Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Angle between two vectors Date: Sat, 29 Dec 2007 05:53:08 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 96 Message-ID: <fl4nc4$47i$1@fred.mathworks.com> References: <ef5ce9c.-1@webcrossing.raydaftYaTP> <fjj9nj$fia$1@fred.mathworks.com> <fjk0tg$jli$1@fred.mathworks.com> <fl377q$4ip$1@fred.mathworks.com> <fl3moi$pvc$1@fred.mathworks.com> <fl3nhq$3tc$1@fred.mathworks.com> <fl3r7p$5nt$1@fred.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: webapp-02-blr.mathworks.com Content-Type: text/plain; charset="ISO-8859-1" Content-Transfer-Encoding: 8bit X-Trace: fred.mathworks.com 1198907588 4338 172.30.248.37 (29 Dec 2007 05:53:08 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Sat, 29 Dec 2007 05:53:08 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1223357 Xref: news.mathworks.com comp.soft-sys.matlab:443788 "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <fl3r7p$5nt$1@fred.mathworks.com>... > "baris kazar" <mbkazar.nospam@gmail.com> wrote in message <fl3nhq$3tc > $1@fred.mathworks.com>... > > Hi Roger,- > > yes, this is one step closer to what i need but not > > exactly. Let's take a numeric example: > > x=(1,0,0); y=(1,0,1) and z=(1,0-1) > > let's call the angle between x and y theta. > > Then i wanna get 2pi-theta for the angle between x and z. > > i dont have access y and z at the same time. > > hope that this problem statement is clear. > > Thanks much for your reply > > Best regards > -------- > You will have to try harder to explain your problem, if I am to understand > you, Baris. The example you gave has x, y, and z all in the same plane. You > didn't state previously that all your vectors are coplanar. Are they? But > whether they are or not, this doesn't explain how you would define the angle > theta between x and y. It could be plus pi/4 or it could be minus pi/4 (or > 7/4*pi.) Which one would you choose and according to what criterion? It > would depend on which side of the plane, in this case the x-z plane, is > regarded as its positive side - along the plus y-axis, or along the negative > side of the y-axis. Also it depends on whether you are moving from x > towards y or from y towards x if you are adhering to right-hand cross > product direction conventions. > > Select two arbitrary vectors in three-dimensional space (x1,y1,z1) and > (x2,y2,z2) and try to think of a consistent way of defining the angle between > them without reference to any other vector that would allow this quantity to > range over the full four quadrants, 0 to 2*pi. I think you will find this a > difficult thing to do in any way that could reasonably be considered canonical. > In two dimensions, there is a clearly defined counterclockwise direction from > vector x to vector y which would give you the range you desire. In three > dimensions, you lose the sense of what is a "counterclockwise" direction. You > can go from x to y along either of two great circle paths and one direction will > give the supplement angle to the opposite direction. If you always select the > shortest path, then your angle range is restricted to [0,pi]. > > If one vector of each pair is restricted to a particular fixed vector, as your > previous wordage seemed to imply, I still don't see what criterion you wish to > use to define these angles. For example, you can move from the fixed vector > by one degree in all possible directions giving a cone, but which half of these > angles should be adjusted so as to be the supplements (that is 359 degrees,) > of those on the opposite side? If you are restricting your vectors to all be > coplanar, then which side of such a plane is to be considered its "positive" > side? > > Roger Stafford > Hi Roger,- these are excellent questions. one can choose to look at the plane from +ive infinity (above) or from -ive infinity (below). I know that in 3D there is no notion of CW or CCW. However, you can define it locally wrt a point/vector as long as they are on a plane by defining from which side you are looking at. Anyways, i dont wanna confuse anyone any more. I will find another solution to this problem. Best regards