How to determine the equation of plane through origin?












0












$begingroup$


How to determine the equation of plane which passes through the origin and contains the line x = 2+3t , y = 1-4t, z= 6-t.



I set t = 1 and t = 0 to find the another point, so I can get (0,0,0), (5,-3,5) and (2,1,6)



then do the next step



Is it correct ?






linear-algebra vectors plane-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Both $(-5,-3,5)-(0,0,0)$ and $(2,1,6)-(0,0,0)$ are parallel to the plane. The vector product of them is a normal vector to the plane.
    $endgroup$
    – CY Aries
    May 23 '17 at 16:52
















0












$begingroup$


How to determine the equation of plane which passes through the origin and contains the line x = 2+3t , y = 1-4t, z= 6-t.



I set t = 1 and t = 0 to find the another point, so I can get (0,0,0), (5,-3,5) and (2,1,6)



then do the next step



Is it correct ?






linear-algebra vectors plane-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Both $(-5,-3,5)-(0,0,0)$ and $(2,1,6)-(0,0,0)$ are parallel to the plane. The vector product of them is a normal vector to the plane.
    $endgroup$
    – CY Aries
    May 23 '17 at 16:52














0












0








0





$begingroup$


How to determine the equation of plane which passes through the origin and contains the line x = 2+3t , y = 1-4t, z= 6-t.



I set t = 1 and t = 0 to find the another point, so I can get (0,0,0), (5,-3,5) and (2,1,6)



then do the next step



Is it correct ?






linear-algebra vectors plane-geometry







share|cite|improve this question









$endgroup$




How to determine the equation of plane which passes through the origin and contains the line x = 2+3t , y = 1-4t, z= 6-t.



I set t = 1 and t = 0 to find the another point, so I can get (0,0,0), (5,-3,5) and (2,1,6)



then do the next step



Is it correct ?






linear-algebra vectors plane-geometry




linear-algebra vectors plane-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 23 '17 at 16:42









Mohd AzimuddinMohd Azimuddin

164




164












  • $begingroup$
    Both $(-5,-3,5)-(0,0,0)$ and $(2,1,6)-(0,0,0)$ are parallel to the plane. The vector product of them is a normal vector to the plane.
    $endgroup$
    – CY Aries
    May 23 '17 at 16:52


















  • $begingroup$
    Both $(-5,-3,5)-(0,0,0)$ and $(2,1,6)-(0,0,0)$ are parallel to the plane. The vector product of them is a normal vector to the plane.
    $endgroup$
    – CY Aries
    May 23 '17 at 16:52
















$begingroup$
Both $(-5,-3,5)-(0,0,0)$ and $(2,1,6)-(0,0,0)$ are parallel to the plane. The vector product of them is a normal vector to the plane.
$endgroup$
– CY Aries
May 23 '17 at 16:52




$begingroup$
Both $(-5,-3,5)-(0,0,0)$ and $(2,1,6)-(0,0,0)$ are parallel to the plane. The vector product of them is a normal vector to the plane.
$endgroup$
– CY Aries
May 23 '17 at 16:52










2 Answers
2






active

oldest

votes


















0












$begingroup$

The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So



$$(3,-4,-1)times(2,1,6)=(23,20,-11)$$



is normal to the plane.



The plane is $23x+20y-11z=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It contains the vectors you mention in the first line
    $endgroup$
    – Bernard
    May 23 '17 at 17:07










  • $begingroup$
    You're welcome! I'll remove my comments in a moment, since they're useless now.
    $endgroup$
    – Bernard
    May 23 '17 at 17:11










  • $begingroup$
    how you get (3, -4, -1) ?
    $endgroup$
    – Mohd Azimuddin
    May 23 '17 at 17:17










  • $begingroup$
    The direction vector of the line.
    $endgroup$
    – CY Aries
    May 23 '17 at 17:19



















