Find the general equation formed by three points












0












$begingroup$


I have to find the general equation defined by these 3 points:



What I have tried so far:



Unfortunately, I was told that my solution is wrong and I am not sure how else to solve/approach this problem. Any help is appreciated.






vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you mean 'general equation' defined by 3 points? Do you perhaps mean the equation of their (affine) plane? Or perhaps the area of the triangle $ABC$?
    $endgroup$
    – Berci
    Jan 13 at 18:01












  • $begingroup$
    Can you please check whether there is a typo? Over here, you mention the $z$-coordinate of point C to be $2$ but over the included picture it's $-2$.
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:02












  • $begingroup$
    @Berci I think it's the first one. I used rough translation, not English native.
    $endgroup$
    – user634364
    Jan 13 at 18:03










  • $begingroup$
    @WingedBladesofGodric it's -2. C=[0,-2,-2].
    $endgroup$
    – user634364
    Jan 13 at 18:04










  • $begingroup$
    There's also a sign error in the calculation of the value of your second determinant (the one with $j$)
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:10
















0












$begingroup$


I have to find the general equation defined by these 3 points:



What I have tried so far:



Unfortunately, I was told that my solution is wrong and I am not sure how else to solve/approach this problem. Any help is appreciated.






vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you mean 'general equation' defined by 3 points? Do you perhaps mean the equation of their (affine) plane? Or perhaps the area of the triangle $ABC$?
    $endgroup$
    – Berci
    Jan 13 at 18:01












  • $begingroup$
    Can you please check whether there is a typo? Over here, you mention the $z$-coordinate of point C to be $2$ but over the included picture it's $-2$.
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:02












  • $begingroup$
    @Berci I think it's the first one. I used rough translation, not English native.
    $endgroup$
    – user634364
    Jan 13 at 18:03










  • $begingroup$
    @WingedBladesofGodric it's -2. C=[0,-2,-2].
    $endgroup$
    – user634364
    Jan 13 at 18:04










  • $begingroup$
    There's also a sign error in the calculation of the value of your second determinant (the one with $j$)
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:10














0












0








0





$begingroup$


I have to find the general equation defined by these 3 points:



What I have tried so far:



Unfortunately, I was told that my solution is wrong and I am not sure how else to solve/approach this problem. Any help is appreciated.






vectors







share|cite|improve this question











$endgroup$




I have to find the general equation defined by these 3 points:



What I have tried so far:



Unfortunately, I was told that my solution is wrong and I am not sure how else to solve/approach this problem. Any help is appreciated.






vectors




vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 21:41







user634364

















asked Jan 13 at 17:54









user634364user634364

44




44












  • $begingroup$
    How do you mean 'general equation' defined by 3 points? Do you perhaps mean the equation of their (affine) plane? Or perhaps the area of the triangle $ABC$?
    $endgroup$
    – Berci
    Jan 13 at 18:01












  • $begingroup$
    Can you please check whether there is a typo? Over here, you mention the $z$-coordinate of point C to be $2$ but over the included picture it's $-2$.
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:02












  • $begingroup$
    @Berci I think it's the first one. I used rough translation, not English native.
    $endgroup$
    – user634364
    Jan 13 at 18:03










  • $begingroup$
    @WingedBladesofGodric it's -2. C=[0,-2,-2].
    $endgroup$
    – user634364
    Jan 13 at 18:04










  • $begingroup$
    There's also a sign error in the calculation of the value of your second determinant (the one with $j$)
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:10


















  • $begingroup$
    How do you mean 'general equation' defined by 3 points? Do you perhaps mean the equation of their (affine) plane? Or perhaps the area of the triangle $ABC$?
    $endgroup$
    – Berci
    Jan 13 at 18:01












  • $begingroup$
    Can you please check whether there is a typo? Over here, you mention the $z$-coordinate of point C to be $2$ but over the included picture it's $-2$.
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:02












  • $begingroup$
    @Berci I think it's the first one. I used rough translation, not English native.
    $endgroup$
    – user634364
    Jan 13 at 18:03










  • $begingroup$
    @WingedBladesofGodric it's -2. C=[0,-2,-2].
    $endgroup$
    – user634364
    Jan 13 at 18:04










  • $begingroup$
    There's also a sign error in the calculation of the value of your second determinant (the one with $j$)
    $endgroup$
    – Winged Blades of Godric
    Jan 13 at 18:10
















$begingroup$
How do you mean 'general equation' defined by 3 points? Do you perhaps mean the equation of their (affine) plane? Or perhaps the area of the triangle $ABC$?
$endgroup$
– Berci
Jan 13 at 18:01






