 & 0 & 0 & 1\\
4 & 0 & 0 & 1\\
0 & 6 & 0 & 1\\
0 & 0 & -3 & 1
\end{matrix}\right\rvert = \frac16\cdot72 = 12$
Hello, Apprentice123!
The volume of a pyramid with base area $\displaystyle B$ and height $\displaystyle h$ is: .$\displaystyle V \;=\;\frac{1}{3}Bh$
The vertices are: .$\displaystyle (0,0,0),\:(4,0,0),\;(0,6,0,),\:(0,0,-3)$Calculate the volume of the tetrahedron limited by the plane $\displaystyle n_1:\;\;3x+2y-4z-12\:=\:0$
and the coordinate planes.
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: .$\displaystyle V \;=\;\frac{1}{3}(12)(3) \;=\;12\text{ units}^3$