Re: help with mathematica 5.2-Vector

*To*: mathgroup at smc.vnet.net*Subject*: [mg84500] Re: help with mathematica 5.2-Vector*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Sat, 29 Dec 2007 20:02:55 -0500 (EST)*References*: <fl4unn$9o5$1@smc.vnet.net>

Hi, a) you should use the Mathematica Syntax, i.e., a={1,2,3}; b={2,3,4}; b) Graphics[Point[a-b]]//Show will show you the point. In 3d coordinates is not possible to decide if {x,y,z} is a point or a vector, you need homogene coordinates to make a point p={x,y,z,1} or a vector v={x,y,z,0} .. Regards Jens george wrote: > Hello! I started recently working with Mathematica 5.2 and I face a problem that I'm trying to solve but I can't find the solution: I have 2 vectors, let's say: > a=(1,2,3) > b=(2,3,4) > I want to make the graph for c=a-b(it should be an arrow, at 3-Dimensions, right?). > I use the command ListPlot(c), nevertheless it produces cartesian axes with 2 points(I suppose these are the heads of the arrows, but I'm not sure). > Is there anything to do in order to make the correct graph? > (if the graph is not supposed to be an arrow, is not supposed to be at 3-Dimensions, it would be really helpful if someone could explain me why). > Thank you in advance. >