Solving this formula, could anyone help??

Can anyone solve this, to find X, Y and Z please?? a,b,c,d,e,f,v are known values. This has been sending me mad trying to solve it :confused::confused:.

VB Code:

2ad + 2be + 2cf = X(a+d) + Y(b+e) +Z(c+f)
 
'Other known facts are:
u = Sqr(a^2 + b^2 + c^2)
n = Sqr(d^2 + e^2 + f^2)
u = n
'So:
a^2 + b^2 + c^2 = d^2 + e^2 + f^2
'Also:
v^2 = X^2 + Y^2 + Z^2

Any help would be greatfully appreciated.
This all came from this by the way:

VB Code:

0 = aX + bY + cZ
0 = dX + eY + fZ
0 = ad + be + cf
v^2 = X^2 + Y^2 + Z^2

There should be two answers for this.
Thanx for any help.

hey there mate,

i think its possible to solve for x,y and z from the initial equations:

Quote:

---

0 = aX + bY + cZ
0 = dX + eY + fZ
0 = ad + be + cf
v^2 = X^2 + Y^2 + Z^2

---

If a,b,c,d,e,f are known, then its a simple matter of substituting their values and solving three simultanaeous eqns. Is this what you wanted?

Yea thanx thats what I was after but I posted at the top how far I'd got, Then I realised I might have gone down the wrong track so thort it'd be a good idea to post what I started from aswell.

I still need to keep a,b,c,d,e,f,v as a,b,c,d,e,f,v because i'm putting this into code so I need to solve this algebraically:(

He's a pic to help explain it:
http://www.vbforums.com/attachment.p...postid=1489807

Sorry to make this the 4th post in a row but I've managed to add this to it:

Code:

---

dX + eY + fZ = rd + se + tf

---

r, s and t are more known values.
This is starting to get really bad:(

Ill think about it over lunch and get back to you

k, Thanx Loads

Do you have to solve it generally?
It would make things a whole lot simpler it you could give us the known values

No, I need to put it into code so the values will change for different times the procedure is run:(
Don't forget there'll be two results as well, probably just a +/- thing on a SqRt

Rearrange the first equation to
X=-(bY+cZ)/a
and the second to:
X=-(eY+fZ)/d

This gives
-[(bY+cZ)/a]=-[(eY+fZ)/d]

Then
Y(db-ae)=Z(af-dc)

You then have a relationship for Z based on Y and X based on Z and Y.

You can then substitute for X and Y into the equation for V^2 to give you values of V^2 based on Z

I hope you can follow this because it is a lot of typing if you need me to write it out in full!

hmm, got someone else to look at it, not so clear huh? :p

define a value K=(dc-af)/(ae-db)
then
V^2=Z^2[ (K+c)^2 /a^2 + K^2 +1]

define another value
M=1/[ (K+c)^2 / a^2 + K^2 +1 ]

Gives you

X^2 = V^2 (1-K^2 M -M)

Y^2 = V^2 K^2 M

Z^2 = V^2 M

Dunno if that helps or not? :D

thats great thanx, I'll check them out and let you know how it goes :) :) :) :).

just one problem: In the second solution theres this line that I don't understand:

VB Code:

X^2 = V^2 (1-K^2 M -M)
 
'Is it ment to be:
X^2 = V^2 (1 - K^2 * M * (-M)) '??
'I ask this because I don't think this would make sense:
X^2 = V^2 (1 - K^2 * (M - M))

Just to be sure :)

maybe its just
X^2 = V^2 (1 - K^2 * M - M)

Yea but:

VB Code:

X^2 = V^2 (1 - K^2 * M - M)
'Is the same as:
X^2 = V^2 (1 - K^2 * (M - M))
'And that would simplify to:
X^2 = V^2 (1 - K^2 * 0)
' =>
X^2 = V^2

I'm goint to have to assume its X^2 = V^2 (1 - K^2 * M * (-M)) and once i've done that the 3D thing thing be finished cyborg :D :D :D :D :D

I suppose it could be X^2 = V^2 (1 - (K^2 * M) - M) ??

wouldn't this:
X^2 = V^2 (1 - K^2 * M - M)
be the same as this:
X^2 = V^2 (1 - (K^2 * M) - M)
??

Yea, sorry cyborg.

I have a problems now tho, the second solution doesn't like it if some of the variables are zero :( when I first tested it most of them were zero and the VB deguger had a field day :D :D :D. I'm now gona try the first solution you said and see what happens

agmorgan: You any good with vectors??

Hiya, I think I have it :D do you reckon this works, its what I got after finishing off your method agmorgan:

VB Code:

p = (af - dc) / (db - ae)
q = 1 / p
 
Z^2 = V^2 / [1 + p^2 - (b^2 p^2 + c^2 + 2bpc)/a^2]
Y^2 = V^2 / [1 + q^2 - (c^2 q^2 + b^2 + 2bpc)/a^2]
X^2 = V^2 - Y^2 - Z^2

It seems zero friendly aswell :D
Thanx loads, I'll see if it work with what I want it to now. Fingers crossed.

EDIT: Nope, it doesn't like zeros, the p and q bit causes the problem as far as zero goes. Also it just doesn't seem to be working. I'm going to have to see if I can find where the problem is comming from and maybe try something else. Dam ArcCos(Cos(x)) = x for 0 < x < 180, why not 0 < x < 360 then I wouldn't have had to do this in the first place. :mad: :mad:

Not sure if this helps but here is a link to a simultaneous equation solver.

If you contact them maybe they can help you :)

http://web.ask.com/redir?bpg=http%3a...%2falgebra.htm