How do we decide the direction of vector, that is orthogonal to some other vector?












1












$begingroup$


assume that two vector given with relationship below: $$vec{n}cdot vec{u} = 0 $$ Then $vec{n}$ and $vec{u}$ are orthogonal vectors. Assume that we now have the direction of $vec{n}$



Then there are infinitely many possible direction for $vec{u}$
Coordinate System



How do we choose the vector $vec{u}$?



When we consider vector valued functions, we choose Normal vector to be directed inner side of the curve.



When we consider grad $nabla$ we know that it is perpendicular to any level curve. Then we choose $nabla$ to be directed towards outward of the curve.



How do we decide direction? What is the algorithm behind that? What am I missing? Or is it application dependent?






vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    Actually there are infinitely many possible direction, and also magnitude for $vec{u}$. The set of all these $vec{u}$ can be called as the orthogonal complement of $vec{n}$. The dimension of this set is n-1 (assuming $vec{n}$ is a $mathbb{R}^{n}$ vector).
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:29












  • $begingroup$
    A.P warned me about that issue. I have corrected it, thanks for your interest
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:33










  • $begingroup$
    Welcome XD. Concerning the grad, I think it points to the direction such that the scalar valued function increased fastest?
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:41
















1












$begingroup$


assume that two vector given with relationship below: $$vec{n}cdot vec{u} = 0 $$ Then $vec{n}$ and $vec{u}$ are orthogonal vectors. Assume that we now have the direction of $vec{n}$



Then there are infinitely many possible direction for $vec{u}$
Coordinate System



How do we choose the vector $vec{u}$?



When we consider vector valued functions, we choose Normal vector to be directed inner side of the curve.



When we consider grad $nabla$ we know that it is perpendicular to any level curve. Then we choose $nabla$ to be directed towards outward of the curve.



How do we decide direction? What is the algorithm behind that? What am I missing? Or is it application dependent?






vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    Actually there are infinitely many possible direction, and also magnitude for $vec{u}$. The set of all these $vec{u}$ can be called as the orthogonal complement of $vec{n}$. The dimension of this set is n-1 (assuming $vec{n}$ is a $mathbb{R}^{n}$ vector).
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:29












  • $begingroup$
    A.P warned me about that issue. I have corrected it, thanks for your interest
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:33










  • $begingroup$
    Welcome XD. Concerning the grad, I think it points to the direction such that the scalar valued function increased fastest?
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:41














1












1








1





$begingroup$


assume that two vector given with relationship below: $$vec{n}cdot vec{u} = 0 $$ Then $vec{n}$ and $vec{u}$ are orthogonal vectors. Assume that we now have the direction of $vec{n}$



Then there are infinitely many possible direction for $vec{u}$
Coordinate System



How do we choose the vector $vec{u}$?



When we consider vector valued functions, we choose Normal vector to be directed inner side of the curve.



When we consider grad $nabla$ we know that it is perpendicular to any level curve. Then we choose $nabla$ to be directed towards outward of the curve.



How do we decide direction? What is the algorithm behind that? What am I missing? Or is it application dependent?






vectors







share|cite|improve this question











$endgroup$




assume that two vector given with relationship below: $$vec{n}cdot vec{u} = 0 $$ Then $vec{n}$ and $vec{u}$ are orthogonal vectors. Assume that we now have the direction of $vec{n}$



Then there are infinitely many possible direction for $vec{u}$
Coordinate System



How do we choose the vector $vec{u}$?



When we consider vector valued functions, we choose Normal vector to be directed inner side of the curve.



When we consider grad $nabla$ we know that it is perpendicular to any level curve. Then we choose $nabla$ to be directed towards outward of the curve.



How do we decide direction? What is the algorithm behind that? What am I missing? Or is it application dependent?






vectors




vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 9:25









Glorfindel

3,41981830




3,41981830










asked Jul 10 '15 at 10:06









SalihcyilmazSalihcyilmaz

556514




556514












  • $begingroup$
    Actually there are infinitely many possible direction, and also magnitude for $vec{u}$. The set of all these $vec{u}$ can be called as the orthogonal complement of $vec{n}$. The dimension of this set is n-1 (assuming $vec{n}$ is a $mathbb{R}^{n}$ vector).
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:29












  • $begingroup$
    A.P warned me about that issue. I have corrected it, thanks for your interest
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:33










  • $begingroup$
    Welcome XD. Concerning the grad, I think it points to the direction such that the scalar valued function increased fastest?
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:41


















  • $begingroup$
    Actually there are infinitely many possible direction, and also magnitude for $vec{u}$. The set of all these $vec{u}$ can be called as the orthogonal complement of $vec{n}$. The dimension of this set is n-1 (assuming $vec{n}$ is a $mathbb{R}^{n}$ vector).
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:29












