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Unexpect answer for A x B = C x B where A, B, C are R3

Posted 2 years ago

My apologies for this dumb question but I have never gotten such a strange answer from Mathematica before. I was hoping that someone could tell me what Mathematica is doing and why. This actually started with a physics problem: The torque on a dipole (M) in a magnetic field (B) is given by torque (T) T = M x B and the torque applied to dipole particle at a distance (R) from the center of rotation is
T = R x F where (F) acts on the particle then we get a vector algebra problem: R x F = M x B where each capital letter is a 3 vector in R3.

(1) The first question is how do you solve for the three components of (F)?

(2) I tried right and left multiplications with the other three vectors, but Mathematica always returns a null set.

(3) Next, I tried reducing the complexity by creating what should be a very simple problem:

Solve R x F = R x B. This should have the answer F = B. Mathematica doesn't think so. It returns this silliness (below). Would someone please tell me why my Mathematica code is wrong and what happened here??? So, what happened to (fx) and why doesn't F = B appear as an answer?

In[1523]:= Clear[m, b, r, f, t, mx, my, mz, bx, by, bz, rx, ry, rz,
fx, fy, fz, tx, ty, tz]

mx =.;
my =.;
mz =.;
bx =.;
by =.;
bz =.;
fx =.;
fy =.;
fz =.;
rx =.;
ry =.;
rz =.;
tx =.;
ty =.;
tz =.;

m = {mx, my, mz};
b = {bx, by, bz};
r = {rx, ry, rz};
f = {fx, fy, fz};
t = {tx, ty, tz};

ss = Solve[{Simplify[Cross[r, f] == Cross[r, b]]}, {fx, fy, fz}]

Simplify[ss]

During evaluation of In[1523]:= Solve::svars: Equations may not give solutions for all "solve" variables.

Out[1544]= {{fy -> (fx ry)/rx - (-by rx + bx ry)/rx, 
  fz -> (fx rz)/rx - (-bz rx + bx rz)/rx}}

Out[1545]= {{fy -> (by rx - bx ry + fx ry)/rx, 
  fz -> (bz rx - bx rz + fx rz)/rx}}

Many thanks to all the great users that took the the time to read and respond to my questions!!! Muchas Gracias!!

POSTED BY: Gerald Kamin
6 Replies
Posted 2 years ago

Gerald, I think it's pretty easy to find this information. A quick search should reveal plenty of adequate resources.

But to give a brief answer, think about what cross product means geometrically. It's a vector perpendicular to the input vectors and with length equal to the area enclosed by the parallelogram defined by the input vectors. How many ways are there to generate a parallelogram with a given area and with one side/vector fixed? Infinitely many! Furthermore, as Daniel explained, for the non-fixed vector, the constraint is simply that its endpoint falls on a particular line.

POSTED BY: Eric Rimbey

Exactly!

Given a and m

a = {a1, a2, a3}
m = {m1, m2, m3}

Define with an arbitrary parameter x

f = x a - m

Then

Cross[a, m]
Cross[f, a]
POSTED BY: Hans Dolhaine

As noted previously, the solution set is one dimensional. That means two of the three varianles get solutions in terms of the third one.

POSTED BY: Daniel Lichtblau
Posted 2 years ago

Hi Eric and Daniel, Thank you for your kind replies.

I should have been more precise in my statement of the problem.

What I was looking for was a general solution for F when R x F = M x B.

(Hi Eric): The special case R x F = R x B apparently does not have a solution F = B according to what you were saying. Would you be so kind as to explain this ?? Is there a good reference book on vector algebra that discusses these operations in detail?? (Lang??)

So, Is there a general solution for F when R x F = M x B ??

If yes, how do I code this in Mathematica??

When I coded the three sets of equations for { fx, fy, fz } Mathematica only gave TWO answers:

fy --> (by rx - bx ry + fx ry)/rx

fz --> (bz rx - bx rz + fx rz)/rx}}

Why did Mathematica not provide an answer for (fx)??

Again, Thank you kindly gentleman for your replies!!!

POSTED BY: Gerald Kamin

The solution set is one dimensional. Set fx=bx and you recover the special solution of interest.

POSTED BY: Daniel Lichtblau
Posted 2 years ago

Solve R x F = R x B. This should have the answer F = B

Not an answer to your larger question, but this assertion is false for cross products.

POSTED BY: Eric Rimbey
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