# Cartesian coordinate help

#### drogba

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.

#### earboth

MHF Hall of Honor
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 $$\displaystyle r = \frac{11}{26} \sqrt{1379}$$

• 1 person

#### drogba

Hi,

Do you mean the value of cartesian coordinate of Y is M?

#### earboth

MHF Hall of Honor

#### drogba

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.

#### drogba

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?

#### earboth

MHF Hall of Honor
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?
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.

#### drogba

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?

#### earboth

MHF Hall of Honor
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?
Nearly - and I'll suspend the oil treatment (Happy)

$$\displaystyle \left|\begin{array}{rcl}-3+4x-2y& =& 0 \\ -34+8x-2y+6z &=& 0 \\ -94+18x-4y+2z &=& 0 \end{array} \right.$$ ..... $$\displaystyle \implies$$ ..... $$\displaystyle \left|\begin{array}{rcl}4x-2y& =& 3 \\ 8x-2y+6z &=& 34 \\ 18x-4y+2z &=& 94 \end{array} \right.$$