  • $begingroup$
    A.P warned me about that issue. I have corrected it, thanks for your interest
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:33










  • $begingroup$
    Welcome XD. Concerning the grad, I think it points to the direction such that the scalar valued function increased fastest?
    $endgroup$
    – Brian Cheung
    Jul 10 '15 at 10:41
















$begingroup$
Actually there are infinitely many possible direction, and also magnitude for $vec{u}$. The set of all these $vec{u}$ can be called as the orthogonal complement of $vec{n}$. The dimension of this set is n-1 (assuming $vec{n}$ is a $mathbb{R}^{n}$ vector).
$endgroup$
– Brian Cheung
Jul 10 '15 at 10:29






$begingroup$
Actually there are infinitely many possible direction, and also magnitude for $vec{u}$. The set of all these $vec{u}$ can be called as the orthogonal complement of $vec{n}$. The dimension of this set is n-1 (assuming $vec{n}$ is a $mathbb{R}^{n}$ vector).
$endgroup$
– Brian Cheung
Jul 10 '15 at 10:29














$begingroup$
A.P warned me about that issue. I have corrected it, thanks for your interest
$endgroup$
– Salihcyilmaz
Jul 10 '15 at 10:33




$begingroup$
A.P warned me about that issue. I have corrected it, thanks for your interest
$endgroup$
– Salihcyilmaz
Jul 10 '15 at 10:33












$begingroup$
Welcome XD. Concerning the grad, I think it points to the direction such that the scalar valued function increased fastest?
$endgroup$
– Brian Cheung
Jul 10 '15 at 10:41




$begingroup$
Welcome XD. Concerning the grad, I think it points to the direction such that the scalar valued function increased fastest?
$endgroup$
– Brian Cheung
Jul 10 '15 at 10:41










2 Answers
2






active

oldest

votes


















1












$begingroup$

It is very much application dependent. Note that in $Bbb{R}^3$ there are infinitely many vectors orthogonal to a given vector, not just $4$. Indeed, if $vec{n} = (a,b,c)$ and $vec{u} = (x,y,z)$, then
$$
0 = vec{n}cdotvec{u} = ax + by + cz
$$
is the equation of a plane through the origin.




In the case of the normal vector $vec{n}$ to a curve $gamma$ at some point $p$, you can describe the choice in the following (impractical) way:




  1. Consider the tangent vector $vec{t}$ at $gamma$ in $p$.

  2. The unit vectors orthogonal to $vec{t}$ parametrise all the planes containing $vec{t}$. Consider the osculating plane $pi$ of $gamma$ at $p$ (which is uniquely determined).

  3. Unless $vec{n} = 0$ (in which case $gamma$ has a flex point at $p$), the line through $p$ with direction $vec{t}$ divides $pi$ in two half-planes. The projection of $gamma$ on $pi$ is contained in one of them, which we call $H$.

  4. The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for correcting me, your right. There are infinitely many vectors.
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:17










  • $begingroup$
    As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
    $endgroup$
    – A.P.
    Jul 10 '15 at 10:46












  • $begingroup$
    I didnt understand osculating plane part. Can you elaborate on that?
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 11:05










  • $begingroup$
    It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
    $endgroup$
    – A.P.
    Jul 10 '15 at 11:19



















0












$begingroup$

There is a special case for twice-differentiable curves acting in $R^3$ For any point of such a curve you can explicitly define an orthonormal basis in $R^3$ that includes the normalized direction and two orthonormal vectors.



The trick is to realize that if $f''$ exists and is and not zero



$(f'(t) cdot f'(t))' = 0 = 2f''(t)cdot f'(t) = 0$



and it follows that



$(frac{f'(t)}{||f'(t)||}, frac{f''(t)}{||f''(t)||}, frac{f'(t)}{||f'(t)||} times frac{f''(t)}{||f''(t)||})$



is an orthonormal basis in $R^3$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    It is very much application dependent. Note that in $Bbb{R}^3$ there are infinitely many vectors orthogonal to a given vector, not just $4$. Indeed, if $vec{n} = (a,b,c)$ and $vec{u} = (x,y,z)$, then
    $$
    0 = vec{n}cdotvec{u} = ax + by + cz
    $$
    is the equation of a plane through the origin.




    In the case of the normal vector $vec{n}$ to a curve $gamma$ at some point $p$, you can describe the choice in the following (impractical) way:




    1. Consider the tangent vector $vec{t}$ at $gamma$ in $p$.

    2. The unit vectors orthogonal to $vec{t}$ parametrise all the planes containing $vec{t}$. Consider the osculating plane $pi$ of $gamma$ at $p$ (which is uniquely determined).

