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I'm trying to solve a system of vector equations. I know this is solvable since I've done it myself by hand (I reduced it to solving the root of a quartic anyway). I'm trying to do it in Mathematica because I have another similar one I want to solve that I'm getting the same error for.

$Assumptions = (p | q | u | v | a | b) ∈ Vectors[3, Reals];
    $Assumptions = (t | k) ∈ Reals;
Reduce[{{1/2 a t^2 + u t + p == 1/2 b t^2 + v t + q}, {Norm[a] == k}}, {a, t}];

Evaluating this shows:

Solve::naqs: "{p+(a\t^2)/2+t\ u==q+(b\t^2)/2+t\ v}&&{Norm[a]==k} is not a quantified system of equations and inequalities."

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  • $\begingroup$ here, and in your answer, your second $Assumptions overwrites the first one. Sort of moot though because I'm pretty sure Reduce cant make any use of the vector assumption. $\endgroup$
    – george2079
    Jun 4, 2015 at 21:12
  • $\begingroup$ @george2079 Thanks for mentioning it... both those points answer my other question, which then that brings up the question of how do you solve a system of vector equations? $\endgroup$
    – mentics
    Jun 4, 2015 at 21:31
  • $\begingroup$ Closely related to the futility of $Assumptions here : Inequalities with assumptions and constraints. $\endgroup$
    – Jens
    Jun 4, 2015 at 21:36
  • $\begingroup$ The brute force approach is to express all of your vectors as lists with symbolic components. The results are usually pretty unwieldy however. $\endgroup$
    – george2079
    Jun 4, 2015 at 21:48

1 Answer 1

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Well, I'm not sure if the syntax I used is supposed to work for the set of equations, but if I switch to using && between the equations, it appears to work fine:

$Assumptions = (p | q | u | v | a | b) \[Element] Vectors[3, Reals]
    $Assumptions = (t | k) \[Element] Reals
Reduce[
 1/2*a*t^2 + u*t + p == 1/2 * b * t^2 + v*t + q &&
  Norm[a] == k, {a, t}, Reals]
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