Re: unit vector is dimensionless, how to draw when coordinates for length?

In article <1139521062.209375.248640@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
i.love.jeevitha@xxxxxxxxx wrote:

Virgil wrote:
In article <1139456528.812781.109060@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
i.love.jeevitha@xxxxxxxxx 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.

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?

A "unit vector" is of length 1 unit using whatever are the current units
of distance.


Sorry I realized that I probably should have asked this question in
sci.physics and later started a thread there. As for your answer, if
that was true I would be very happy. Unfortunately the formula for the
unit vector takes out the effect of any unit.

unit vector = v / ||v||

Whatever unit vector v has, so will it's norm. Thus the univ vector in
direction of v has a magnitude of 1, but is dimensionless. Makes sense
except when I put down my linear algebra book and pick up my mechanics
book. Axes are often length based, and sometimes we normalize a
positoin vector (make it dimensionless) then use it to introduce a
vector with different units (say, Force) into the picture/axes. The
dimensionless unit vector defies my thinking for cases like this,
because I can't wrap my mind around "magnitude 1" for this case.

But someone explained that the unit vector in this case would be a
"pure direction" which relieves most of my headache (better than
ibuprofen).

Why not merely define a vector, v, as being a unit vector if and only if
||v|| = 1, in whatever units one is using?

In "pure" vector considerations, where no units are involved anyway, the
two definitions coincide, but in "applied" considerations, it avoids the
problem of having "unit vectors" which are not part of the vector space
under consideration.
.

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