General method to find the perpendicular distance between a plane and a point.












2












$begingroup$


Here's the question I'm puzzling over:



$textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$



I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:



$text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$



This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?






vector-spaces vectors 3d







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Here's the question I'm puzzling over:



    $textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$



    I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:



    $text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$



    This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?






    vector-spaces vectors 3d







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Here's the question I'm puzzling over:



      $textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$



      I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:



      $text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$



      This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?






      vector-spaces vectors 3d







      share|cite|improve this question









      $endgroup$




      Here's the question I'm puzzling over:



      $textbf{Find the perpendicular distance of the point } (p, q, r) textbf{ from the plane } \ax + by + cz = d.$



      I tried bringing in the idea of a dot product and attempted to get going with solving the problem, but I'm heading nowhere. That is:



      $text{The direction vector of the normal of the plane } = (atextbf{i}+btextbf{j}+ctextbf{k}) text{, where } \textbf{i}, textbf{ j},textbf{ k } text{ are unit vectors.}$



      This dotted with the direction vector of the point (position vector, precisely) should equal 0. Am I right? And will this method work even?






      vector-spaces vectors 3d




      vector-spaces vectors 3d






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 14 '18 at 4:56









      RamanaRamana

      17210




      17210






















          2 Answers
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          $begingroup$

          Let
          $$bbox {
          vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
          vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
          vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
          where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
          $$bbox{ vec{n} cdot vec{w} = d }$$



          The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
          $$bbox{ l = vec{n} cdot vec{v} - d }$$
          because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.



          To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
          $$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$



          For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.






            share|cite|improve this answer









            $endgroup$














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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

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              active

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              2












              $begingroup$

              Let
              $$bbox {
              vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
              vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
              vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
              where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
              $$bbox{ vec{n} cdot vec{w} = d }$$



              The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
              $$bbox{ l = vec{n} cdot vec{v} - d }$$
              because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.



              To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
              $$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$



              For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Let
                $$bbox {
                vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
                vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
                vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
                where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
                $$bbox{ vec{n} cdot vec{w} = d }$$



                The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
                $$bbox{ l = vec{n} cdot vec{v} - d }$$
                because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.



                To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
                $$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$



                For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let
                  $$bbox {
                  vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
                  vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
                  vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
                  where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
                  $$bbox{ vec{n} cdot vec{w} = d }$$



                  The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
                  $$bbox{ l = vec{n} cdot vec{v} - d }$$
                  because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.



                  To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
                  $$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$



                  For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.






                  share|cite|improve this answer









                  $endgroup$



                  Let
                  $$bbox {
                  vec{n} = ( a , b , c ) = left [ begin{matrix} a \ b \ c end{matrix} right ] }, quad bbox {
                  vec{v} = ( p , q , r ) = left [ begin{matrix} p \ q \ r end{matrix} right ] }, quad bbox {
                  vec{w} = ( x , y , z ) = left [ begin{matrix} x \ y \ z end{matrix} right ] }$$
                  where $vec{n}$ is the plane normal vector, $vec{v}$ is the point those distance $L$ to the plane we want to find out, and $vec{w}$ is just an example point. The equation of the plane is then
                  $$bbox{ vec{n} cdot vec{w} = d }$$



                  The signed distance $l$ between the plane and point $vec{v}$, in units of plane normal length, is
                  $$bbox{ l = vec{n} cdot vec{v} - d }$$
                  because both $d$ (the signed distance between origin and the plane) and $l$ (the signed distance between the point and the plane) are measured in the same direction, $vec{n}$.



                  To find the actual distance $L$ between the point $vec{v}$ and the plane, we must divide the absolute value of $l$ by the length of the normal vector $vec{n}$:
                  $$bbox[#fffff7, 1em] { L = frac{leftlvert vec{n} cdot vec{v} - d rightrvert }{leftlVert vec{n} rightrVert} = frac{ leftlvert a p + b q + c r - d rightrvert}{sqrt{a^2 + b^2 + c^2}} }$$



                  For a more detailed derivation, look at the Wolfram Mathworld Point-Plane Distance article.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 13:47









                  Nominal AnimalNominal Animal

                  7,1332617




                  7,1332617























                      2












                      $begingroup$

                      If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.






                          share|cite|improve this answer









                          $endgroup$



                          If $(x,y,z)$ is the point in the plane which has shortest distance to $(p,q,r)$ then $(x-p,y-q,z-r)$ must be along the normal, i.e. $(x-p,y-q,z-r)=alpha (a,b,c)$ for some $alpha$. Thus $ x=p+alpha a$ etc. Use the equation of the plane to find $alpha$ and then compute $sqrt {(x-p)^{2}+(y-q)^{2}+(z-r)^{2}}$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 14 '18 at 5:55









                          Kavi Rama MurthyKavi Rama Murthy

                          73.3k53170




                          73.3k53170






























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