$begingroup$
How do you mean 'general equation' defined by 3 points? Do you perhaps mean the equation of their (affine) plane? Or perhaps the area of the triangle $ABC$?
$endgroup$
– Berci
Jan 13 at 18:01














$begingroup$
Can you please check whether there is a typo? Over here, you mention the $z$-coordinate of point C to be $2$ but over the included picture it's $-2$.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:02






$begingroup$
Can you please check whether there is a typo? Over here, you mention the $z$-coordinate of point C to be $2$ but over the included picture it's $-2$.
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:02














$begingroup$
@Berci I think it's the first one. I used rough translation, not English native.
$endgroup$
– user634364
Jan 13 at 18:03




$begingroup$
@Berci I think it's the first one. I used rough translation, not English native.
$endgroup$
– user634364
Jan 13 at 18:03












$begingroup$
@WingedBladesofGodric it's -2. C=[0,-2,-2].
$endgroup$
– user634364
Jan 13 at 18:04




$begingroup$
@WingedBladesofGodric it's -2. C=[0,-2,-2].
$endgroup$
– user634364
Jan 13 at 18:04












$begingroup$
There's also a sign error in the calculation of the value of your second determinant (the one with $j$)
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:10




$begingroup$
There's also a sign error in the calculation of the value of your second determinant (the one with $j$)
$endgroup$
– Winged Blades of Godric
Jan 13 at 18:10










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint: Write $$vec{x}=vec{OA}+svec{AB}+tvec{AC}$$ where $s,t$ are real numbers.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
    $endgroup$
    – user634364
    Jan 13 at 18:04












  • $begingroup$
    See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 13 at 18:09



















0












$begingroup$

Hint:



Solve



$$begin{cases}
1cdot a+1cdot b+2cdot c+d=0,
\3cdot a+1cdot b-2cdot c+d=0,
\0cdot a-2cdot b-2cdot c+d=0.
end{cases}$$



You can set one of the coefficients arbitrarily and solve for the others.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Assuming we're after the equation of the plane $ABC$, you're on the right direction.



    You already found a normal vector of that plane, though missed a sign, it should be $n=(-12, {pmb +}12, -6)$, so the equation will indeed contain $-12x+12y-6y$ on one side, but the other side, which is going to be a constant, is still to be found.

    For that, simply put the coordinates of a point on the $ABC$ plane (say, either $A$, $B$ or $C$) into this expression, and that will give you the constant.

    [If they happen not to match for any choice of $A$, $B$ or $C$, that would mean you made a mistake in your previous calculations.]



    So, now it's
    $$-12x+12y-6y = -12cdot{bf1}+12cdot{bf1}-6cdot{bf2} = -12$$
    or, equivalently,
    $$2x-2y+z=2$$
    You can verify that each of $A,B,C$ indeed satisfies it.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
      $endgroup$
      – Winged Blades of Godric
      Jan 13 at 18:11












    • $begingroup$
      Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
      $endgroup$
      – Berci
      Jan 13 at 18:16










    • $begingroup$
      The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
      $endgroup$
      – user634364
      Jan 13 at 18:20












    • $begingroup$
      We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
      $endgroup$
      – Berci
      Jan 13 at 18:22










    • $begingroup$
      Ohh, that makes more sense, thanks a lot!
      $endgroup$
      – user634364
      Jan 13 at 18:25












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Hint: Write $$vec{x}=vec{OA}+svec{AB}+tvec{AC}$$ where $s,t$ are real numbers.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
      $endgroup$
      – user634364
      Jan 13 at 18:04












    • $begingroup$
      See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 13 at 18:09
















    0












    $begingroup$

    Hint: Write $$vec{x}=vec{OA}+svec{AB}+tvec{AC}$$ where $s,t$ are real numbers.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
      $endgroup$
      – user634364
      Jan 13 at 18:04












    • $begingroup$
      See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 13 at 18:09














    0












    0








    0





    $begingroup$

    Hint: Write $$vec{x}=vec{OA}+svec{AB}+tvec{AC}$$ where $s,t$ are real numbers.






    share|cite|improve this answer









    $endgroup$



    Hint: Write $$vec{x}=vec{OA}+svec{AB}+tvec{AC}$$ where $s,t$ are real numbers.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 13 at 17:59









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    79.1k42867




    79.1k42867












    • $begingroup$
      Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
      $endgroup$
      – user634364
      Jan 13 at 18:04












    • $begingroup$
      See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 13 at 18:09


















    • $begingroup$
      Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
      $endgroup$
      – user634364
      Jan 13 at 18:04












    • $begingroup$
      See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 13 at 18:09
















