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











up vote
1
down vote

favorite












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




























    up vote
    1
    down vote

    favorite












    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


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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♦

      36.8k95093




      36.8k95093










      asked Nov 22 at 16:51









      Gödel

      1,393319




      1,393319






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          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


















          up vote
          4
          down vote













          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






























            up vote
            2
            down vote













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






            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',
              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








              up vote
              2
              down vote



              accepted










              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















              up vote
              2
              down vote



              accepted










              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













              up vote
              2
              down vote



              accepted







              up vote
              2
              down vote



              accepted






              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

              92.7k94495




              92.7k94495












              • 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










              up vote
              4
              down vote













              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



























                up vote
                4
                down vote













                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

























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  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

                  2,399622




                  2,399622






















                      up vote
                      2
                      down vote













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






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













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






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









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






                          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

                          97.7k1463116




                          97.7k1463116






























                              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

                              Ellipse (mathématiques)

                              Image
                              Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                              Read more

                              Quarter-circle Tiles

                              Image
                              up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                              Read more

                              Mont Emei

                              Image
                              Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                              Read more