Science > Physics > unit vector of a length(position vector) graphed in length based axes?
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Science > Physics |
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"" |
| Date: |
08 Feb 2006 10:01:19 PM |
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unit vector of a length(position vector) graphed in length based axes? |
Say there are x, y, z coordinates set up for "some space" on earth,
where the coordinates represent lengths. Say the space is a
playground or a space around some buildings in downtown new york.
If there is a position vector between 2 points in this space, say
between two buildings or something, then the magnitude of this vector
is a length (metres, or whatever). That is the dimension of the
position vector or any vector which this coordinate system is really
set up for is length.
Now if we find the unit vector of the said position vector, it is
dimensionless. How would one graph the unit vector on this coordinate
system? How would one go about "thinking" about what it really means
to say that this unit vector has magnitude 1? Is that 1m? No. Then what
is it (geometrically) ?
The issues gets even more muddled if we consider forces. Sometimes one
finds the unit vector of a position vector between two points (along a
rope or something) which has a force acting along it. The force vector
can then be determined by multiplying the unit vector by the magnitude
of the force. This obviously means that the unit vector is
dimensionless and can be used to bring about vectors with different
units into the same "x y z" frame. Anyone have an idea about what it
means to say a unit vector has length 1, with respect to thise
coordinate system (which measures lengths)? How can it be graphed in
this xyz frame?
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| User: "Hexenmeister" |
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| Title: Re: unit vector of a length(position vector) graphed in length based axes? |
11 Feb 2006 09:15:33 AM |
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<i.love.jeevitha@gmail.com> wrote in message
news:1139457679.765790.178410@g47g2000cwa.googlegroups.com...
Say there are x, y, z coordinates set up for "some space" on earth,
where the coordinates represent lengths. Say the space is a
playground or a space around some buildings in downtown new york.
If there is a position vector between 2 points in this space, say
between two buildings or something, then the magnitude of this vector
is a length (metres, or whatever). That is the dimension of the
position vector or any vector which this coordinate system is really
set up for is length.
Now if we find the unit vector of the said position vector, it is
dimensionless.
Hardly. You've defined its dimension by "unit" (scalar, value 1) and by
orientation as a vector.
How would one graph the unit vector on this coordinate
system? How would one go about "thinking" about what it really means
to say that this unit vector has magnitude 1? Is that 1m? No.
Err, yes. It is so by definition of unit vector. You can use any units
you like, but once you've chosen inches, miles, meters, strangs, boobles
or whatever you want to call a unit, that's a unit.
A lightyear (ly) is a unit of distance, an Astronomical Unit (AU) is a
unit if distance, an Angstrom is a unit of distance. We choose units
to make the manipulation of numbers easy. It serves no useful purpose
to speak of 1,000,000,000,000,000,000,000,000,000,000,000 Angstroms.
Androcles.
Then what
is it (geometrically) ?
The issues gets even more muddled if we consider forces. Sometimes one
finds the unit vector of a position vector between two points (along a
rope or something) which has a force acting along it. The force vector
can then be determined by multiplying the unit vector by the magnitude
of the force. This obviously means that the unit vector is
dimensionless and can be used to bring about vectors with different
units into the same "x y z" frame. Anyone have an idea about what it
means to say a unit vector has length 1, with respect to thise
coordinate system (which measures lengths)? How can it be graphed in
this xyz frame?
.
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| User: "Sam Wormley" |
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| Title: Re: unit vector of a length(position vector) graphed in length basedaxes? |
08 Feb 2006 10:10:37 PM |
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wrote:
Say there are x, y, z coordinates set up for "some space" on earth,
where the coordinates represent lengths. Say the space is a
playground or a space around some buildings in downtown new york.
Restate you problem with clearer definitions....
Unit Vector
http://mathworld.wolfram.com/UnitVector.html
Radius (Position) Vector
http://mathworld.wolfram.com/RadiusVector.html
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| User: "" |
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| Title: Re: unit vector of a length(position vector) graphed in length based axes? |
08 Feb 2006 10:46:32 PM |
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Sam Wormley wrote:
i.love.jeevitha@gmail.com wrote:
Say there are x, y, z coordinates set up for "some space" on earth,
where the coordinates represent lengths. Say the space is a
playground or a space around some buildings in downtown new york.
Restate you problem with clearer definitions....
Unit Vector
http://mathworld.wolfram.com/UnitVector.html
Radius (Position) Vector
http://mathworld.wolfram.com/RadiusVector.html
Unit vector defined exactly like mathworld (vector multiplied the
recirprocal of it's norm).
