perpendicular distance between midpoint and a line

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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

























    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























      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 4 at 11:41









      Bob Ueland

      1243




      1243




















          1 Answer
          1






          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










          Your Answer




          StackExchange.ifUsing("editor", function ()
          return StackExchange.using("mathjaxEditing", function ()
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          );
          );
          , "mathjax-editing");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904940%2fperpendicular-distance-between-midpoint-and-a-line%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          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

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904940%2fperpendicular-distance-between-midpoint-and-a-line%23new-answer', 'question_page');

          );

          Post as a guest
































































          -

          這個網誌中的熱門文章

          Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

          圖片
          Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
          閱讀完整內容

          Why am i infinitely getting the same tweet with the Twitter Search API?

          圖片
          up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
          閱讀完整內容

          Carbon dioxide

          圖片
          "CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
          閱讀完整內容