On Tue, 22 Jan 2013 07:58:20 -0800 (PST), "Tim Golden BandTech.com" <email@example.com> wrote:
>On Tuesday, January 22, 2013 12:47:11 AM UTC-5, hbe...@gmail.com wrote: >> Say we have a 1-D subspace L (i.e., a line thru the origin) in R^3 , and >> >> let q=(x,y,z) be a point in R^3 which is not on the line. Then I don't see >> >> how to project q orthogonally onto L; > >The answer is simply the dot product. The projection from q to L is > dotProduct( q, L ) .
That's nonsense in several ways. First, L is not a vector, so there's no such thing as this dot product. Second, if L _were_ a vector this dot product would be a scalar, while the orthogonal prohection of a vector onto a subspace is a vector.
Your answer does have the virtue of simplicity, though...
> >To round this out, you can multiply this value by L hat(the unit vector along L) but this then would be back in the q space, which is not technically what you are after; the dot product provides a real value which is the value along L; this is your one dimensional projection, and this is how simple multidimensional projection is, though few seem to appreciate it. The main challenge in going multidimensional is ensuring that you have an orthogonal system. Even this is unnecessary, but this is typically what people want in a multiD projection. Clearly you do not have this issue in the 1D case. Still, the multiD projection is simply a series of these 1D projections. > >Projection can grow much more complicated from this basic form, but this is the most primitive form that is general. For instance I am growing interested in implementing occlusion, but for the bland Euclidean sense everything is black and white so this doesn't matter much. > > - Tim http://bandtechnology.com > > >> I can see how to project q into a >> >> 2-D subspace P : the projection of q into P is the ortho complement, and >> >> every vector in P is in the orthogonal complement of the ortho projected >> >> line. But, the same is not the case with q and L. Sorry for the rambling; >> >> my question is then actually: >> >> >> >> If L is a 1-D subspace of R^3, and q=(x,y,z) is a point not on L. Can we >> >> define the orthogonal projection of q into L, or do we need to have a plane >> >> P (as subspace) , to define an ortho projection of q? >> >> >> >> Thanks.