Positoin vector is just the vector representing the position of some
point relative to the other (example, if a cube has a side length of
3m, the position vector representing one of the diagonals would be
(3m,3m,3m) - the unit vector associated with this position vector would
be (1,1,1) -dimensionless) How would i think about this unit vector on
the same length-based axes where the cube is drawn? What does it mean
that the unit vector has magnitude 1?
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| User: "Timo Nieminen" |
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| Title: Re: unit vector of a length(position vector) graphed in length basedaxes? |
08 Feb 2006 10:57:29 PM |
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On Thu, 8 Feb 2006 wrote:
Positoin vector is just the vector representing the position of some
point relative to the other (example, if a cube has a side length of
3m, the position vector representing one of the diagonals would be
(3m,3m,3m) - the unit vector associated with this position vector would
be (1,1,1) -dimensionless) How would i think about this unit vector on
the same length-based axes where the cube is drawn?
Your length-based axes are not dimensionless, your unit vector is. Given a
2D graph with length-based axes, where would you locate (3,$2000)?
What does it mean
that the unit vector has magnitude 1?
Your unit vector is pure direction; to make it mean something in your
chosen coordinate system, multiply it by a magnitude with the same
dimensions as your axes.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: unit vector of a length(position vector) graphed in length based axes? |
09 Feb 2006 02:51:05 PM |
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Timo Nieminen wrote:
On Thu, 8 Feb 2006 wrote:
Positoin vector is just the vector representing the position of some
point relative to the other (example, if a cube has a side length of
3m, the position vector representing one of the diagonals would be
(3m,3m,3m) - the unit vector associated with this position vector would
be (1,1,1) -dimensionless) How would i think about this unit vector on
the same length-based axes where the cube is drawn?
Your length-based axes are not dimensionless, your unit vector is. Given a
2D graph with length-based axes, where would you locate (3,$2000)?
What does it mean
that the unit vector has magnitude 1?
Your unit vector is pure direction; to make it mean something in your
chosen coordinate system, multiply it by a magnitude with the same
dimensions as your axes.
So the unit vector in this situation really doesnt' have a magnitude.
So one cannot draw it by itself on these axes?
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| User: "Timo Nieminen" |
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| Title: Re: unit vector of a length(position vector) graphed in length basedaxes? |
09 Feb 2006 02:53:01 PM |
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On Fri, 9 Feb 2006 wrote:
Timo Nieminen wrote:
On Thu, 8 Feb 2006 wrote:
Positoin vector is just the vector representing the position of some
point relative to the other (example, if a cube has a side length of
3m, the position vector representing one of the diagonals would be
(3m,3m,3m) - the unit vector associated with this position vector would
be (1,1,1) -dimensionless) How would i think about this unit vector on
the same length-based axes where the cube is drawn?
Your length-based axes are not dimensionless, your unit vector is. Given a
2D graph with length-based axes, where would you locate (3,$2000)?
What does it mean
that the unit vector has magnitude 1?
Your unit vector is pure direction; to make it mean something in your
chosen coordinate system, multiply it by a magnitude with the same
dimensions as your axes.
So the unit vector in this situation really doesnt' have a magnitude.
So one cannot draw it by itself on these axes?
It doesn't have a meaningful length on those axes. You can always
arbitrarily choose a length, so thaat the drawing looks OK. It would just
be an arrow of whatever length you choose, just indicating a direction.
--
Timo
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| User: "" |
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| Title: Re: unit vector of a length(position vector) graphed in length based axes? |
09 Feb 2006 03:28:49 PM |
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Timo Nieminen wrote:
On Fri, 9 Feb 2006 wrote:
Timo Nieminen wrote:
On Thu, 8 Feb 2006 wrote:
Positoin vector is just the vector representing the position of some
point relative to the other (example, if a cube has a side length of
3m, the position vector representing one of the diagonals would be
(3m,3m,3m) - the unit vector associated with this position vector would
be (1,1,1) -dimensionless) How would i think about this unit vector on
the same length-based axes where the cube is drawn?
Your length-based axes are not dimensionless, your unit vector is. Given a
2D graph with length-based axes, where would you locate (3,$2000)?
What does it mean
that the unit vector has magnitude 1?
Your unit vector is pure direction; to make it mean something in your
chosen coordinate system, multiply it by a magnitude with the same
dimensions as your axes.
