orthogonal projection of a vector onto a plane












1












$begingroup$


I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?






linear-algebra projection







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  • $begingroup$
    This plane does not pass through the origin.
    $endgroup$
    – amd
    Dec 28 '18 at 1:26
















1












$begingroup$


I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?






linear-algebra projection







share|cite|improve this question











$endgroup$












  • $begingroup$
    This plane does not pass through the origin.
    $endgroup$
    – amd
    Dec 28 '18 at 1:26














1












1








1





$begingroup$


I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?






linear-algebra projection







share|cite|improve this question











$endgroup$




I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?






linear-algebra projection




linear-algebra projection






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share|cite|improve this question













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edited Dec 28 '18 at 5:32









Andrei

12.4k21128




12.4k21128










asked Dec 28 '18 at 1:18









husein mbhusein mb

82




82












  • $begingroup$
    This plane does not pass through the origin.
    $endgroup$
    – amd
    Dec 28 '18 at 1:26


















  • $begingroup$
    This plane does not pass through the origin.
    $endgroup$
    – amd
    Dec 28 '18 at 1:26
















$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26




$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26










2 Answers
2






active

oldest

votes


















0












$begingroup$

It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      ,Thank you for your answer it was very helpfull .
      $endgroup$
      – husein mb
      Dec 28 '18 at 1:53










    • $begingroup$
      The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
      $endgroup$
      – Andrei
      Dec 28 '18 at 4:36










    • $begingroup$
      @Andrei yes, you're right, you need to pick two coplanar vectors.
      $endgroup$
      – John Doe
      Dec 28 '18 at 4:38






    • 1




      $begingroup$
      Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
      $endgroup$
      – amd
      Dec 28 '18 at 4:39












    • $begingroup$
      @amd yep, thats right, thanks
      $endgroup$
      – John Doe
      Dec 28 '18 at 4:49











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.






        share|cite|improve this answer











        $endgroup$



        It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 28 '18 at 5:57

























        answered Dec 28 '18 at 5:37









        Kavi Rama MurthyKavi Rama Murthy

        62.7k42262




        62.7k42262























            0












            $begingroup$

            You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              ,Thank you for your answer it was very helpfull .
              $endgroup$
              – husein mb
              Dec 28 '18 at 1:53










            • $begingroup$
              The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
              $endgroup$
              – Andrei
              Dec 28 '18 at 4:36










            • $begingroup$
              @Andrei yes, you're right, you need to pick two coplanar vectors.
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:38






            • 1




              $begingroup$
              Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
              $endgroup$
              – amd
              Dec 28 '18 at 4:39












            • $begingroup$
              @amd yep, thats right, thanks
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:49
















            0












            $begingroup$

            You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              ,Thank you for your answer it was very helpfull .
              $endgroup$
              – husein mb
              Dec 28 '18 at 1:53










            • $begingroup$
              The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
              $endgroup$
              – Andrei
              Dec 28 '18 at 4:36










            • $begingroup$
              @Andrei yes, you're right, you need to pick two coplanar vectors.
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:38






            • 1




              $begingroup$
              Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
              $endgroup$
              – amd
              Dec 28 '18 at 4:39












            • $begingroup$
              @amd yep, thats right, thanks
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:49














            0












            0








            0





            $begingroup$

            You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.






            share|cite|improve this answer











            $endgroup$



            You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 28 '18 at 4:48

























            answered Dec 28 '18 at 1:47









            John DoeJohn Doe

            11.2k11239




            11.2k11239












            • $begingroup$
              ,Thank you for your answer it was very helpfull .
              $endgroup$
              – husein mb
              Dec 28 '18 at 1:53










            • $begingroup$
              The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
              $endgroup$
              – Andrei
              Dec 28 '18 at 4:36










            • $begingroup$
              @Andrei yes, you're right, you need to pick two coplanar vectors.
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:38






            • 1




              $begingroup$
              Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
              $endgroup$
              – amd
              Dec 28 '18 at 4:39












            • $begingroup$
              @amd yep, thats right, thanks
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:49


















            • $begingroup$
              ,Thank you for your answer it was very helpfull .
              $endgroup$
              – husein mb
              Dec 28 '18 at 1:53










            • $begingroup$
              The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
              $endgroup$
              – Andrei
              Dec 28 '18 at 4:36










            • $begingroup$
              @Andrei yes, you're right, you need to pick two coplanar vectors.
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:38






            • 1




              $begingroup$
              Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
              $endgroup$
              – amd
              Dec 28 '18 at 4:39












            • $begingroup$
              @amd yep, thats right, thanks
              $endgroup$
              – John Doe
              Dec 28 '18 at 4:49
















            $begingroup$
            ,Thank you for your answer it was very helpfull .
            $endgroup$
            – husein mb
            Dec 28 '18 at 1:53




            $begingroup$
            ,Thank you for your answer it was very helpfull .
            $endgroup$
            – husein mb
            Dec 28 '18 at 1:53












            $begingroup$
            The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
            $endgroup$
            – Andrei
            Dec 28 '18 at 4:36




            $begingroup$
            The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
            $endgroup$
            – Andrei
            Dec 28 '18 at 4:36












            $begingroup$
            @Andrei yes, you're right, you need to pick two coplanar vectors.
            $endgroup$
            – John Doe
            Dec 28 '18 at 4:38




            $begingroup$
            @Andrei yes, you're right, you need to pick two coplanar vectors.
            $endgroup$
            – John Doe
            Dec 28 '18 at 4:38




            1




            1




            $begingroup$
            Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
            $endgroup$
            – amd
            Dec 28 '18 at 4:39






            $begingroup$
            Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
            $endgroup$
            – amd
            Dec 28 '18 at 4:39














            $begingroup$
            @amd yep, thats right, thanks
            $endgroup$
            – John Doe
            Dec 28 '18 at 4:49




            $begingroup$
            @amd yep, thats right, thanks
            $endgroup$
            – John Doe
            Dec 28 '18 at 4:49


















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