    3. Unless $vec{n} = 0$ (in which case $gamma$ has a flex point at $p$), the line through $p$ with direction $vec{t}$ divides $pi$ in two half-planes. The projection of $gamma$ on $pi$ is contained in one of them, which we call $H$.

    4. The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for correcting me, your right. There are infinitely many vectors.
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 10:17










    • $begingroup$
      As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
      $endgroup$
      – A.P.
      Jul 10 '15 at 10:46












    • $begingroup$
      I didnt understand osculating plane part. Can you elaborate on that?
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 11:05










    • $begingroup$
      It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
      $endgroup$
      – A.P.
      Jul 10 '15 at 11:19
















    1












    $begingroup$

    It is very much application dependent. Note that in $Bbb{R}^3$ there are infinitely many vectors orthogonal to a given vector, not just $4$. Indeed, if $vec{n} = (a,b,c)$ and $vec{u} = (x,y,z)$, then
    $$
    0 = vec{n}cdotvec{u} = ax + by + cz
    $$
    is the equation of a plane through the origin.




    In the case of the normal vector $vec{n}$ to a curve $gamma$ at some point $p$, you can describe the choice in the following (impractical) way:




    1. Consider the tangent vector $vec{t}$ at $gamma$ in $p$.

    2. The unit vectors orthogonal to $vec{t}$ parametrise all the planes containing $vec{t}$. Consider the osculating plane $pi$ of $gamma$ at $p$ (which is uniquely determined).

    3. Unless $vec{n} = 0$ (in which case $gamma$ has a flex point at $p$), the line through $p$ with direction $vec{t}$ divides $pi$ in two half-planes. The projection of $gamma$ on $pi$ is contained in one of them, which we call $H$.

    4. The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for correcting me, your right. There are infinitely many vectors.
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 10:17










    • $begingroup$
      As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
      $endgroup$
      – A.P.
      Jul 10 '15 at 10:46












    • $begingroup$
      I didnt understand osculating plane part. Can you elaborate on that?
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 11:05










    • $begingroup$
      It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
      $endgroup$
      – A.P.
      Jul 10 '15 at 11:19














    1












    1








    1





    $begingroup$

    It is very much application dependent. Note that in $Bbb{R}^3$ there are infinitely many vectors orthogonal to a given vector, not just $4$. Indeed, if $vec{n} = (a,b,c)$ and $vec{u} = (x,y,z)$, then
    $$
    0 = vec{n}cdotvec{u} = ax + by + cz
    $$
    is the equation of a plane through the origin.




    In the case of the normal vector $vec{n}$ to a curve $gamma$ at some point $p$, you can describe the choice in the following (impractical) way:




    1. Consider the tangent vector $vec{t}$ at $gamma$ in $p$.

    2. The unit vectors orthogonal to $vec{t}$ parametrise all the planes containing $vec{t}$. Consider the osculating plane $pi$ of $gamma$ at $p$ (which is uniquely determined).

    3. Unless $vec{n} = 0$ (in which case $gamma$ has a flex point at $p$), the line through $p$ with direction $vec{t}$ divides $pi$ in two half-planes. The projection of $gamma$ on $pi$ is contained in one of them, which we call $H$.

    4. The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.






    share|cite|improve this answer











    $endgroup$



    It is very much application dependent. Note that in $Bbb{R}^3$ there are infinitely many vectors orthogonal to a given vector, not just $4$. Indeed, if $vec{n} = (a,b,c)$ and $vec{u} = (x,y,z)$, then
    $$
    0 = vec{n}cdotvec{u} = ax + by + cz
    $$
    is the equation of a plane through the origin.




    In the case of the normal vector $vec{n}$ to a curve $gamma$ at some point $p$, you can describe the choice in the following (impractical) way:




    1. Consider the tangent vector $vec{t}$ at $gamma$ in $p$.

    2. The unit vectors orthogonal to $vec{t}$ parametrise all the planes containing $vec{t}$. Consider the osculating plane $pi$ of $gamma$ at $p$ (which is uniquely determined).

    3. Unless $vec{n} = 0$ (in which case $gamma$ has a flex point at $p$), the line through $p$ with direction $vec{t}$ divides $pi$ in two half-planes. The projection of $gamma$ on $pi$ is contained in one of them, which we call $H$.

    4. The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jul 10 '15 at 11:46

























    answered Jul 10 '15 at 10:14









    A.P.A.P.