So the unit vector in this situation really doesnt' have a magnitude.
So one cannot draw it by itself on these axes?
It doesn't have a meaningful length on those axes. You can always
arbitrarily choose a length, so thaat the drawing looks OK. It would just
be an arrow of whatever length you choose, just indicating a direction.
Makes sense thanks.
I also have a follow up. Say one wants to find the scalar projection
of a F onto a position(length based) vector. Essentially, trying to
find the component of that force in the direction of the position
vector (which could represent a rope for instance).
proj = F dot r
----------
r dot r
The units turn out to be = N/m. The book I'm self studying about
doesn't mention units or anything in this case, but says the results it
the component of the force in the direction of the position vector. It
make sense, EXCEPT when one considers the units involved. N/m ??
Should I just disregard this and take it for what it is? Is there a
mathematical explanation that doesn't lead to this odd result?
Thank you so much Mr Nieminen
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| User: "Greg Neill" |
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| Title: Re: unit vector of a length(position vector) graphed in length based axes? |
09 Feb 2006 03:53:23 PM |
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<i.love.jeevitha@gmail.com> wrote in message
news:1139520528.952346.267390@z14g2000cwz.googlegroups.com...
Makes sense thanks.
I also have a follow up. Say one wants to find the scalar projection
of a F onto a position(length based) vector. Essentially, trying to
find the component of that force in the direction of the position
vector (which could represent a rope for instance).
proj = F dot r
----------
r dot r
The units turn out to be = N/m. The book I'm self studying about
doesn't mention units or anything in this case, but says the results it
the component of the force in the direction of the position vector. It
make sense, EXCEPT when one considers the units involved. N/m ??
Should I just disregard this and take it for what it is? Is there a
mathematical explanation that doesn't lead to this odd result?
F dot r has units of energy. If r were a displacement vector
(distance) through which the force was applied, you'd have
the work done by that force.
But you're after the magnitude of the projection of F in the
direction of r. You want to divide the work by the magnitide
of the distance, that is,
(F dot r)/|r| where |r| is the magnitude of r
Or alternatively,
(F dot r)/sqrt(r dot r)
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| User: "" |
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| Title: Re: unit vector of a length(position vector) graphed in length based axes? |
10 Feb 2006 11:33:12 PM |
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Greg Neill wrote:
<i.love.jeevitha@gmail.com> wrote in message
news:1139520528.952346.267390@z14g2000cwz.googlegroups.com...
Makes sense thanks.
I also have a follow up. Say one wants to find the scalar projection
of a F onto a position(length based) vector. Essentially, trying to
find the component of that force in the direction of the position
vector (which could represent a rope for instance).
proj = F dot r
----------
r dot r
The units turn out to be = N/m. The book I'm self studying about
doesn't mention units or anything in this case, but says the results it
the component of the force in the direction of the position vector. It
make sense, EXCEPT when one considers the units involved. N/m ??
Should I just disregard this and take it for what it is? Is there a
mathematical explanation that doesn't lead to this odd result?
F dot r has units of energy. If r were a displacement vector
(distance) through which the force was applied, you'd have
the work done by that force.
But you're after the magnitude of the projection of F in the
direction of r. You want to divide the work by the magnitide
of the distance, that is,
(F dot r)/|r| where |r| is the magnitude of r
Or alternatively,
(F dot r)/sqrt(r dot r)
Ahh my mistake. So units work out fine for both vector and scalar
projects. Excellent, now the dimensionless unit vector doesn't even
need to be mentioned (I can just project different types of vectors
onto each other).
Thanks everyone
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| User: "Timo Nieminen" |
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| Title: Re: unit vector of a length(position vector) graphed in length basedaxes? |
09 Feb 2006 04:01:01 PM |
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On Fri, 9 Feb 2006 wrote:
I also have a follow up. Say one wants to find the scalar projection
of a F onto a position(length based) vector. Essentially, trying to
find the component of that force in the direction of the position
vector (which could represent a rope for instance).
proj = F dot r
----------
r dot r
Note that the above is a _scalar_. If you want to get a vector as the
result, try ((F.r)/(r.r)] * r. If you want to use r as a basis vector, and
find that vector component of F, then you want a scalar result, but you
use (F.r)/|r|, which is equal to (F.r)/sqrt(r.r). Notice the difference.
Now, if F has different units from r, the second operation, using r as a
basis vector doesn't make sense. The first does, but notice that the
result has the same units as F.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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