Consider the solution set S of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$ . Calculate the...












0












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Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.






linear-algebra







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  • $begingroup$
    Welcome to Maths SX! There exists a general formula for that.
    $endgroup$
    – Bernard
    Jan 14 at 0:03










  • $begingroup$
    I cannot find this formula anywhere. Do you know this formula?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:09
















0












$begingroup$


Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Maths SX! There exists a general formula for that.
    $endgroup$
    – Bernard
    Jan 14 at 0:03










  • $begingroup$
    I cannot find this formula anywhere. Do you know this formula?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:09














0












0








0


1



$begingroup$


Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.






linear-algebra







share|cite|improve this question











$endgroup$




Question: Consider the solution set $S$ of the linear equation $x_1 + 2x_2 + x_3 = 1$ in $Bbb R^3$. Calculate the distance of the point $(1, 1, 1)$ from $S$.
I thought at first that you can make a matrix of the linear equation of set $S$ and then calculate the outcomes of $x_1, x_2$ and $x_3$. When this is done you can calculate the distance between the two points with the square root of the sum of the squares of the differences between corresponding coordinates.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 0:00









Bernard

124k742117




124k742117










asked Jan 13 at 23:42









hessel roodenburghessel roodenburg

31




31












  • $begingroup$
    Welcome to Maths SX! There exists a general formula for that.
    $endgroup$
    – Bernard
    Jan 14 at 0:03










  • $begingroup$
    I cannot find this formula anywhere. Do you know this formula?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:09


















  • $begingroup$
    Welcome to Maths SX! There exists a general formula for that.
    $endgroup$
    – Bernard
    Jan 14 at 0:03










  • $begingroup$
    I cannot find this formula anywhere. Do you know this formula?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:09
















$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03




$begingroup$
Welcome to Maths SX! There exists a general formula for that.
$endgroup$
– Bernard
Jan 14 at 0:03












$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09




$begingroup$
I cannot find this formula anywhere. Do you know this formula?
$endgroup$
– hessel roodenburg
Jan 14 at 0:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane



In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:20








  • 1




    $begingroup$
    The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
    $endgroup$
    – user289143
    Jan 14 at 0:23



















0












$begingroup$

Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:



Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
$$M=M_0+t,vec n.$$



The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
$$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$



Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
$$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane



    In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
      $endgroup$
      – hessel roodenburg
      Jan 14 at 0:20








    • 1




      $begingroup$
      The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
      $endgroup$
      – user289143
      Jan 14 at 0:23
















    1












    $begingroup$

    It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane



    In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
      $endgroup$
      – hessel roodenburg
      Jan 14 at 0:20








    • 1




      $begingroup$
      The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
      $endgroup$
      – user289143
      Jan 14 at 0:23














    1












    1








    1





    $begingroup$

    It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane



    In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$






    share|cite|improve this answer











    $endgroup$



    It's just how to calculate the distance from a point to an hyperplane: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane



    In your case: let $underline{x}=(x,y,z)=(1, 1, 1)$ and $S: ax_1+bx_2+cx_3=d$. Now $$d(x,S)=frac{|ax+by+cz-d|}{sqrt{a^2+b^2+c^2}}=frac{|1*1+2*1+1*1-1|}{sqrt{1^2+2^2+1^2}}=frac{3}{sqrt{6}}=frac{sqrt{6}}{2}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 14 at 0:19









    Bernard

    124k742117




    124k742117










    answered Jan 14 at 0:10









    user289143user289143

    1,069313




    1,069313












    • $begingroup$
      But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
      $endgroup$
      – hessel roodenburg
      Jan 14 at 0:20








    • 1




      $begingroup$
      The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
      $endgroup$
      – user289143
      Jan 14 at 0:23


















    • $begingroup$
      But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
      $endgroup$
      – hessel roodenburg
      Jan 14 at 0:20








    • 1




      $begingroup$
      The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
      $endgroup$
      – user289143
      Jan 14 at 0:23
















    $begingroup$
    But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:20






    $begingroup$
    But how is it possible that you have 3 free variables in the set S and only one final answer? I mean in order to get 1 you can have multiple sets of S right?
    $endgroup$
    – hessel roodenburg
    Jan 14 at 0:20






    1




    1




    $begingroup$
    The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
    $endgroup$
    – user289143
    Jan 14 at 0:23




    $begingroup$
    The set $S$ represents a plane in $mathbb{R}^3$. The distance from a point to a plane is defined as the distance from the closest point in the plane to our point
    $endgroup$
    – user289143
    Jan 14 at 0:23











    0












    $begingroup$

    Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:



    Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
    $$M=M_0+t,vec n.$$



    The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
    $$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
    so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$



    Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
    $$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:



      Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
      $$M=M_0+t,vec n.$$



      The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
      $$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
      so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$



      Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
      $$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:



        Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
        $$M=M_0+t,vec n.$$



        The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
        $$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
        so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$



        Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
        $$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$






        share|cite|improve this answer









        $endgroup$



        Here is an easy way to obtain the general formula for the distance of a point $M_0=(x_0,y_0,z_0)$ to the plane $Pi$ with equation $;ax+by+cz=d$:



        Remember the vector $vec n=(a,b,c)$ is normal to the plane $Pi$. If $H$ is the orthogonal projection of $M_0$ onto $Pi$, we'll use the parametric equation of the perpendicular line to $Pi$, $(M_0H)$: it is (in vector notation):
        $$M=M_0+t,vec n.$$



        The value of $t$ for $H$ is characterised by the fact that $H$ satisfies the vector equation of $Pi$:
        $$vec ncdot overrightarrow{OH}=vec ncdot overrightarrow{OM}_0+ t ,vec ncdotvec n=ax_0+by_0+cz_0+t|vec n|^2=d,$$
        so that $;t=dfrac{d-ax_0-by_0-cz_0}{|vec n|^2}.$



        Now the sought distance is $;|overrightarrow{M_0H}|=|t,vec n|=|t||n|$, and eventually
        $$d(M_0,Pi)=frac{|d-ax_0-by_0-cz_0|}{|vec n|^2},|vec n|=frac{|ax_0+by_0+cz_0-d|}{|vec n|}. $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 0:45









        BernardBernard

        124k742117




        124k742117






























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