    $begingroup$
    Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
    $endgroup$
    – user634364
    Jan 13 at 18:04






    $begingroup$
    Thanks for the hint! But I'm not really sure how to use this. What is OA and where do I get the s and t from? I'd appreciate a more detailed explanation as I'm not very advanced.
    $endgroup$
    – user634364
    Jan 13 at 18:04














    $begingroup$
    See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 13 at 18:09




    $begingroup$
    See here tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 13 at 18:09











    0












    $begingroup$

    Hint:



    Solve



    $$begin{cases}
    1cdot a+1cdot b+2cdot c+d=0,
    \3cdot a+1cdot b-2cdot c+d=0,
    \0cdot a-2cdot b-2cdot c+d=0.
    end{cases}$$



    You can set one of the coefficients arbitrarily and solve for the others.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint:



      Solve



      $$begin{cases}
      1cdot a+1cdot b+2cdot c+d=0,
      \3cdot a+1cdot b-2cdot c+d=0,
      \0cdot a-2cdot b-2cdot c+d=0.
      end{cases}$$



      You can set one of the coefficients arbitrarily and solve for the others.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:



        Solve



        $$begin{cases}
        1cdot a+1cdot b+2cdot c+d=0,
        \3cdot a+1cdot b-2cdot c+d=0,
        \0cdot a-2cdot b-2cdot c+d=0.
        end{cases}$$



        You can set one of the coefficients arbitrarily and solve for the others.






        share|cite|improve this answer









        $endgroup$



        Hint:



        Solve



        $$begin{cases}
        1cdot a+1cdot b+2cdot c+d=0,
        \3cdot a+1cdot b-2cdot c+d=0,
        \0cdot a-2cdot b-2cdot c+d=0.
        end{cases}$$



        You can set one of the coefficients arbitrarily and solve for the others.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 18:12









        Yves DaoustYves Daoust

        133k676232




        133k676232























            0












            $begingroup$

            Assuming we're after the equation of the plane $ABC$, you're on the right direction.



            You already found a normal vector of that plane, though missed a sign, it should be $n=(-12, {pmb +}12, -6)$, so the equation will indeed contain $-12x+12y-6y$ on one side, but the other side, which is going to be a constant, is still to be found.

            For that, simply put the coordinates of a point on the $ABC$ plane (say, either $A$, $B$ or $C$) into this expression, and that will give you the constant.

            [If they happen not to match for any choice of $A$, $B$ or $C$, that would mean you made a mistake in your previous calculations.]



            So, now it's
            $$-12x+12y-6y = -12cdot{bf1}+12cdot{bf1}-6cdot{bf2} = -12$$
            or, equivalently,
            $$2x-2y+z=2$$
            You can verify that each of $A,B,C$ indeed satisfies it.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
              $endgroup$
              – Winged Blades of Godric
              Jan 13 at 18:11












            • $begingroup$
              Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
              $endgroup$
              – Berci
              Jan 13 at 18:16










            • $begingroup$
              The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
              $endgroup$
              – user634364
              Jan 13 at 18:20












            • $begingroup$
              We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
              $endgroup$
              – Berci
              Jan 13 at 18:22










            • $begingroup$
              Ohh, that makes more sense, thanks a lot!
              $endgroup$
              – user634364
              Jan 13 at 18:25
















            0












            $begingroup$

            Assuming we're after the equation of the plane $ABC$, you're on the right direction.



            You already found a normal vector of that plane, though missed a sign, it should be $n=(-12, {pmb +}12, -6)$, so the equation will indeed contain $-12x+12y-6y$ on one side, but the other side, which is going to be a constant, is still to be found.

            For that, simply put the coordinates of a point on the $ABC$ plane (say, either $A$, $B$ or $C$) into this expression, and that will give you the constant.

            [If they happen not to match for any choice of $A$, $B$ or $C$, that would mean you made a mistake in your previous calculations.]



            So, now it's
            $$-12x+12y-6y = -12cdot{bf1}+12cdot{bf1}-6cdot{bf2} = -12$$
            or, equivalently,
            $$2x-2y+z=2$$
            You can verify that each of $A,B,C$ indeed satisfies it.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
              $endgroup$
              – Winged Blades of Godric
              Jan 13 at 18:11












            • $begingroup$
              Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
              $endgroup$
              – Berci
              Jan 13 at 18:16










            • $begingroup$
              The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
              $endgroup$
              – user634364
              Jan 13 at 18:20












            • $begingroup$
              We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
              $endgroup$
              – Berci
              Jan 13 at 18:22










            • $begingroup$
              Ohh, that makes more sense, thanks a lot!
              $endgroup$
              – user634364
              Jan 13 at 18:25














            0












            0








            0





            $begingroup$

            Assuming we're after the equation of the plane $ABC$, you're on the right direction.



