Forming an orthonormal vector when you already have two perpendicular vectors

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So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:



Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$.



Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector?



I have included a picture just to make it more clear.



Thank you in advance :)



enter image description here






linear-algebra orthonormal cross-product







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  • Yes, for two vectors in $mathbb R^3$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal.
    – Kavi Rama Murthy
    3 hours ago















up vote
4
down vote

favorite












So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:



Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$.



Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector?



I have included a picture just to make it more clear.



Thank you in advance :)



enter image description here






linear-algebra orthonormal cross-product







share|cite|improve this question























  • Yes, for two vectors in $mathbb R^3$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal.
    – Kavi Rama Murthy
    3 hours ago













up vote
4
down vote

favorite









up vote
4
down vote

favorite











So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:



Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$.



Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector?



I have included a picture just to make it more clear.



Thank you in advance :)



enter image description here






linear-algebra orthonormal cross-product







share|cite|improve this question















So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:



Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$.



Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector?



I have included a picture just to make it more clear.



Thank you in advance :)



enter image description here






linear-algebra orthonormal cross-product




linear-algebra orthonormal cross-product






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edited 2 hours ago









mechanodroid

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asked 3 hours ago









MathNoob123

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311











  • Yes, for two vectors in $mathbb R^3$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal.
    – Kavi Rama Murthy
    3 hours ago

















  • Yes, for two vectors in $mathbb R^3$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal.
    – Kavi Rama Murthy
    3 hours ago
















Yes, for two vectors in $mathbb R^3$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal.
– Kavi Rama Murthy
3 hours ago





Yes, for two vectors in $mathbb R^3$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal.
– Kavi Rama Murthy
3 hours ago











3 Answers
3






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up vote
2
down vote













Here we are using the property of cross product which is defined only for $vin mathbbR^3$.



enter image description here



The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.






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    up vote
    2
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    The cross-product $u_1times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$lVert u_1times u_2rVert=lVert u_1rVert.lVert u_2rVert.sintheta,$$where $theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).






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      Yes, you can check it by direct calculation.



      Assume that $u = beginbmatrix u_1 \ u_2 \ u_3 endbmatrix$ and $v = beginbmatrix v_1 \ v_2 \ v_3 endbmatrix$ are orthonormal vectors. Their cross product is defined as
      $$u times v = beginbmatrix u_2v_3 - u_3v_2 \ u_3v_1-u_1v_3 \ u_1v_2 - u_2v_1 endbmatrix$$



      We have



      $$langle u, u times vrangle = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1 = 0$$



      $$langle v, u times vrangle = u_2v_1v_3 - u_3v_1v_2 + u_2v_1v_2 - u_1v_2v_3 + u_1v_2v_3 - u_2v_1v_3 = 0$$



      beginalign
      |utimes v|^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1-u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \
      &= u_2^2v_3^2 - 2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 - 2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 - 2u_1u_2v_1v_2 + u_2^2v_1^2 \
      &= u_1^2(v_2^2 + v_3^2) + u_2^2(v_1^2 + v_3^2) + u_3^2(v_1^2 + v_2^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
      &= u_1^2(1-v_1^2) + u_2^2(1 -v_2^2) + u_3^2(1-v_3^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
      &= (u_1^2+u_2^2+u_3^2) - (u_1^2v_1^2 + u_2^2v_2^2+u_3^2v_3^2 + 2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3)\
      &= |u|^2 - langle u,vrangle ^2\
      &= 1
      endalign



      so $u, v, utimes v$ is an orthonormal basis for $mathbbR^3$.






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        3 Answers
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        3 Answers
        3






        active

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        active

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        active

        oldest

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        up vote
        2
        down vote













        Here we are using the property of cross product which is defined only for $vin mathbbR^3$.



        enter image description here



        The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.






        share|cite|improve this answer
























          up vote
          2
          down vote













          Here we are using the property of cross product which is defined only for $vin mathbbR^3$.



          enter image description here



          The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.






          share|cite|improve this answer






















            up vote
            2
            down vote










            up vote
            2
            down vote









            Here we are using the property of cross product which is defined only for $vin mathbbR^3$.



            enter image description here



            The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.






            share|cite|improve this answer












            Here we are using the property of cross product which is defined only for $vin mathbbR^3$.



            enter image description here



            The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            gimusi

            71.8k73888




            71.8k73888




















                up vote
                2
                down vote













                The cross-product $u_1times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$lVert u_1times u_2rVert=lVert u_1rVert.lVert u_2rVert.sintheta,$$where $theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  The cross-product $u_1times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$lVert u_1times u_2rVert=lVert u_1rVert.lVert u_2rVert.sintheta,$$where $theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    The cross-product $u_1times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$lVert u_1times u_2rVert=lVert u_1rVert.lVert u_2rVert.sintheta,$$where $theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).






                    share|cite|improve this answer












                    The cross-product $u_1times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$lVert u_1times u_2rVert=lVert u_1rVert.lVert u_2rVert.sintheta,$$where $theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    José Carlos Santos

                    121k16101185




                    121k16101185




















                        up vote
                        1
                        down vote













                        Yes, you can check it by direct calculation.



