Isolating vector b from equation in form (axb)xa

[MannyZanny]

I am working on developing a physics engine and I'm having trouble isolating a vector variable from an equation. The entire equation is quite lengthy but I know how to figure out the rest if I find a way to move some variables around. I am much stronger in developing equations in calculus but not so much in linear algebra. Here is a small snippet:

j/m + ((r x j) x r)/I

j/m is vector j divided by scalar m, x is a cross product and I is a 3x3 matrix that is diagonal, sorry not sure of the proper term.

ie. I =
[x 0 0]
[0 x 0]
[0 0 x]

I am trying to factor out j to get something like j ( 1/m + (r x r)/I) so I can move that part to the other side of the equation ending in j=... I realize it's not a full equation which may make it more difficult for you to help me. I have worked out a simpler version of this equation but I can't figure this one out and I've spent hours trying to use identities and trying other things. I have read a couple papers that show these equations but one is expecting to solve component wise (jx = ... jy = ... jz = ...) which I know how to do but I wanted to get the matrix and vector solution. Another one shows the matrix solution but not how to arrive at that answer and uses some math notation I'm not sure of. Thanks for any help and if it is too hard without seeing the complete equation or you would like to see the papers I mentioned, let me know. I'll check the forum in the morning once I get some sleep :)

 

[arkajad]

Look at the Wikipedia entry "Cross product", "triple product expansion". From this you will find:

(1/m-r^2)j=(r.j)r

Since it is only a part of your equation, no further advice can be given. From this part alone you see that, unless r^2=1/m, j is parallel to r, its magnitude is arbitrary.

P.S. I ignored I as I am not sure whether it is a multiple of the identity or less trivial diagonal matrix. If it is just a multiple of the identity, it is just a number.

 

[D H]

I am trying to factor out j to get something like j ( 1/m + (r x r)/I) so I can move that part to the other side of the equation ending in j=...

You can't. That vector triple product only carries information about the portion of j that is orthogonal to r. In other words,

$$(\vec r\times \vec j)\times \vec r = (\vec r\times(\alpha \vec r + \vec j))\times \vec r$$

for any and all values of alpha.

 

[MannyZanny]

I decided to take the answer of the one I found in the paper and try to work backwards from once I got as forward as possible in my solution. The equations were very similar so I knew I was on the right track but there was one part of the equation I couldn't match with the known solution and figured these two different parts must be the same. So, I wrote a program to solve each part and found that they produced the same values within 0.000033 due to rounding errors. What I found was:

((r x j) x r)*I^-1 = j (r*^T I^-1 r*)

where r* is the cross product matrix:

[0 Rz -Ry]
[-Rz 0 Rx]
[Ry -Rx 0 ]

This now allows me to "factor" out j and find an equation. But does anyone know of this equality, it seems to hold true even though I only tested a few sample values and I have no idea how to prove it. Also, the paper I read calls r* the cross product matrix but the cross product matrix I saw on wikipedia looked much different and contained 2 vectors. I stumbled upon the above matrix by noticing a similarity to other works. Is there another name for the cross product matrix above? I found the formula under a heading of Exterior Product or Grassman Product, wikipedia says its part of exterior algebra, anyone know what that is and how it applies here, it seems as though it's part of geometry?

 

[arkajad]

You should tell us what are Rx,Ry,Rz and what exactly is I. Then maybe someone will answer your question.

 

[MannyZanny]

r is a vector from the center of mass to the point of collision so r* contains the components of vector r. Rx=x component of r, Ry=y component of r, Rz=z component of r. Sorry, I put a capital to emphasize that it was a component of vector r.
I is the inertial tensor and examples can be seen here http://en.wikipedia.org/wiki/List_of_moment_of_inertia_tensors

I is a 3x3 diagonal matrix but not necessarily a muliple of the identity matrix. ie (scalar)*(Identity Matrix)=I cannot be assumed.

But I tried arbitrary values and found that in general

((a x b) x a)*C^-1 = b (a*^T C^-1 a*)

where a and b are 3D vectors, C is a 3x3 diagonal matrix. And a* is defined as above.
So, if a=(1,2,3) then a*=
[0 3 -2]
[-3 0 1]
[2 -1 0]