Determining a vector orthogonal to $q_1=(1,1,1)$ and $q_3=(1,1,-2)$, why I'm wrong with my calculations?












1














Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.



My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$



And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?






linear-algebra vectors orthogonality







share|cite|improve this question





























    1














    Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.



    My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$



    And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?






    linear-algebra vectors orthogonality







    share|cite|improve this question



























      1












      1








      1







      Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.



      My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$



      And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?






      linear-algebra vectors orthogonality







      share|cite|improve this question















      Consider the vectors $q_1=(1,1,1)$ and $q_3=(1,1,-2)$. I need to find a third vector $q_2$ such that ${q_1,q_2,q_3}$ is a arthogonal basis for $mathbb{R}^3$.



      My problem is the following: I did take $v=(1,0,0)$ and I did verify that ${q_1,q_3,v}$ is a basis for $mathbb{R}^3$. Then I did take $$q_2=v-langle v|q_1rangle q_1-langle v|q_3rangle q_3=(-1,-2,1)$$



      And, by Gram-Schmidt process, $q_2$ must be orthogonal to $q_1$ and $q_3$. But, as we can see, it does not happen. So, where is my mistake?






      linear-algebra vectors orthogonality




      linear-algebra vectors orthogonality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 18:50









      quid♦

      36.9k95093




      36.9k95093










      asked Nov 22 at 16:51









      Gödel

      1,393319




      1,393319






















          3 Answers
          3






          active

          oldest

          votes


















          2














          HINT



          Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by



          $$q_2=q_1times q_3$$



          As an alternative by GS we have



          $$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$






          share|cite|improve this answer























          • I don't know why I don't use cross product before, It is a simple way to solve the problem.
            – Gödel
            Nov 22 at 17:10










          • @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
            – gimusi
            Nov 22 at 17:18



















          4














          None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
          $$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.






          share|cite|improve this answer































            2














            $$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$






            share|cite|improve this answer





















              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',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              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%2f3009351%2fdetermining-a-vector-orthogonal-to-q-1-1-1-1-and-q-3-1-1-2-why-im-wro%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              HINT



              Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by



              $$q_2=q_1times q_3$$



              As an alternative by GS we have



              $$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$






              share|cite|improve this answer























              • I don't know why I don't use cross product before, It is a simple way to solve the problem.
                – Gödel
                Nov 22 at 17:10










              • @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
                – gimusi
                Nov 22 at 17:18
















              2














              HINT



              Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by



              $$q_2=q_1times q_3$$



              As an alternative by GS we have



              $$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$






              share|cite|improve this answer























              • I don't know why I don't use cross product before, It is a simple way to solve the problem.
                – Gödel
                Nov 22 at 17:10










              • @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
                – gimusi
                Nov 22 at 17:18














              2












              2








              2






              HINT



              Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by



              $$q_2=q_1times q_3$$



              As an alternative by GS we have



              $$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$






              share|cite|improve this answer














              HINT



              Since $q_1$ and $q_3$ are orthogonal it suffices to find $q_3$ by



              $$q_2=q_1times q_3$$



              As an alternative by GS we have



              $$q_2=v-langle v|hat q_1rangle hat q_1-langle v| hat q_3rangle hat q_3=(1,0,0)-frac13(1,1,1)-frac16(1,1,-2)=left(frac12,-frac12,0right)$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 22 at 17:06

























              answered Nov 22 at 16:54









              gimusi

              1




              1












              • I don't know why I don't use cross product before, It is a simple way to solve the problem.
                – Gödel
                Nov 22 at 17:10










              • @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
                – gimusi
                Nov 22 at 17:18


















              • I don't know why I don't use cross product before, It is a simple way to solve the problem.
                – Gödel
                Nov 22 at 17:10










              • @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
                – gimusi
                Nov 22 at 17:18
















              I don't know why I don't use cross product before, It is a simple way to solve the problem.
              – Gödel
              Nov 22 at 17:10




