3dvectors
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?
$2\phi$ is the angle of the cone (spherical sector)
From Wikipedia: $V=\frac{2\pi r^2 h}{3}$ ...1
From the cone:
$\frac{r-h}{r}=cos\phi$
After simplifying
$h=r(1-cos\phi)$
Replace in 1 to get
$V=\frac{2\pi r^3}{3}(1-cos\phi)$
Here is one way to think of it. A tetrahedron is $\dfrac{1}{6}$ of the volume of the parallelipiped formed by $\vec{a},\vec{b},\vec{c}$. The volume of the parallelepiped is the scalar triple product $|(a \times b) \cdot c|$. Thus, the volume of a tetrahedron is $\dfrac{1}{6} |(a \times b) \cdot c|$
In order to solve the question like you are trying to, notice that by $V = \dfrac{1}{3}Bh = \dfrac{1}{6}||a \times b|| \cdot h$. Then, $h = ||c|| \cdot |\cos(\theta)|$. Thus, we have $V = \dfrac{1}{6}||a \times b|| \cdot ||c|| \cdot |\cos(\theta)|$. Now see that $|c \cdot (a \times b)| = ||c|| \cdot ||(a \times b)|| \cdot |\cos(\theta)|$ and thus $V = \dfrac{1}{6}|(a \times b) \cdot c|$.
I'd also like to say that the notation you are using is a little weird. In order to avoid confusion, $|x|$ denotes absolute value and $||x||$ denotes magnitude.
Best Answer
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 $$V=\frac{1}{6}|\overrightarrow{AB}\cdot\left(\overrightarrow{AC}\times\overrightarrow{AD}\right)|=\frac{1}{6}\begin{vmatrix}1 & -1 & 1 \\ -1 & 1 & 1 \\ 1 & 1 & -1\end{vmatrix}=\frac{|-4|}{6}=\frac{2}{3}$$