       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).