Inverse of cross product?

Well, since Chronos already took care of it, I’ll go ahead and add a couple of things:

Transposing a vector isn’t the same as inverting a vector. In fact, you can’t invert a vector–only a square matrix.

Chronos’s suggestion about considering the boundaries of your model is right on the money. I’ll also point out that the right hand rule can be…ahem…handy, particularly for spatially challenging problems.

Boundary case: a 1-dimensional vector is a square matrix.

Oh, fine, get me on a technicality!

::grumble:: I knew I was somehow going to get nailed on that post ::grumble:: :stuck_out_tongue:

Well, I suppose it depends what you mean by “inverse”. Given any unital ring R, morphisms from m to n in the category of matrices over R may have a left inverse if n>m, and may have a right inverse if m>n. If m=n, and one exists, it is the other as well, which is what we usually refer to as “the” matrix inverse.

I would add one comment to Chronos’s correct explanation. The fact that r is not unique does not limit one’s ability to solve physical problems involving torque. Of the three quantities in r X F = T, F is a bona fide vector. So it’s the same quantity regardless of what coordinate system you represent it in, i.e. what coordinates you assign the point of application. What you are doing in using Chronos’s suggestion to choose the answer in which the force and “displacement vector” are perpendicular is choosing a particularly simple coordinate system in which to define the point of application of the force.

I put the term “displacement vector” in quotes because it is not really a vector. It is not invariant under coordinate transformations. However, once you choose a coordinate system to express F in, r is well-defined. Also, torque, like any cross product, is not quite a vector either. It is invariant under inversion of its coordinate axes, whereas a true vector is inverted if you invert the coordinate axes (exchange x and y).