Question to other math geeks out there:
If I have vectors u and v, it’s trivial to compute the cross product, u x v. Unfortunately, what I have is u x v and v, and I need u. Is there a property of cross products that allows one to determine one of the components, or to otherwise “solve for u”?
I know one definition of cross product is |u||v|sin q, but solving that isn’t going to work because I don’t know the angle, and I need the direction as well as magnitude.
Any suggestions?
Your problem doesn’t have a unique solution. Since v x v = 0, you’ll have (u + tv) x v = u x v for any real number t.
I’ll think about this a little more and see if I can come up with some natural way of “setting t to zero” or something, and let you kno what I come up with.
Here you go: If you know u.v as well (period = dot product), then you can determine u. We have the identity
(u x v) x w = (u.w) v - (v.w) u;
so this implies that
(u x v) x v = (u.v) v - (v.v) u.
Rearranging gives us
u = ((u.v) v - (u x v) x v) / (v.v).
In particular, if you want to demand that (u.v) = 0, i.e. u and v are orthogonal, then u = v x (u x v) / (v.v).
There is another way you can do it.
Given u = (A, B, C), v = (X, Y, Z) and u x v = (i, j, k)
BZ - CY = i
CX - AZ = j
AY - BX = k
You know i, j, k, x, y, z and so you can solve a, b, c.
Trigonal, we’ve tried that method, but the answer doesn’t make physical sense.
Basically, what we’re after is a solution to the torque equation, T = r X F, where we know torque and force, but not the point of application. Solving three simultaneous linear equations gives a solution, but not one that makes sense in our physical model. I’m thinking maybe MikeG was right the first time when he said there was no unique solution.
Because of transmissivity of forces, any force of magnitude |F| will produce the same torque as long as it acts in the direction of F. Thus there are infinitely many vectors r that will solve the equation. Am I correct in this, or am I missing something important?
I must be reading this wrong. If a force has magnitude |F| and acts in the direction of F, then it is F, yes?
I think I was trying to be too careful with my nomenclature. I think the answer is, yes, a force with a given magnitude, acting in the direction of F, is simply F. I’ve worked out a few specific cases and gotten the same cross-product each time, so transmissibility is most likely getting in the way.
Someone on another message board told me to multiply both sides of the torque equation by F[sup]-1[/sup], helpfully forgetting that vectors don’t have inverses. Unless there’s something I’ve forgotten about vectors.
Er, if
F=(1)
(2)
(3)
Then, isn’t F[sup]-1[/sup]=(1,2,3)
Whether a vector is a column or a row is important in this case, since the vector product is simply a generalisation of the outer product.
D’Oh! The outer product is a generalisation of the vector product. Not the other way round. :smack:
I am intelligent. Honest.
Again, this is a consequence of the non-uniqueness of the solution. You’re essentially trying to find a solution to the matrix equation (forgive me if you’re not familiar with linear algebra)
[ 0 -C B ][ x ] [ i ]
[ C 0 -A ][ y ] = [ j ]
[ -B A 0 ][ z ] = [ k ]
But the determinant of the matrix on the left-hand side of the equation is 0; so the solution you get isn’t unique. Indeed, a solution may not necessarily exist: if u x v isn’t perpendicular to v, then there can’t be a solution for u since u x v is always perpendicular to both u and v.
Nope, and that’s exactly what I was saying above. If you add a multiple of F to your position vector r, you get the same torque since (r + t F) x F = r x F.
Oh, and it’s MikeS.
Sorry about getting your last initial wrong, MikeS. 
Angua, I admit I don’t know much about linear algebra, is a transpose the same thing as an inverse for a vector? I’m not clear on how one inverts anything other than a square matrix, and MathWorld doesn’t have any information to suggest otherwise.
No, its not. My mistake. However, I think that your contact may have meant the transpose, as that may well work…
But of course, that won’t make much difference at all. :smack:
I reckon its do-able. Just not sure how.
Shouldn’t this work:
T=(T[sub]1[/sub],T[sub]2[/sub],T[sub]3[/sub])
r=(r[sub]1[/sub],r[sub]2[/sub],r[sub]3[/sub])
F=(F[sub]1[/sub],F[sub]2[/sub],F[sub]3[/sub])
Now,
T[sub]1[/sub]=r[sub]2[/sub]F[sub]3[/sub] - r[sub]3[/sub]F[sub]2[/sub] ---(1)
T[sub]2[/sub]=r[sub]3[/sub]F[sub]1[/sub] - r[sub]1[/sub]F[sub]3[/sub] ---(2)
T[sub]3[/sub]=r[sub]1[/sub]F[sub]2[/sub] - r[sub]2[/sub]F[sub]1[/sub] ---(3)
Rearrange (1) to give us:
r[sub]2[/sub]=(T[sub]1[/sub]+r[sub]3[/sub]F[sub]2[/sub])/F[sub]3[/sub] ---- (4)
Rearrange (3) to give us:
r[sub]2[/sub]=(F[sub]2[/sub]r[sub]1[/sub]-T[sub]3[/sub])/F[sub]1[/sub] ---- (5)
Equate (4) and (5) to give an equation in terms of r[sub]1[/sub] and r[sub]3[/sub]. Use (2) to get either r[sub]1[/sub] or r[sub]3[/sub] in terms of the other. Solve.
Or am I missing something? 3 unknowns, 3 simultaneous equations, simple(ish) surely.
To illustrate, with a couple of ASCII diagrams. Arrows represent forces, and the X is the pivot point about which we’re calculating torque. We know the force, and we know what we want the torque to be, but we don’t know where to apply the force. Both of the diagrams below have the same appplied force and the same torque, but different points of application:
Case A Case B
---> --->
X X
In fact, any point along the line of the arrow will also give the same torque.
Do you have any other constraints on where the force is applied? If, for example, the object you’re torquing is a rod, then the force must be applied somewhere on the rod, and you know the direction of the displacement vector. If you don’t have any such constraints, then there are multiple correct answers, and you could give any of them. In that case, I would recommend the one with the force at right angles to the displacement vector, as that’s the simplest case.
Angua, that’s the same method I suggested. The problem is, if you actually go through and solve it, you’ll see that (as also pointed out by MikeS) you end up with zero, or “trivial” solution.
Arse. And of course! Since we get r[sub]whatever[/sub]=F.T. Which by definition is zero!
Curses! Beaten by a maths problem! I’m going to have to concede defeat here. Which I do so reluctantly.
Chronos would be correct. I just discovered in my statics book what he’s talking about, and they describe it as “transmissibility of forces.” You can have infinitely many vectors r, all pointing to different forces along the same line, and they all generate the same torque. In fact, all those vectors determine a plane, and the torque vector acts normal to that plane.
For my real-world problem, I can somehow determine the boundaries of the object these forces act on, and make sure my force line is somewhere intersecting it. Of course it’s not an easily defined three-dimensional object.
Angua, you concede defeat much more easily than my manager does. I just have to convince him what he wants won’t work.
WAIT A SECOND!
This assumes you can solve a linear system
( 0 Z -Y) (A) (i)
(-Z 0 X) (B) = (j)
( Y -X 0) (C) (k)
(damned lack of matrices…) Anyhow, notice that the matrix on the left is always singular. This system cannot be solved in general.
Hey, I might be a scientist, but I’m a pragmatic scientist. 