3dalgebra-precalculusanalytic geometrygeometry
I found a following question in 1st year undergrad course:
Find the equation of sphere circumscribing the tetrahedron bounded by the planes
$$x+y=0 \qquad y+z=0 \qquad z+x=0 \qquad x+y+z=1$$
Can anyone tell me in detail how to find the equation of sphere ?
I know if i get four vertices of tetrahedron then i can able find out the equation. But how to find the vertices of tetrahedron?
I happened to have a suitable image on my laptop (for my freshman course), so I couldn't resist
It is supposed to be a methane molecule as opposed to a tetrahedron, so think of the hydrogen atoms (green) as vertices of the tetrahedron. As you see, this is the second arrangement in Old John's (+1) answer. One vertex is at the point $(1,1,1)$ and the other three at the points $(\pm1,\pm1,\pm1)$ with exactly two minus signs occuring. In other words, four selected corners of a cube.
I second Old John's opinion that this is very well suited for 3D-rotations, because you surely know the effect of a rotation on the cube...
Given any tetrahedron $T$ with vertices $p_1, p_2, p_3, p_4$. Let
We know that for each $i$, $\displaystyle\;\alpha_i = \frac{r}{h_i}$, together with the fact: $$ h_1A_1 = h_2A_2 = h_3 A_3 = h_4 A_4 = 3V $$ We have $\displaystyle\;\alpha_i = \frac{r}{h_i} = \frac{rA_i}{3V}$ and $\displaystyle\;\sum_{i=1}^4 \alpha_i = 1$ reduces to $\displaystyle\;\frac{r}{3V}\sum_{i=1}^4 A_i = 1$. As a result, $$u = \sum_{i=1}^4 \alpha_i p_i = \frac{r}{3V}\sum_{i=1}^4 A_i p_i = \frac{\sum\limits_{i=1}^4 A_i p_i}{\sum\limits_{i=1}^4 A_i} $$ i.e the in-center is the area weighted average of the vertices.
The actual computation of the coordinates of in-center for this problem is left as an exercise.
Best Answer
There are $4$ planes
$$\begin{cases} P_0 &: x + y + z = 1\\ P_1 &: y + z = 0\\ P_2 &: x + z = 0\\ P_3 &: x + y = 0 \end{cases}$$ To find the vertices, select $3$ planes from $P_0,P_1,P_2,P_3$ and intersect them. Each vertex corresponds to a different way to select the planes and there are totally $4$ selections. For $i = 0, \ldots, 3$, let $v_i = (x_i,y_i,z_i)$ be the vertex belongs to $\bigcap\limits_{j=0,\ne i}^3 P_j$.
For $v_0 \in P_1\cap P_2 \cap P_3$, it is clear $v_0 = (0,0,0)$.
For $v_1 \in P_0\cap P_2 \cap P_3$, since it belongs to $P_2 \cap P_3$, we have $$x_1 + z_1 = 0, x_1 + y_1 = 0 \implies v_1 = (x_1, -x_1, -x_1)$$ Substitute this into the equation of $P_0$, we get $x_1 = -1 \implies v_1 = (-1,1,1)$
By a similar manner, we can determine the coordinates of $v_2, v_3$. The end result is $$\begin{cases} v_0 = (0,0,0)\\ v_1 = (-1,1,1)\\ v_2 = (1,-1,1)\\ v_3 = (1,1,-1) \end{cases}$$
Given coordinates of the four vertices, the circumsphere is given by following equation:
$$\left|\begin{matrix} 1 & x & y & z & x^2 + y^2 + z^2\\ 1 & x_0 & y_0 & z_0 & x_0^2 + y_0^2 + z_0^2\\ 1 & x_1 & y_1 & z_1 & x_1^2 + y_1^2 + z_1^2\\ 1 & x_2 & y_2 & z_2 & x_2^2 + y_2^2 + z_2^2\\ 1 & x_3 & y_3 & z_3 & x_3^2 + y_3^2 + z_3^2\\ \end{matrix}\right| = 0$$ Using the coordinates of $v_i$ derived above, we get $$ \left|\begin{matrix} 1 & x & y & z & x^2 + y^2 + z^2\\ 1 & 0 & 0 & 0 & 0\\ 1 & -1 & 1 & 1 & 3\\ 1 & 1 & -1 & 1 & 3\\ 1 & 1 & 1 & -1 & 3\\ \end{matrix}\right| = - \left|\begin{matrix} x & y & z & x^2 + y^2 + z^2\\ -1 & 1 & 1 & 3\\ 1 & -1 & 1 & 3\\ 1 & 1 & -1 & 3\\ \end{matrix}\right| = 0$$ This can be further simplified to $$\bbox[border:1px solid blue;padding: 1em;]{x^2+y^2+z^2 - 3(x+y+z) = 0}$$
If you don't like evaluating a $5 \times 5$ determinant, here is another way to determine the circumsphere.
Just like circle inversion in the plane, if you perform a sphere inversion with respect to the unit sphere in $\mathbb{R}^3$, any sphere passing through the origin will get mapped to a plane. Since one of the vertex $v_0$ is origin, the circumsphere we seek will get mapped to a plane.
Under the sphere inversion, $v_1, v_2, v_3$ get mapped to $\frac13 v_1$, $\frac13 v_2$ and $\frac13 v_3$. It is easy to see they are lying on the plane $x + y + z = \frac13$. If we reverse the sphere inversion, we obtain following equation for the circumsphere.
$$\frac{x+y+z}{x^2+y^2+z^2} = \frac13 \iff (x^2+y^2+z^2) - 3(x+y+z) = 0$$ This is the same equation we obtained before using a $5\times 5$ determinant.