              I don't know why I don't use cross product before, It is a simple way to solve the problem.
              – Gödel
              Nov 22 at 17:10












              @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
              – gimusi
              Nov 22 at 17:18




              @Gödel Yes indeed it is the faster way in that case if we can't see that by inspection.
              – gimusi
              Nov 22 at 17:18











              4














              None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
              $$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.






              share|cite|improve this answer




























                4














                None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
                $$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.






                share|cite|improve this answer


























                  4












                  4








                  4






                  None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
                  $$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.






                  share|cite|improve this answer














                  None of $q_1$ and $q_2$ are normalized. Hence, the formula would be
                  $$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 22 at 17:03

























                  answered Nov 22 at 16:56









                  SinTan1729

                  2,462622




                  2,462622























                      2














                      $$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$






                      share|cite|improve this answer


























                        2














                        $$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$






                        share|cite|improve this answer
























                          2












                          2








                          2






                          $$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$






                          share|cite|improve this answer












                          $$q_2=v-frac{langle v|q_1rangle}{langle q_1|q_1rangle} q_1-frac{langle v|q_3rangle}{langle q_3|q_3rangle} q_3$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 at 16:54









                          Siong Thye Goh

                          98.9k1464116




                          98.9k1464116






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009351%2fdetermining-a-vector-orthogonal-to-q-1-1-1-1-and-q-3-1-1-2-why-im-wro%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              What visual should I use to simply compare current year value vs last year in Power BI desktop

                              Image
                              0 I have column SubToBind that shows current year ratio and column PY SubToBind which shows previous year ratio. How can I simply create a card that would be displaying them both in a nice way? Something like that: I cannot achieve that with regular card, I also tried Card with States by OKViz but also unable to see both values. In KPI visual - I dont like the word "Goal", it will confuse end users. Are there any cards or other visuals that would do the job? Thanks powerbi dax powerbi-embedded share | improve this question edited Jun 11 '18 at 20:11 Oleg
                              Read more

                              Alexandru Averescu

                              Image
                              Alexandre Averescu Alexandru Averescu Naissance 3 avril 1859 Babele, Ismail  (ro) près de Izmaïl (principauté de Moldavie) Décès 2 octobre 1938 (à 79 ans) [ 1 ] Bucarest Origine   Roumanie Arme cavalerie Grade Maréchal Années de service 1901-1917 Commandement chef d'état-major général 2 e  armée 3 e  armée Groupe d'armée sud armée du Danube Conflits Guerre russo-turque de 1877-1878 Deuxième guerre balkanique Première Guerre mondiale Faits d'armes Bataille de Turtucaia/Tutrakan Bataille de Mărăști Distinctions Autres fonctions 3 fois président du Conseil des ministres ministre des Affaires étrangères ministre sans portefeuille conseiller de la Couronne modifier   Alexandre Averescu (en roumain : Alexandru Averescu .mw-parser-output .prononciation>a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Loudspeaker.svg/11px-Loudspeaker.svg.png")center
                              Read more

                              Trompette piccolo

                              Image
                              Ne doit pas être confondu avec trompette de poche. Trompette piccolo en si bémol, avec des branches d'embouchure différentes pour le si bémol (la plus courte) ou la (la plus longue) Trompette piccolo La trompette piccolo est la plus petite de la famille des trompettes. Elle joue une octave plus haut que la trompette standard en si bémol. La plupart des trompettes piccolo sont conçues pour jouer dans les deux tonalités de si bémol ou la, en utilisant une branche d'embouchure différente pour chaque tonalité. Le tube de la trompette piccolo en si bémol a une longueur moitié de celle de la trompette normale en si bémol. Des trompettes piccolo en sol, fa, ut sont également fabriquées, mais elles sont rares. La trompette soprano en ré, aussi connu comme la « trompette Bach », a été inventée vers 1890 par le facteur belge Victor-Charles Mahillon pour jouer les parties hautes de la trompette dans la musique de Bach et Haendel. La trompette piccolo moderne permet aux
                              Read more