# Cartesian coordinate help

• April 10th 2012, 09:20 AM
drogba
Cartesian 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.
• April 10th 2012, 09:47 AM
earboth
Re: Cartesian coordinate help
Quote:

Originally Posted by 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.

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 $M(x_M,y_M, z_M)$ and radius r is:

$(x-x_M)^2+(y-y_M)^2+(z-z_M)^2 = r^2$

Plug in the coordinates of the given points and solve the system of equations for $(x_M,y_M, z_M, r)$

3. For confirmation only: $M\left(\frac{233}{26} , \frac{427}{26} , -\frac{21}{26} \right)$ and $r = \frac{11}{26} \sqrt{1379}$
• April 10th 2012, 10:06 AM
drogba
Re: Cartesian coordinate help
Hi,

Do you mean the value of cartesian coordinate of Y is M?
• April 10th 2012, 10:23 AM
earboth
Re: Cartesian coordinate help
Quote:

Originally Posted by drogba
Hi,

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

Yes
• April 10th 2012, 08:18 PM
drogba
Re: 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.
• April 10th 2012, 08:29 PM
drogba
Re: 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?
• April 10th 2012, 10:37 PM
earboth
Re: Cartesian coordinate help
Quote:

Originally Posted by 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?

No! (In those dark ages you would have been boiled in hot oil for this kind of calculations (Nod) )

From

$(1-X_m)^2 + (3-Y_m)^2 + (1-Z_m)^2 = r^2$

you'll get

$1-2X_m+X_m^2 + 9-6Y_m+Y_m^2+1-2Z_m+Z_m^2=r^2$

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 $X_m, Y_m \ \text{and}\ Z_m$. 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.
• April 10th 2012, 11:13 PM
drogba
Re: 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?
• April 10th 2012, 11:46 PM
earboth
Re: Cartesian coordinate help
Quote:

Originally Posted by 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 <--- 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)

Your system of equations becomes:

$\left|\begin{array}{rcl}-3+4x-2y& =& 0 \\ -34+8x-2y+6z &=& 0 \\ -94+18x-4y+2z &=& 0 \end{array} \right.$ ..... $\implies$ ..... $\left|\begin{array}{rcl}4x-2y& =& 3 \\ 8x-2y+6z &=& 34 \\ 18x-4y+2z &=& 94 \end{array} \right.$
• April 10th 2012, 11:49 PM
drogba
Re: 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 ?
• April 11th 2012, 10:53 AM
earboth
Re: Cartesian coordinate help
Quote:

Originally Posted by 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 ?

Yes, of course you can change the order of the equations. The target of these subtractions is to get rid of the squares.

This rigid order of operations is only one of my habits so that I'm sure I don't forget to use one of the equations.
• April 14th 2012, 06:47 AM
drogba
Re: 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.!!!