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Find the volume of the tetrahedron formed by the planes whose equations are $x+y=0$, $y+z=0$, $z+x=0$ and $x+y+z-1=0$.

I am trying to use the formula $\frac{1}{3}$ Area of base $\times$ Height. I achieved vertex as $(0,0,0)$ and $a=\sqrt2$ as side of base but I am achieving volume as $\frac{1}{6}$ which is wrong as per the given answer (which is $\frac{2}{3}$).

Could someone help me with this?

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  • $\begingroup$ Could you write the vertices of the tethrahedron? $\endgroup$
    – mfl
    Commented Oct 11, 2016 at 19:09

1 Answer 1

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Find the vertices, which are the planes intersections, i.e., $$\left(\pi_{1}\right)x+y=0\;;\left(\pi_{2}\right)y+z=0\;;\left(\pi_{3}\right)z+x=0\;\rightarrow A\left(0,0,0\right)$$ $$\left(\pi_{1}\right)x+y=0\;;\left(\pi_{2}\right)y+z=0\;;\left(\pi_{4}\right)x+y+z-1=0\;\rightarrow B\left(1,-1,1\right)$$ $$\left(\pi_{1}\right)x+y=0\;;\left(\pi_{3}\right)z+x=0\;;\left(\pi_{4}\right)x+y+z-1=0\;\rightarrow C\left(-1,1,1\right)$$ $$\left(\pi_{2}\right)y+z=0\;;\left(\pi_{3}\right)z+x=0\;;\left(\pi_{4}\right)x+y+z-1=0\;\rightarrow D\left(1,1,-1\right)$$ Thus, you have the tetrahedron edge vectors $$\overrightarrow{AB}=\left(1,-1,1\right)\;;\;\overrightarrow{AC}=\left(-1,1,1\right)\;;\;\overrightarrow{AD}=\left(1,1,-1\right)$$ The volume is given by the triple product $$V=\frac{1}{6}|\overrightarrow{AB}\cdot\left(\overrightarrow{AC}\times\overrightarrow{AD}\right)|=\frac{1}{6}\begin{vmatrix}1 & -1 & 1 \\ -1 & 1 & 1 \\ 1 & 1 & -1\end{vmatrix}=\frac{|-4|}{6}=\frac{2}{3}$$

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