    8,12021740




    8,12021740












    • $begingroup$
      Thanks for correcting me, your right. There are infinitely many vectors.
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 10:17










    • $begingroup$
      As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
      $endgroup$
      – A.P.
      Jul 10 '15 at 10:46












    • $begingroup$
      I didnt understand osculating plane part. Can you elaborate on that?
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 11:05










    • $begingroup$
      It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
      $endgroup$
      – A.P.
      Jul 10 '15 at 11:19


















    • $begingroup$
      Thanks for correcting me, your right. There are infinitely many vectors.
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 10:17










    • $begingroup$
      As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
      $endgroup$
      – A.P.
      Jul 10 '15 at 10:46












    • $begingroup$
      I didnt understand osculating plane part. Can you elaborate on that?
      $endgroup$
      – Salihcyilmaz
      Jul 10 '15 at 11:05










    • $begingroup$
      It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
      $endgroup$
      – A.P.
      Jul 10 '15 at 11:19
















    $begingroup$
    Thanks for correcting me, your right. There are infinitely many vectors.
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:17




    $begingroup$
    Thanks for correcting me, your right. There are infinitely many vectors.
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 10:17












    $begingroup$
    As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
    $endgroup$
    – A.P.
    Jul 10 '15 at 10:46






    $begingroup$
    As an example I added a "synthetic" description of the choice of direction for the normal vector to a curve at a point. Does it help?
    $endgroup$
    – A.P.
    Jul 10 '15 at 10:46














    $begingroup$
    I didnt understand osculating plane part. Can you elaborate on that?
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 11:05




    $begingroup$
    I didnt understand osculating plane part. Can you elaborate on that?
    $endgroup$
    – Salihcyilmaz
    Jul 10 '15 at 11:05












    $begingroup$
    It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
    $endgroup$
    – A.P.
    Jul 10 '15 at 11:19




    $begingroup$
    It is the plane which has the largest order of contact with $gamma$ at $p$. Intuitively, you can think of it as the plane through $p$ that is closest to $gamma$. It's a bit like with tangent lines: even though there are infinitely many lines that intersect $gamma$ exactly at $p$, there is only one which has order of contact (i.e. intersection multiplicity) $geq 2$ with $gamma$ at $p$.
    $endgroup$
    – A.P.
    Jul 10 '15 at 11:19











    0












    $begingroup$

    There is a special case for twice-differentiable curves acting in $R^3$ For any point of such a curve you can explicitly define an orthonormal basis in $R^3$ that includes the normalized direction and two orthonormal vectors.



    The trick is to realize that if $f''$ exists and is and not zero



    $(f'(t) cdot f'(t))' = 0 = 2f''(t)cdot f'(t) = 0$



    and it follows that



    $(frac{f'(t)}{||f'(t)||}, frac{f''(t)}{||f''(t)||}, frac{f'(t)}{||f'(t)||} times frac{f''(t)}{||f''(t)||})$



    is an orthonormal basis in $R^3$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      There is a special case for twice-differentiable curves acting in $R^3$ For any point of such a curve you can explicitly define an orthonormal basis in $R^3$ that includes the normalized direction and two orthonormal vectors.



      The trick is to realize that if $f''$ exists and is and not zero



      $(f'(t) cdot f'(t))' = 0 = 2f''(t)cdot f'(t) = 0$



      and it follows that



      $(frac{f'(t)}{||f'(t)||}, frac{f''(t)}{||f''(t)||}, frac{f'(t)}{||f'(t)||} times frac{f''(t)}{||f''(t)||})$



      is an orthonormal basis in $R^3$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        There is a special case for twice-differentiable curves acting in $R^3$ For any point of such a curve you can explicitly define an orthonormal basis in $R^3$ that includes the normalized direction and two orthonormal vectors.



        The trick is to realize that if $f''$ exists and is and not zero



        $(f'(t) cdot f'(t))' = 0 = 2f''(t)cdot f'(t) = 0$



        and it follows that



        $(frac{f'(t)}{||f'(t)||}, frac{f''(t)}{||f''(t)||}, frac{f'(t)}{||f'(t)||} times frac{f''(t)}{||f''(t)||})$



        is an orthonormal basis in $R^3$.






        share|cite|improve this answer











        $endgroup$



        There is a special case for twice-differentiable curves acting in $R^3$ For any point of such a curve you can explicitly define an orthonormal basis in $R^3$ that includes the normalized direction and two orthonormal vectors.



        The trick is to realize that if $f''$ exists and is and not zero



        $(f'(t) cdot f'(t))' = 0 = 2f''(t)cdot f'(t) = 0$



        and it follows that



        $(frac{f'(t)}{||f'(t)||}, frac{f''(t)}{||f''(t)||}, frac{f'(t)}{||f'(t)||} times frac{f''(t)}{||f''(t)||})$



        is an orthonormal basis in $R^3$.







        share|cite|improve this answer














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        edited Aug 22 '16 at 10:31

























        answered Aug 22 '16 at 10:25









        saldukoosaldukoo

        1286




        1286






























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