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.
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.
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.
2. The general equation of sphere with $\displaystyle M(x_M,y_M, z_M)$ and radius r is:
$\displaystyle (x-x_M)^2+(y-y_M)^2+(z-z_M)^2 = r^2$
Plug in the coordinates of the given points and solve the system of equations for $\displaystyle (x_M,y_M, z_M, r)$
3. For confirmation only: $\displaystyle M\left(\frac{233}{26} , \frac{427}{26} , -\frac{21}{26} \right)$ and $\displaystyle r = \frac{11}{26} \sqrt{1379}$
No! (In those dark ages you would have been boiled in hot oil for this kind of calculations )
From
$\displaystyle (1-X_m)^2 + (3-Y_m)^2 + (1-Z_m)^2 = r^2$
you'll get
$\displaystyle 1-2X_m+X_m^2 + 9-6Y_m+Y_m^2+1-2Z_m+Z_m^2=r^2$
You have finally a system of 4 equations which look all alike. Then subtract the 1st equation columnwise from the 2nd, then from the 3rd and finally from the 4th. All squares will vanish.
Now you have a system of 3 linear simultaneous equations which contain $\displaystyle X_m, Y_m \ \text{and}\ Z_m$. Solve this system using the method which is most convenient for you.
When you have determined the coordinates of M (in the text of the question this point is labeled Y) you can use any of the 4 equations to get the length of r.
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
(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).
And i will have: (5) -3+4x-2y
(6) -34+8x-2y+6z
(7) -94+18x-4y+2z
Am i doing it right?
Nearly - and I'll suspend the oil treatment
Your system of equations becomes:
$\displaystyle \left|\begin{array}{rcl}-3+4x-2y& =& 0 \\ -34+8x-2y+6z &=& 0 \\ -94+18x-4y+2z &=& 0 \end{array} \right.$ ..... $\displaystyle \implies$ ..... $\displaystyle \left|\begin{array}{rcl}4x-2y& =& 3 \\ 8x-2y+6z &=& 34 \\ 18x-4y+2z &=& 94 \end{array} \right.$
Hi,
Thanks for the guidance.Btw i have another question. Is it possible to have orthogonal equation with only -1 and 1 values only.
So that A.B, B.C and A.C will be 0 by just using -1 and 1? (eg. A = -1 1 , B = 1 1) ? Thanks.!!!