0












$begingroup$

The directing vector of the line in the plane is $vec v(3,-4,1)$



A point on the line is $A(2,1,6)$



Now get the vector $vec{OA}(2,1,6)$



Do the vector product $vec v ×vec{OA}$ and you will get a normal vector to the plane which is $vec n_P=vec v ×vec{OA}= (-25,-16,11)$



The equation of the plane will then be
$$-25x-16y+11z+r=0$$
Substitute the coordinates of $A$ in this equation and you'll get that $r=0$



Finally, the equation of the plane (P) is:
$$-25x-16y+11z=0$$
Or equally,
$$25x+16y-11z=0$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So



    $$(3,-4,-1)times(2,1,6)=(23,20,-11)$$



    is normal to the plane.



    The plane is $23x+20y-11z=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It contains the vectors you mention in the first line
      $endgroup$
      – Bernard
      May 23 '17 at 17:07










    • $begingroup$
      You're welcome! I'll remove my comments in a moment, since they're useless now.
      $endgroup$
      – Bernard
      May 23 '17 at 17:11










    • $begingroup$
      how you get (3, -4, -1) ?
      $endgroup$
      – Mohd Azimuddin
      May 23 '17 at 17:17










    • $begingroup$
      The direction vector of the line.
      $endgroup$
      – CY Aries
      May 23 '17 at 17:19
















    0












    $begingroup$

    The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So



    $$(3,-4,-1)times(2,1,6)=(23,20,-11)$$



    is normal to the plane.



    The plane is $23x+20y-11z=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It contains the vectors you mention in the first line
      $endgroup$
      – Bernard
      May 23 '17 at 17:07










    • $begingroup$
      You're welcome! I'll remove my comments in a moment, since they're useless now.
      $endgroup$
      – Bernard
      May 23 '17 at 17:11










    • $begingroup$
      how you get (3, -4, -1) ?
      $endgroup$
      – Mohd Azimuddin
      May 23 '17 at 17:17










    • $begingroup$
      The direction vector of the line.
      $endgroup$
      – CY Aries
      May 23 '17 at 17:19














    0












    0








    0





    $begingroup$

    The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So



    $$(3,-4,-1)times(2,1,6)=(23,20,-11)$$



    is normal to the plane.



    The plane is $23x+20y-11z=0$.






    share|cite|improve this answer











    $endgroup$



    The plane is parallel to $(3,-4,-1)$ and $(2,1,6)-(0,0,0)=(2,1,6)$. So



    $$(3,-4,-1)times(2,1,6)=(23,20,-11)$$



    is normal to the plane.



    The plane is $23x+20y-11z=0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 23 '17 at 17:09

























    answered May 23 '17 at 16:51









    CY AriesCY Aries

    16.8k11744




    16.8k11744












    • $begingroup$
      It contains the vectors you mention in the first line
      $endgroup$
      – Bernard
      May 23 '17 at 17:07










    • $begingroup$
      You're welcome! I'll remove my comments in a moment, since they're useless now.
      $endgroup$
      – Bernard
      May 23 '17 at 17:11










    • $begingroup$
      how you get (3, -4, -1) ?
      $endgroup$
      – Mohd Azimuddin
      May 23 '17 at 17:17










    • $begingroup$
      The direction vector of the line.
      $endgroup$
      – CY Aries
      May 23 '17 at 17:19


















    • $begingroup$
      It contains the vectors you mention in the first line
      $endgroup$
      – Bernard
      May 23 '17 at 17:07










    • $begingroup$
      You're welcome! I'll remove my comments in a moment, since they're useless now.
      $endgroup$
      – Bernard
      May 23 '17 at 17:11










    • $begingroup$
      how you get (3, -4, -1) ?
      $endgroup$
      – Mohd Azimuddin
      May 23 '17 at 17:17










    • $begingroup$
      The direction vector of the line.
      $endgroup$
      – CY Aries
      May 23 '17 at 17:19
















