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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.
>>
>>
>


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