perpendicular distance between midpoint and a line

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There is a well known formula for projection of a point b onto a line a. The sought point p can be expressed as p = x a, where x is a dot b divided by a dot a.
projection



I have a somewhat similar situation.
enter image description here



I have a midpoint m of a line segment starting at b and ending at c. I draw a line through m, that is perpendicular to the line segment. This line intersects the line a at some point p. Since p is on the line a it can be expressed as p = x a . So I want to find x expressed in the terms of a, b and c. The formula has to be similar (expressed in vectors) to what I have for projection. I am trying to avoid using the cartesian coordinates of type (x1,y1), because there are 3 points involved and using coordinates would give a very complicated formula.






linear-algebra projection







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    up vote
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    down vote

    favorite












    There is a well known formula for projection of a point b onto a line a. The sought point p can be expressed as p = x a, where x is a dot b divided by a dot a.
    projection



    I have a somewhat similar situation.
    enter image description here



    I have a midpoint m of a line segment starting at b and ending at c. I draw a line through m, that is perpendicular to the line segment. This line intersects the line a at some point p. Since p is on the line a it can be expressed as p = x a . So I want to find x expressed in the terms of a, b and c. The formula has to be similar (expressed in vectors) to what I have for projection. I am trying to avoid using the cartesian coordinates of type (x1,y1), because there are 3 points involved and using coordinates would give a very complicated formula.






    linear-algebra projection







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      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      There is a well known formula for projection of a point b onto a line a. The sought point p can be expressed as p = x a, where x is a dot b divided by a dot a.
      projection



      I have a somewhat similar situation.
      enter image description here



      I have a midpoint m of a line segment starting at b and ending at c. I draw a line through m, that is perpendicular to the line segment. This line intersects the line a at some point p. Since p is on the line a it can be expressed as p = x a . So I want to find x expressed in the terms of a, b and c. The formula has to be similar (expressed in vectors) to what I have for projection. I am trying to avoid using the cartesian coordinates of type (x1,y1), because there are 3 points involved and using coordinates would give a very complicated formula.






      linear-algebra projection







      share|cite|improve this question













      There is a well known formula for projection of a point b onto a line a. The sought point p can be expressed as p = x a, where x is a dot b divided by a dot a.
      projection



      I have a somewhat similar situation.
      enter image description here



      I have a midpoint m of a line segment starting at b and ending at c. I draw a line through m, that is perpendicular to the line segment. This line intersects the line a at some point p. Since p is on the line a it can be expressed as p = x a . So I want to find x expressed in the terms of a, b and c. The formula has to be similar (expressed in vectors) to what I have for projection. I am trying to avoid using the cartesian coordinates of type (x1,y1), because there are 3 points involved and using coordinates would give a very complicated formula.






      linear-algebra projection




      linear-algebra projection






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      asked Sep 4 at 11:41









      Bob Ueland

      1243




      1243




















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          enter image description here



          Step 1: Project $frac12(mathbf b + mathbf c)$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf q$.



          Step 2: Project $mathbf a$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf r$.



          Then $x = dfraclVert mathbf qrVertlVert mathbf rrVert.$






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          • What a beautiful solution!
            – Bob Ueland
            Sep 4 at 12:24










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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









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          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          enter image description here



          Step 1: Project $frac12(mathbf b + mathbf c)$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf q$.



          Step 2: Project $mathbf a$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf r$.



          Then $x = dfraclVert mathbf qrVertlVert mathbf rrVert.$






          share|cite|improve this answer




















          • What a beautiful solution!
            – Bob Ueland
            Sep 4 at 12:24














          up vote
          1
          down vote



          accepted










          enter image description here



          Step 1: Project $frac12(mathbf b + mathbf c)$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf q$.



          Step 2: Project $mathbf a$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf r$.



          Then $x = dfraclVert mathbf qrVertlVert mathbf rrVert.$






          share|cite|improve this answer




















          • What a beautiful solution!
            – Bob Ueland
            Sep 4 at 12:24












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          enter image description here



          Step 1: Project $frac12(mathbf b + mathbf c)$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf q$.



          Step 2: Project $mathbf a$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf r$.



          Then $x = dfraclVert mathbf qrVertlVert mathbf rrVert.$






          share|cite|improve this answer












          enter image description here



          Step 1: Project $frac12(mathbf b + mathbf c)$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf q$.



          Step 2: Project $mathbf a$ onto $mathbf c - mathbf b,$
          producing a vector $mathbf r$.



          Then $x = dfraclVert mathbf qrVertlVert mathbf rrVert.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 4 at 12:19









          David K

          49k340109




          49k340109











          • What a beautiful solution!
            – Bob Ueland
            Sep 4 at 12:24
















          • What a beautiful solution!
            – Bob Ueland
            Sep 4 at 12:24















          What a beautiful solution!
          – Bob Ueland
          Sep 4 at 12:24




          What a beautiful solution!
          – Bob Ueland
          Sep 4 at 12:24

















           

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