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Re: Solving Equation Set

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"Thomas D. Dean" <tomdean@speakeasy.org> writes:

> From an email about another CAS, Maxima
>
> ##
> ## In the following sequence, I'm trying to prove that the length of the
> ## cross product of two orthogonal unit vectors is =1.
> ##
> ## The normal of a vector is the length of the vector
> ## |A| = sqrt(a1^2+a2^2+a3^2)
> ## if |A| = 1 then
> ## |A|^2 = sqrt(a1^2+a2^2+a3^2)^2 = 1^2
> ## so that
> ## a1^2+a2^2+a3^2 = 1
> ##
> ## N(A)=|A|^2=a1^2+a2^2+a3^2 see above
> ## N(B)=|B|^2=b1^2+b2^2+b3^2 see above
> ## N(AxB) is the norm of AxB.
> ## A.B=a1*b1+a2*b2+a3*b3=0.
> ##
> restart;
> A:=<a1,a2,a3>;
> B:=<b1,b2,b3>;
> tmp:=LinearAlgebra[CrossProduct](A,B);
> eq1:=tmp . tmp = NAxB;  ## NAxB is the cross product
> eq2:=A . A = 1;         ## A is a unit vector
> eq3:=B . B = 1;         ## B is a unit vector
> eq4:=A . B = 0;         ## A is perpendicular ot B
>
> solve({eq1,eq2,eq3,eq4},NAxB);
>
> sol1:=lhs(eq1)+lhs(eq4)^2 = rhs(eq1)+rhs(eq4);
> sol2:=eliminate([sol1,eq2,eq3],b3);
> sol3:=eliminate(sol2[2][2],a3);
> sol4:=solve(sol3[2][2],NAxB);
>
> I can peek and poke to get a solution.  There should be an easier way.
> How do I do this?
>
> Tom Dean

simplify(tmp.tmp, [A.A=1,B.B=1,A.B=0]) assuming real;

            1

-- 
Joe Riel

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Solving Equation Set "Thomas D. Dean" <tomdean@speakeasy.org> - 2014-06-08 20:20 -0700
  Re: Solving Equation Set Joe Riel <joer@san.rr.com> - 2014-06-09 11:17 -0700
  Re: Solving Equation Set Axel Vogt <&noreply@axelvogt.de> - 2014-06-09 20:20 +0200

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