How do we decide the direction of vector, that is orthogonal to some other vector?
$begingroup$
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 $vec{u}$
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
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
$endgroup$
add a comment |
$begingroup$
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 $vec{u}$
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
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $vec{u}$
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
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
$endgroup$
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 $vec{u}$
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
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
edited Dec 24 '18 at 9:25
Glorfindel
3,41981830
3,41981830
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
asked Jul 10 '15 at 10:06
SalihcyilmazSalihcyilmaz
556514
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
- Consider the tangent vector $vec{t}$ at $gamma$ in $p$.
- 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).
- 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 $H$.
- The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $R^3$.
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
- Consider the tangent vector $vec{t}$ at $gamma$ in $p$.
- 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).
- 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 $H$.
- The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
- Consider the tangent vector $vec{t}$ at $gamma$ in $p$.
- 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).
- 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 $H$.
- The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
- Consider the tangent vector $vec{t}$ at $gamma$ in $p$.
- 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).
- 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 $H$.
- The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
$endgroup$
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:
- Consider the tangent vector $vec{t}$ at $gamma$ in $p$.
- 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).
- 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 $H$.
- The vector $vec{n}/|vec{n}|$ is uniquely determined as the unit vector orthogonal to $vec{t}$ which is contained in $H$.
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
edited Jul 10 '15 at 11:46
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
answered Jul 10 '15 at 10:14
A.P.A.P.
8,12021740
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $R^3$.
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
$endgroup$
add a comment |
$begingroup$
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 $R^3$.
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
$endgroup$
add a comment |
$begingroup$
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 $R^3$.
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
$endgroup$
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 $R^3$.
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
edited Aug 22 '16 at 10:31
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
answered Aug 22 '16 at 10:25
saldukoosaldukoo
1286
1286
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$begingroup$
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
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– 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?
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– Brian Cheung
Jul 10 '15 at 10:41