    $begingroup$
    It contains the vectors you mention in the first line
    $endgroup$
    – Bernard
    May 23 '17 at 17:07




    $begingroup$
    It contains the vectors you mention in the first line
    $endgroup$
    – Bernard
    May 23 '17 at 17:07












    $begingroup$
    You're welcome! I'll remove my comments in a moment, since they're useless now.
    $endgroup$
    – Bernard
    May 23 '17 at 17:11




    $begingroup$
    You're welcome! I'll remove my comments in a moment, since they're useless now.
    $endgroup$
    – Bernard
    May 23 '17 at 17:11












    $begingroup$
    how you get (3, -4, -1) ?
    $endgroup$
    – Mohd Azimuddin
    May 23 '17 at 17:17




    $begingroup$
    how you get (3, -4, -1) ?
    $endgroup$
    – Mohd Azimuddin
    May 23 '17 at 17:17












    $begingroup$
    The direction vector of the line.
    $endgroup$
    – CY Aries
    May 23 '17 at 17:19




    $begingroup$
    The direction vector of the line.
    $endgroup$
    – CY Aries
    May 23 '17 at 17:19











    0












    $begingroup$

    The directing vector of the line in the plane is $vec v(3,-4,1)$



    A point on the line is $A(2,1,6)$



    Now get the vector $vec{OA}(2,1,6)$



    Do the vector product $vec v ×vec{OA}$ and you will get a normal vector to the plane which is $vec n_P=vec v ×vec{OA}= (-25,-16,11)$



    The equation of the plane will then be
    $$-25x-16y+11z+r=0$$
    Substitute the coordinates of $A$ in this equation and you'll get that $r=0$



    Finally, the equation of the plane (P) is:
    $$-25x-16y+11z=0$$
    Or equally,
    $$25x+16y-11z=0$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The directing vector of the line in the plane is $vec v(3,-4,1)$



      A point on the line is $A(2,1,6)$



      Now get the vector $vec{OA}(2,1,6)$



      Do the vector product $vec v ×vec{OA}$ and you will get a normal vector to the plane which is $vec n_P=vec v ×vec{OA}= (-25,-16,11)$



      The equation of the plane will then be
      $$-25x-16y+11z+r=0$$
      Substitute the coordinates of $A$ in this equation and you'll get that $r=0$



      Finally, the equation of the plane (P) is:
      $$-25x-16y+11z=0$$
      Or equally,
      $$25x+16y-11z=0$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The directing vector of the line in the plane is $vec v(3,-4,1)$



        A point on the line is $A(2,1,6)$



        Now get the vector $vec{OA}(2,1,6)$



        Do the vector product $vec v ×vec{OA}$ and you will get a normal vector to the plane which is $vec n_P=vec v ×vec{OA}= (-25,-16,11)$



        The equation of the plane will then be
        $$-25x-16y+11z+r=0$$
        Substitute the coordinates of $A$ in this equation and you'll get that $r=0$



        Finally, the equation of the plane (P) is:
        $$-25x-16y+11z=0$$
        Or equally,
        $$25x+16y-11z=0$$






        share|cite|improve this answer









        $endgroup$



        The directing vector of the line in the plane is $vec v(3,-4,1)$



        A point on the line is $A(2,1,6)$



        Now get the vector $vec{OA}(2,1,6)$



        Do the vector product $vec v ×vec{OA}$ and you will get a normal vector to the plane which is $vec n_P=vec v ×vec{OA}= (-25,-16,11)$



        The equation of the plane will then be
        $$-25x-16y+11z+r=0$$
        Substitute the coordinates of $A$ in this equation and you'll get that $r=0$



        Finally, the equation of the plane (P) is:
        $$-25x-16y+11z=0$$
        Or equally,
        $$25x+16y-11z=0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 14 '18 at 15:22









        Fareed AFFareed AF

        52612




        52612






























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