1. Four non-complanar points in 3D define a sphere around M with radius r. The radius is the equal distance you are looking for.Hi guys,
I have a question.
Lets say i have cartesian coordinate of X1 = (1,3,1), X2 = (3,2,1) andX 3 = (5,2,4) , X4 = (10,1,2) and there is Y has equal distance to each of the X. how can i find the cartesian coordinate of Y? thanks.
No! (In those dark ages you would have been boiled in hot oil for this kind of calculations (Nod) )When i try to follow the formula,
i plug in these value.
for X1 , i put (1-Xm)^2 + (3-Ym)^2 + (1-Zm)^2 = r^2 and i have this : 1-Xm + 3-Ym + 1-Zm = r
am i right?
Nearly - and I'll suspend the oil treatment (Happy)Ops i think i need to be boiled in hot oil that.
I have already these equations : (1) 1-2x+x^2 + 9-6y+y^2 + 1-2z+z^2 = r^2 <--- I didn't check if you have done these calculations correctly
(2) 9-6x+x^2 + 4-4y+y^2 + 1-2z+z^2 = r^2
(3) 25-10x+x^2 + 4-4y+y^2 + 1-2z+z^2 = r^2
(4) 100-20x+x^2 + 1-2y+y^2 + 4-4z+z^2 = r^2
So i subtract (1) with (2) , (1) with (3) and (1) with (4).
<--- at the RHS of the equations you have r^2 - r^2 = 0
And i will have: (5) -3+4x-2y
(6) -34+8x-2y+6z
(7) -94+18x-4y+2z
Am i doing it right?