# Is a unit vector really unitless and dimensionless?

According to my textbooks, a unit vector has no units and no dimensions, but is only used to specify direction. It only shows the orientation of a corresponding vector in space. I think it's true, or that's what it looks like. It makes sense because a unit vector is defined as 'a vector' divided by its magnitude. Since we have the same numerical value in numerator and the denominator, a unit vector has a magnitude of 1 unit. Likewise, we have the same unit in both numerator and the denominator, that makes a unit vector 'unitless', and hence dimensionless. That's why I think a unit vector has no dimensions. Please correct me if I'm wrong.

But, another question naturally comes to our mind. Why if I say, "a force of 1 N due east" or "a displacement of 1m, 30° NOE"?

Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors"? That's what I'm struggling to understand. There's no reason why we can't call these two unit vectors. Because both have a magnitude of 1 unit, and both are vectors. However, both have units, and hence both are not dimensionless.

Something to realize is that your vector of magnitude $$1\ \rm N$$ only has "unit" length because you chose to measure or represent your force in Newtons. If you chose some other unit, like pounds, then you would not have $$1$$ pound of force.

On the other hand, your actual unit vectors are indeed unitless$$^*$$. This is because unit vectors are defined as the ratio between two things with the same units. They will always have a (unitless) magnitude of $$1$$. In fact, this is true for any unitless quantity, since they do not depend on your choice of units (which is an intentionally redundant statement).

$$^*$$ I have always found this amusing. Unit vectors are unitless.

• So basically, the quantities in the question having a magnitude of one is wholly dependent on what units you use, whereas the unit vector will have a magnitude of one regardless of the units used?
– user209192
Oct 15, 2018 at 7:38
• The application of unit is quite amusing. Here, two uses are at work: 1) the physical unit, e.g. metres; and 2) the unit-direction, in the sence of due-north. A unit vector, simply spoken, is the 3D analogue of a compass rose on a street map. Go 300 metres to the north, then head west, only has a meaning when the compass-directions are defined. In 3D space, a vector x = (0 1 3) only has a meaning, when the unit-directions are defined. Oct 15, 2018 at 8:40
• @JTPenguin That is exactly right. Oct 15, 2018 at 11:36
• in a way, unit vector are unitless becase they are "the unit" by themselves (you can measure other vectors using them) - and, obviously, a unit itself can't have units.
– user42659
Oct 15, 2018 at 12:25

Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ? Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $$\textbf{F}=(1\ \text{N})\hat{\textbf{x}}$$. (Some people notate $$\hat{\textbf{x}}$$ as $$\hat{\textbf{i}}$$.) Then the unit vector in the direction of $$\textbf{F}$$ is

$$\frac{\textbf{F}}{|\textbf{F}|} = \hat{\textbf{x}} \, ,$$

which is not the same as $$\textbf{F}$$. It has different units, and it is not true that $$|\textbf{F}|=|\hat{\textbf{x}}|=1$$, since things with incompatible units can never be equal.

• Thanks a lot for the quick and precise explanation. I have a question, you said the unit vector ("x cap", don't know how to use mathjax yet, sorry) has different units, than the units of F. But aren't unit vectors supposed to be unitless? What are the units of that 'unit vector', btw?
– 4d_
Oct 14, 2018 at 17:53
• @πtimese It has different units because F has units and $\hat{\mathbf{x}}$ doesn't. Oct 14, 2018 at 18:24
• The units of the unit vector are that it has no units. If you like, you can formalize this sort of dimensional analysis using group theory, in which case the units are isomorphic to a vector space, e.g., kg.m/s2 is (1,1,-2) based on the exponent. Then "unitless" is the identity element of the group, or (0,0,0).
– user4552
Oct 14, 2018 at 19:58
• @πtimese The MathJax syntax for that symbol is \hat{x} , or \bf{\hat{x}} if you want it to be in bold font. Oct 15, 2018 at 1:28
• @Ben Thanks! So the bottom line is : Unit vectors are indeed unitless
– 4d_
Oct 16, 2018 at 2:32

If $$\vec{v}$$ is a vector with a physical unit, then its unit vector is defined as: $$\hat{v}=\frac{\vec{v}}{||\vec{v}||}$$ Where: $$||\vec{v}||=\sqrt{\sum_i v_i^2}$$ where every components $$v_i$$ has the physical unit. This clearly means that the unit vector is dimensionless.

• Correspondingly, $\vec{v}=\lambda \hat{v}$, where $\lambda$ is the magnitude and is the dimensional element Oct 15, 2018 at 4:11

Well, if you have a dimensionless unit vector $$\mathbf{i}$$, that uniquely defines a unit vector for force $$\mathbf{i}_F = \mathbf{i} \cdot 1\,\mathrm{N}$$, same for displacement $$\mathbf{i}_D = \mathbf{i} \cdot 1\,\mathrm{m}$$ and for any other units you care about. Given that all of these are related in a well-defined manner, nothing’s stopping you from expressing your quantities in terms of a force basis should you so choose. General relativity, QFT and certain other branches of physics involve basis changes which are way more complicated than the question of whether $$|\mathbf{i}|$$ equals 1 or 1 N.

• How can we have unit vectors that can describe two different quantities (e.g. for or position)? Oct 31, 2020 at 21:17
• @AntoniosSarikas Different vectors live in different spaces. There is a space containing things like “two steps to the north”, “one kilometre to the east”, “a metre upwards” and all combinations thereof. At the same time, a weight can tug at a rope with a force of 42 N in the direction of the center of the Earth, but this vector lives in a different space from the displacement vectors. Nov 1, 2020 at 1:06