Re: Vector problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg115998] Re: Vector problem*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 28 Jan 2011 06:17:02 -0500 (EST)

The *mathematics* part of the problem is that this is exactly the same as: Solve[{({x, y} - v).Conjugate[v] == 0, ({x, y} - u).Conjugate[u] == 0}, {x, y}] (If the vectors are real you do not need Conjugate). Andrzej Kozlowski On 27 Jan 2011, at 09:41, m.perrin at me.com wrote: > The Mathematica part of the problem was that Projection could be included in the Solve function. I wasn't aware that Solve could take that input > > Solve [ {Projection[ {x,y} , v ] == v, Projection[ {x,y} , u ] == u}, {x,y}] > > (* where {x,y} is unknown vector z, and u and v are the known 2D vectors; spaces are for clarity*) > > But as I posted, I solved the problem (no pun intended). > > > > On Jan 26, 2011, at 05:07 AM, Murray Eisenberg <murray at math.umass.edu> wrote: > > This doesn't at all sound like a Mathematica problem, but rather like a > mathematics problem. > > On 1/25/2011 4:19 AM, Mark Perrin wrote: >> I'm having trouble formulating a problem in Mathematica. >> >> If I have two 2D vectors, u and v that are the known projections of a third unknown 2D vector z, how do I determine z? >> >> My word problem (I think) is find the solution to: Projection[z,u] == u&& Projection [z,v] == v >> >> Any help in starting this problem would be much appreciated. >> >> >