Unit vectors -- How can they be dimensionless?

[mcastillo356]

Hi, what is a unit vector? I mean, it is $\hat{A}=\vec A/|A|$. A dimensionless vector with modulus (absolute value) one, I've read somewhere.
So, dimensionless with modulus. Isn't that a contradiction? I mean, absolute value regardless dimension? Am I out of context?. $\Bbb R^3$ is a three-dimensional space...$\Bbb R^2$ a two-dimensional space, but $\Bbb R$ is not a dimension?
So, why is a unit vector dimensionless?

 

[collinsmark]

When performing the $\hat A = \frac{\vec A}{A}$ operation, the $\hat A$ retains the direction of $\vec A$. So yes, unit vectors all have magnitudes of 1, without units, but they do have separate directions. It's the directional information that they convey.

By the way, this post might be better suited for the General Mathematics subforum, or maybe the Introductory Physics Homework Help subforum. (At present it's in the General Discussion subforum.)

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Edit:

Rather than just leave it at that, allow me to give an example. Suppose that in three dimensions, with unit vectors $\hat {a_x}$, $\hat {a_y}$, and $\hat {a_z}$ (each unit vector representing one of the three Cartesian directions), you are given a displacement of 5 meters.

$s = 5 \ \mathrm{m}$

But with that alone, you have no idea what direction this displacement is.

But it is possible to specify this with unit vectors, such as

$\vec s = (5 \ \mathrm{m}) \hat {a_x}$

Now you know that the displacement is along the x-axis.

Or, as another example, suppose that the 5 meter displacement is on the x-y plane, along the x-y diagonal, you could write:

$\vec s = \left( \frac{5}{\sqrt{2}} \ \mathrm{m} \right) \hat {a_x} + \left( \frac{5}{\sqrt{2}} \ \mathrm{m} \right) \hat {a_y}$

 

[mcastillo356]

Thanks!Understood This was a question I feared to do, because I was afraid of making it the wrong post (magnitude, dimension, modulus... Have a different and eventually more than one meaning in my language). I really feel released.

Greetings