Calculate the value of x, y and z coordinates

Question

System of the equations:

Equation 1:

$$-360 = -6x + x^2 - 40y + y^2 - 20z + z^2$$

Equation 2:

$$-600 = -40x + x^2 - 30y + y^2 - 100z + z^2$$

Equation 3:

$$59.85 = -10x + x^2 - 4y + y^2 - 10z + z^2$$

Answer 1

It looks like you are trying to find the intersection of three spheres (with equation $r^2 = (x-a)^2+(y-b)^2+(z-c)^2$ or $x^2-2ax+y^2-2by+z^2-2cz = r^2-a^2-b^2-c^2$).

Anyway, the equations are

$\begin{align} a &=-360 = -6x + x^2 - 40y + y^2 - 20z + z^2\\ b&= -600 = -40x + x^2 - 30y + y^2 - 100z + z^2\\ c&= 59.85 = -10x + x^2 - 4y + y^2 - 10z + z^2\\ \end{align} $

Subtracting the first two, $a-b = 34x -10y +80z$. Subtracting the last two, $b-c = 30x -26y -90z$.

These will let you get any two of $x, y$, and $z$ in terms of the other. For example, you might get $y = px+q$ and $z = rx+s$ for some $p, q, r, $and $s$.

Substitute these expressions in any of your equations and you will get a quadratic for $x$. For example, if you substitute these in the first equation, you get $\begin{align} a &= -6x + x^2 - 40y + y^2 - 20z + z^2\\ &=-6x + x^2 - 40(px+q) + (px+q)^2 - 20(rx+s) + (rx+s)^2\\ &=-6x + x^2 - 40px-40q + p^2x^2+2pxq+q^2 - 20rx-20s + r^2x^2+2rxs+s^2\\ &=-x(6+40p+20r+2pq+2rs) + x^2(1+p^2+r^2) +q^2 -40q-20s +s^2\\ \end{align} $

Solve this quadratic in $x$ and you will get 0, 1, or 2 real roots depending how the three spheres intersect. Get $y$ and $z$ for each $x$ and you have your solutions.

To see how the spheres can intersect, look at two spheres. They can be disjoint, be tangent, or intersect in a circle. Adding a third sphere allows for no common points, one common point if the sphere passes through the tangent point, or two points if the sphere passes through the circle of intersection.

There is also the possibility of two of the spheres coinciding, in which case the intersection could also be a circle. In this case, the three original equations would not be linearly independent.