Unit vectors -- How can they be dimensionless?

AI Thread Summary
A unit vector is defined as a vector with a magnitude of one, represented mathematically as \(\hat{A} = \vec{A}/|A|\). It is considered dimensionless because it conveys direction without associated units, despite retaining the directional information of the original vector. The discussion highlights that while unit vectors have a magnitude of one, they are essential for specifying direction in space, such as in Cartesian coordinates. Examples illustrate how unit vectors can clarify the direction of a displacement, emphasizing their practical utility in physics and mathematics. Understanding this concept alleviates confusion regarding the relationship between magnitude and dimensionality.
mcastillo356
Gold Member
Messages
639
Reaction score
348
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?
 
Mathematics news on Phys.org
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.)

---------------------
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}
 
Last edited:
  • Like
  • Informative
Likes mfb and 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
 
  • Like
Likes berkeman and collinsmark
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top