Equation in 3d-Space

Question

Consider we have 3 fixed points A,B and C also,We have this Equation :

$MA^2+MB^2+MC^2=30$.

What does this Equation represent in 3d-Space,( where M is such a Non-fixed point in that space)?

Suppose $A(a1,b1,b1),B(a2,b2,c2),C(a3,b3,c3),M(x,y,z)$

any help will be appreciated.

by writing $MA$: it means the distance between point M and point A — Yaser Tarek, Apr 06 '18 at 23:45

John Wayland Bales · Accepted Answer · 2018-04-08T05:22:08.270

Given $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3),M(x,y,z)$ and $r>0$.

What subset of $\mathbb{R}^3$ is specified by the equation

\begin{equation} |MA|^2+|MB|^2+|MC|^2=r^2 \tag{1}\end{equation}

\begin{eqnarray} |MA|^2&=&(x-x_1)^2+(y-y_1)^2+(z-z_1)^2\\ |MB|^2&=&(x-x_2)^2+(y-y_2)^2+(z-z_2)^2\\ |MC|^2&=&(x-x_3)^2+(y-y_3)^2+(z-z_3)^2 \end{eqnarray} So we can re-write equation $(1)$ as

\begin{eqnarray} &&3x^2-2x(x_1+x_2+x_3)+(x_1^2+x_2^2+x_3^2)\\ &+&3y^2-2y(y_1+y_2+y_3)+(y_1^2+y_2^2+y_3^2)\\ &+&3z^2-2z(z_1+z_2+z_3)+(z_1^2+z_2^2+z_3^2)\\ &=&r^2 \end{eqnarray} Define \begin{eqnarray} \bar{x}&=&\frac{x_1+x_2+x_3}{3}\\ \bar{y}&=&\frac{y_1+y_2+y_3}{3}\\ \bar{z}&=&\frac{z_1+z_2+z_3}{3}\\ \end{eqnarray} Then the equation can be written in the form

\begin{eqnarray} (x-\bar{x})^2+(y-\bar{y})^2+(z-\bar{z})^2=R^2 \end{eqnarray}

where $R$ is a constant defined in terms of $r$ and the coordinates of $A,\,B$ and $C$.

This is the equation of a sphere.

Note that OP has amended the question to ask what conditions on $A,\,B$ and $C$ guarantee that the solution will be a sphere.

First let

$$D=\left(\bar{x},\bar{y},\bar{z}\right)$$

The equation can also be written in the form

\begin{equation} x^2-2x\bar{x}+y^2-2y\bar{y}+z^2-2z\bar{z}+\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)=\frac{1}{3}r^2 \end{equation} which can be written \begin{eqnarray} (x-\bar{x})^2+(y-\bar{x})^2+(y-\bar{y})^2&=&\frac{1}{3}r^2+\bar{x}^2+\bar{y}^2+\bar{z}^2-\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)\\ &=&\frac{1}{3}r^2+\vert D\vert^2 -\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)\\ &>&0 \end{eqnarray}

So the condition is

$$ \vert A\vert^2+\vert B\vert^2+\vert C\vert^2-3\vert D\vert^2<r^2 $$

Thus, if we have $A(2,2,2),\,B(3,3,3),\,C(4,4,4)$ and $r^2=30$, then we have $D(3,3,3)$. So

\begin{eqnarray} \vert A\vert^2+\vert B\vert^2+\vert C\vert^2-3\vert D\vert^2&=&12+27+48-3(27)\\ &=&6\\ &<&30 \end{eqnarray} so there will be a sphere for that collection of three points.

Using the previous result

\begin{equation} R^2=\frac{1}{3}r^2+\vert D\vert^2 -\frac{1}{3}(\vert A\vert^2+\vert B\vert^2+\vert C\vert^2)=8 \end{equation}

we get

$$ (x-3)^2+(y-3)^2+(z-3)^2=8 $$

thnaks alot, but R could be a negative value right? i mean for some specific coordinates of$ A , B , C$ — Yaser Tarek, Apr 07 '18 at 06:35
if we choose $A(2,2,2) ,B(3,3,3),C(4,4,4) , r=30$ this will make $R^2$ a negative value — Yaser Tarek, Apr 07 '18 at 06:55
That is correct. There will not be a solution for some values of $A,, B$ and $C$. — John Wayland Bales, Apr 07 '18 at 15:01
thanks alot but can we figure out such a condition for the points $A,B$ and $C$ So the equation will represent a sphere — Yaser Tarek, Apr 07 '18 at 15:33
Actually, there is a sphere for the three points you mentioned in your comment. See my edit to the question. — John Wayland Bales, Apr 07 '18 at 22:04
i see that , thanks for correction your edit was useful for me:) — Yaser Tarek, Apr 08 '18 at 15:01

Equation in 3d-Space

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