How to get vector a from known inner product?

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Assuming that a.b=z,where b=b1,b2,b3 and z=x b1+y b2+z b3, obviously a=x,y,z, how to realize it by mathematica?










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up vote
3
down vote

favorite












Assuming that a.b=z,where b=b1,b2,b3 and z=x b1+y b2+z b3, obviously a=x,y,z, how to realize it by mathematica?










share|improve this question























  • dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
    – kglr
    Sep 6 at 10:16











  • ... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – kglr
    Sep 6 at 10:19












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Assuming that a.b=z,where b=b1,b2,b3 and z=x b1+y b2+z b3, obviously a=x,y,z, how to realize it by mathematica?










share|improve this question















Assuming that a.b=z,where b=b1,b2,b3 and z=x b1+y b2+z b3, obviously a=x,y,z, how to realize it by mathematica?







vector products coefficients






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edited Sep 6 at 9:18









kglr

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asked Sep 6 at 8:08









dnvin

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  • dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
    – kglr
    Sep 6 at 10:16











  • ... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – kglr
    Sep 6 at 10:19
















  • dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
    – kglr
    Sep 6 at 10:16











  • ... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    – kglr
    Sep 6 at 10:19















dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
– kglr
Sep 6 at 10:16





dnvin, welcome to mma.se. Our standard welcome message: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) If you haven't already done so please take the tour! ...
– kglr
Sep 6 at 10:16













... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– kglr
Sep 6 at 10:19




... 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
– kglr
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2 Answers
2






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up vote
4
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accepted










If a.b1,b2,b3 == x b1+y b2+z b3 for all b1,b2,b3:



Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]] 



a3 == z && a2 == y && a1 == x




or



SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3] 



a1 -> x, a2 -> y, a3 -> z







share|improve this answer




















  • Thanks a lot! This is what I want.
    – dnvin
    Sep 6 at 10:09











  • @dnvin, my pleasure. Welcome mma.se.
    – kglr
    Sep 6 at 10:13

















up vote
3
down vote













If I understand correctly, you have a single linear equation with 3 unknowns.



It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    If a.b1,b2,b3 == x b1+y b2+z b3 for all b1,b2,b3:



    Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]] 



    a3 == z && a2 == y && a1 == x




    or



    SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3] 



    a1 -> x, a2 -> y, a3 -> z







    share|improve this answer




















    • Thanks a lot! This is what I want.
      – dnvin
      Sep 6 at 10:09











    • @dnvin, my pleasure. Welcome mma.se.
      – kglr
      Sep 6 at 10:13














    up vote
    4
    down vote



    accepted










    If a.b1,b2,b3 == x b1+y b2+z b3 for all b1,b2,b3:



    Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]] 



    a3 == z && a2 == y && a1 == x




    or



    SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3] 



    a1 -> x, a2 -> y, a3 -> z







    share|improve this answer




















    • Thanks a lot! This is what I want.
      – dnvin
      Sep 6 at 10:09











    • @dnvin, my pleasure. Welcome mma.se.
      – kglr
      Sep 6 at 10:13












    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    If a.b1,b2,b3 == x b1+y b2+z b3 for all b1,b2,b3:



    Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]] 



    a3 == z && a2 == y && a1 == x




    or



    SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3] 



    a1 -> x, a2 -> y, a3 -> z







    share|improve this answer












    If a.b1,b2,b3 == x b1+y b2+z b3 for all b1,b2,b3:



    Reduce[ForAll[b1, b2, b3, Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3]] 



    a3 == z && a2 == y && a1 == x




    or



    SolveAlways[ Dot[a1, a2, a3, b1, b2, b3] == x b1 + y b2 + z b3, b1, b2, b3] 



    a1 -> x, a2 -> y, a3 -> z








    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Sep 6 at 9:19









    kglr

    161k8185384




    161k8185384











    • Thanks a lot! This is what I want.
      – dnvin
      Sep 6 at 10:09











    • @dnvin, my pleasure. Welcome mma.se.
      – kglr
      Sep 6 at 10:13
















    • Thanks a lot! This is what I want.
      – dnvin
      Sep 6 at 10:09











    • @dnvin, my pleasure. Welcome mma.se.
      – kglr
      Sep 6 at 10:13















    Thanks a lot! This is what I want.
    – dnvin
    Sep 6 at 10:09





    Thanks a lot! This is what I want.
    – dnvin
    Sep 6 at 10:09













    @dnvin, my pleasure. Welcome mma.se.
    – kglr
    Sep 6 at 10:13




    @dnvin, my pleasure. Welcome mma.se.
    – kglr
    Sep 6 at 10:13










    up vote
    3
    down vote













    If I understand correctly, you have a single linear equation with 3 unknowns.



    It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.






    share|improve this answer
























      up vote
      3
      down vote













      If I understand correctly, you have a single linear equation with 3 unknowns.



      It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.






      share|improve this answer






















        up vote
        3
        down vote










        up vote
        3
        down vote









        If I understand correctly, you have a single linear equation with 3 unknowns.



        It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.






        share|improve this answer












        If I understand correctly, you have a single linear equation with 3 unknowns.



        It's not possible to back out the 3 unknown values. You have an equation for a 3D hyperplane with an infinite number of solutions.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Sep 6 at 8:16









        GerardF123

        986




        986



























             

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