            You already found a normal vector of that plane, though missed a sign, it should be $n=(-12, {pmb +}12, -6)$, so the equation will indeed contain $-12x+12y-6y$ on one side, but the other side, which is going to be a constant, is still to be found.

            For that, simply put the coordinates of a point on the $ABC$ plane (say, either $A$, $B$ or $C$) into this expression, and that will give you the constant.

            [If they happen not to match for any choice of $A$, $B$ or $C$, that would mean you made a mistake in your previous calculations.]



            So, now it's
            $$-12x+12y-6y = -12cdot{bf1}+12cdot{bf1}-6cdot{bf2} = -12$$
            or, equivalently,
            $$2x-2y+z=2$$
            You can verify that each of $A,B,C$ indeed satisfies it.






            share|cite|improve this answer











            $endgroup$



            Assuming we're after the equation of the plane $ABC$, you're on the right direction.



            You already found a normal vector of that plane, though missed a sign, it should be $n=(-12, {pmb +}12, -6)$, so the equation will indeed contain $-12x+12y-6y$ on one side, but the other side, which is going to be a constant, is still to be found.

            For that, simply put the coordinates of a point on the $ABC$ plane (say, either $A$, $B$ or $C$) into this expression, and that will give you the constant.

            [If they happen not to match for any choice of $A$, $B$ or $C$, that would mean you made a mistake in your previous calculations.]



            So, now it's
            $$-12x+12y-6y = -12cdot{bf1}+12cdot{bf1}-6cdot{bf2} = -12$$
            or, equivalently,
            $$2x-2y+z=2$$
            You can verify that each of $A,B,C$ indeed satisfies it.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 13 at 18:21

























            answered Jan 13 at 18:08









            BerciBerci

            62k23776




            62k23776








            • 1




              $begingroup$
              Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
              $endgroup$
              – Winged Blades of Godric
              Jan 13 at 18:11












            • $begingroup$
              Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
              $endgroup$
              – Berci
              Jan 13 at 18:16










            • $begingroup$
              The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
              $endgroup$
              – user634364
              Jan 13 at 18:20












            • $begingroup$
              We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
              $endgroup$
              – Berci
              Jan 13 at 18:22










            • $begingroup$
              Ohh, that makes more sense, thanks a lot!
              $endgroup$
              – user634364
              Jan 13 at 18:25














            • 1




              $begingroup$
              Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
              $endgroup$
              – Winged Blades of Godric
              Jan 13 at 18:11












            • $begingroup$
              Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
              $endgroup$
              – Berci
              Jan 13 at 18:16










            • $begingroup$
              The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
              $endgroup$
              – user634364
              Jan 13 at 18:20












            • $begingroup$
              We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
              $endgroup$
              – Berci
              Jan 13 at 18:22










            • $begingroup$
              Ohh, that makes more sense, thanks a lot!
              $endgroup$
              – user634364
              Jan 13 at 18:25








            1




            1




            $begingroup$
            Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
            $endgroup$
            – Winged Blades of Godric
            Jan 13 at 18:11






            $begingroup$
            Normal vector ought be $(-12, 12, -6)$; assuming that I've not gone crazy.
            $endgroup$
            – Winged Blades of Godric
            Jan 13 at 18:11














            $begingroup$
            Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
            $endgroup$
            – Berci
            Jan 13 at 18:16




            $begingroup$
            Yes, that makes sense. I actually didn't calculate it, relied on the posted image of the attempted solution..
            $endgroup$
            – Berci
            Jan 13 at 18:16












            $begingroup$
            The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
            $endgroup$
            – user634364
            Jan 13 at 18:20






            $begingroup$
            The part I don't understand is how to find the constant. Could you explain in a better way? With an example if possible (I don't understand all English terms). Thanks!
            $endgroup$
            – user634364
            Jan 13 at 18:20














            $begingroup$
            We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
            $endgroup$
            – Berci
            Jan 13 at 18:22




            $begingroup$
            We're looking for an equation of the form $2x-2y+z=c$, and $A$ must satisfy it, so just put in its coordinates to find $c$. If the calculations are correct, you must get the same $c$ if you put $B$'s or $C$'s coordinates.
            $endgroup$
            – Berci
            Jan 13 at 18:22












            $begingroup$
            Ohh, that makes more sense, thanks a lot!
            $endgroup$
            – user634364
            Jan 13 at 18:25




            $begingroup$
            Ohh, that makes more sense, thanks a lot!
            $endgroup$
            – user634364
            Jan 13 at 18:25


















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