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







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    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
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          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
          $$ q_2=v-{langle v|q_1rangle over langle q_1|q_1rangle} q_1-{langle v|q_3rangle over langle q_3|q_3rangle} q_3 = ({1 over 2},-{1 over 2},0)$$.






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            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





















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              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






























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