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.

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- Apr 10th 2012, 09:20 AMdrogbaCartesian coordinate help
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. - Apr 10th 2012, 09:47 AMearbothRe: Cartesian coordinate help
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 $\displaystyle r = \frac{11}{26} \sqrt{1379}$ - Apr 10th 2012, 10:06 AMdrogbaRe: Cartesian coordinate help
Hi,

Do you mean the value of cartesian coordinate of Y is M? - Apr 10th 2012, 10:23 AMearbothRe: Cartesian coordinate help
- Apr 10th 2012, 08:18 PMdrogbaRe: Cartesian coordinate help
Hi,

May i know how to get the value for Xm , Ym and Zm? and after calculating i still could not get r = 11/26 square root of 1379. Thanks. - Apr 10th 2012, 08:29 PMdrogbaRe: Cartesian coordinate help
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? - Apr 10th 2012, 10:37 PMearbothRe: Cartesian coordinate help
No! (In those dark ages you would have been boiled in hot oil for this kind of calculations (Nod) )

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. - Apr 10th 2012, 11:13 PMdrogbaRe: Cartesian coordinate help
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? - Apr 10th 2012, 11:46 PMearbothRe: Cartesian coordinate help
Nearly - and I'll suspend the oil treatment (Happy)

Your system of equations becomes:

$\displaystyle \left|\begin{array}{rcl}-3+4x-2y& =& 0 \\ -34+8x-2y+6z &=& 0 \\ -94+18x-4y+2z &=& 0 \end{array} \right.$ ..... $\displaystyle \implies$ ..... $\displaystyle \left|\begin{array}{rcl}4x-2y& =& 3 \\ 8x-2y+6z &=& 34 \\ 18x-4y+2z &=& 94 \end{array} \right.$ - Apr 10th 2012, 11:49 PMdrogbaRe: Cartesian coordinate help
Hi,

I think you can suspend the oil treatment already hee.

Btw, must it be always (1) minus (2), (3) and the rest or can it be interchange like (2) minus (1), (3) and so on? thanks. or this is a rule for cartesian ? - Apr 11th 2012, 10:53 AMearbothRe: Cartesian coordinate help
- Apr 14th 2012, 06:47 AMdrogbaRe: Cartesian coordinate help
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.!!!