                        Assume that $u = beginbmatrix u_1 \ u_2 \ u_3 endbmatrix$ and $v = beginbmatrix v_1 \ v_2 \ v_3 endbmatrix$ are orthonormal vectors. Their cross product is defined as
                        $$u times v = beginbmatrix u_2v_3 - u_3v_2 \ u_3v_1-u_1v_3 \ u_1v_2 - u_2v_1 endbmatrix$$



                        We have



                        $$langle u, u times vrangle = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1 = 0$$



                        $$langle v, u times vrangle = u_2v_1v_3 - u_3v_1v_2 + u_2v_1v_2 - u_1v_2v_3 + u_1v_2v_3 - u_2v_1v_3 = 0$$



                        beginalign
                        |utimes v|^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1-u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \
                        &= u_2^2v_3^2 - 2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 - 2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 - 2u_1u_2v_1v_2 + u_2^2v_1^2 \
                        &= u_1^2(v_2^2 + v_3^2) + u_2^2(v_1^2 + v_3^2) + u_3^2(v_1^2 + v_2^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                        &= u_1^2(1-v_1^2) + u_2^2(1 -v_2^2) + u_3^2(1-v_3^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                        &= (u_1^2+u_2^2+u_3^2) - (u_1^2v_1^2 + u_2^2v_2^2+u_3^2v_3^2 + 2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3)\
                        &= |u|^2 - langle u,vrangle ^2\
                        &= 1
                        endalign



                        so $u, v, utimes v$ is an orthonormal basis for $mathbbR^3$.






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          Yes, you can check it by direct calculation.



                          Assume that $u = beginbmatrix u_1 \ u_2 \ u_3 endbmatrix$ and $v = beginbmatrix v_1 \ v_2 \ v_3 endbmatrix$ are orthonormal vectors. Their cross product is defined as
                          $$u times v = beginbmatrix u_2v_3 - u_3v_2 \ u_3v_1-u_1v_3 \ u_1v_2 - u_2v_1 endbmatrix$$



                          We have



                          $$langle u, u times vrangle = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1 = 0$$



                          $$langle v, u times vrangle = u_2v_1v_3 - u_3v_1v_2 + u_2v_1v_2 - u_1v_2v_3 + u_1v_2v_3 - u_2v_1v_3 = 0$$



                          beginalign
                          |utimes v|^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1-u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \
                          &= u_2^2v_3^2 - 2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 - 2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 - 2u_1u_2v_1v_2 + u_2^2v_1^2 \
                          &= u_1^2(v_2^2 + v_3^2) + u_2^2(v_1^2 + v_3^2) + u_3^2(v_1^2 + v_2^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                          &= u_1^2(1-v_1^2) + u_2^2(1 -v_2^2) + u_3^2(1-v_3^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                          &= (u_1^2+u_2^2+u_3^2) - (u_1^2v_1^2 + u_2^2v_2^2+u_3^2v_3^2 + 2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3)\
                          &= |u|^2 - langle u,vrangle ^2\
                          &= 1
                          endalign



                          so $u, v, utimes v$ is an orthonormal basis for $mathbbR^3$.






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Yes, you can check it by direct calculation.



                            Assume that $u = beginbmatrix u_1 \ u_2 \ u_3 endbmatrix$ and $v = beginbmatrix v_1 \ v_2 \ v_3 endbmatrix$ are orthonormal vectors. Their cross product is defined as
                            $$u times v = beginbmatrix u_2v_3 - u_3v_2 \ u_3v_1-u_1v_3 \ u_1v_2 - u_2v_1 endbmatrix$$



                            We have



                            $$langle u, u times vrangle = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1 = 0$$



                            $$langle v, u times vrangle = u_2v_1v_3 - u_3v_1v_2 + u_2v_1v_2 - u_1v_2v_3 + u_1v_2v_3 - u_2v_1v_3 = 0$$



                            beginalign
                            |utimes v|^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1-u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \
                            &= u_2^2v_3^2 - 2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 - 2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 - 2u_1u_2v_1v_2 + u_2^2v_1^2 \
                            &= u_1^2(v_2^2 + v_3^2) + u_2^2(v_1^2 + v_3^2) + u_3^2(v_1^2 + v_2^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                            &= u_1^2(1-v_1^2) + u_2^2(1 -v_2^2) + u_3^2(1-v_3^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                            &= (u_1^2+u_2^2+u_3^2) - (u_1^2v_1^2 + u_2^2v_2^2+u_3^2v_3^2 + 2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3)\
                            &= |u|^2 - langle u,vrangle ^2\
                            &= 1
                            endalign



                            so $u, v, utimes v$ is an orthonormal basis for $mathbbR^3$.






                            share|cite|improve this answer












                            Yes, you can check it by direct calculation.



                            Assume that $u = beginbmatrix u_1 \ u_2 \ u_3 endbmatrix$ and $v = beginbmatrix v_1 \ v_2 \ v_3 endbmatrix$ are orthonormal vectors. Their cross product is defined as
                            $$u times v = beginbmatrix u_2v_3 - u_3v_2 \ u_3v_1-u_1v_3 \ u_1v_2 - u_2v_1 endbmatrix$$



                            We have



                            $$langle u, u times vrangle = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1 = 0$$



                            $$langle v, u times vrangle = u_2v_1v_3 - u_3v_1v_2 + u_2v_1v_2 - u_1v_2v_3 + u_1v_2v_3 - u_2v_1v_3 = 0$$



                            beginalign
                            |utimes v|^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1-u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \
                            &= u_2^2v_3^2 - 2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 - 2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 - 2u_1u_2v_1v_2 + u_2^2v_1^2 \
                            &= u_1^2(v_2^2 + v_3^2) + u_2^2(v_1^2 + v_3^2) + u_3^2(v_1^2 + v_2^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                            &= u_1^2(1-v_1^2) + u_2^2(1 -v_2^2) + u_3^2(1-v_3^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \
                            &= (u_1^2+u_2^2+u_3^2) - (u_1^2v_1^2 + u_2^2v_2^2+u_3^2v_3^2 + 2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3)\
                            &= |u|^2 - langle u,vrangle ^2\
                            &= 1
                            endalign



                            so $u, v, utimes v$ is an orthonormal basis for $mathbbR^3$.







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                            answered 2 hours ago









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                                Confectionery

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                                Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of
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