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  1. #1
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    tetrahedron

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



    Answer:
    12
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    Quote Originally Posted by Apprentice123 View Post
    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

    $\displaystyle V = \pm\frac16\left\lvert\begin{matrix}
    0 & 0 & 0 & 1\\
    4 & 0 & 0 & 1\\
    0 & 6 & 0 & 1\\
    0 & 0 & -3 & 1
    \end{matrix}\right\rvert = \frac16\cdot72 = 12$
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  3. #3
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    Hello, Apprentice123!

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


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

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    Thank you for the help of all
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