# 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)

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

#### drogba

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 ?