# What is the definition of unit vector?

In many physics textbooks it is given the following definition of unit vector: "A unit vector is every vector whose magnitude is 1 unit". I don't like this definition.

On one hand, it is quite common to use a notation for unit vectors (for instance a hat, $$\hat{u}$$) different to the one used for vectors in general (usually, an arrow, $$\vec{A}$$). We could have a vector $$\vec{A}$$ in an exercise or problem and find at the end that it has a magnitude of 1 unit. ¿Would $$\vec{A}$$ be a unit vector? I don't think so.

I think that unit vectors arise as a consecuence of normalizing another vector $$\vec{A}$$, that is, by dividing it by its magnitude,

$$\hat{u} = \frac{\vec{A}}{|\vec{A}|}$$

so that we get a vector $$\hat{u}$$ with just vector $$\vec{A}$$'s direction information.

According to units, at this point we can consider two paths:

1. Suppose that $$|\vec{A}|$$ has the same units as $$\vec{A}$$. Then $$\hat{u}$$ is a dimensionless quantity.

2. Suppose that $$|\vec{A}|$$ is dimensionless. Then $$\hat{u}$$ has the same units as $$\vec{A}$$.

I think the first option is the one that is usually used.

This last thing would also be a reason to not consider every vector with magnitude 1 unit a unit vector. In order to be a unit vector it must be dimensioness.

Is correct my definition of unit vector?

• What do you use for the basis set of your vector space, if you can only define a unit vector with reference to some vectors in the space? To my mind, you must start with the basis vectors and use them to construct the elements of the space, not start with elements of the space and use them to construct the basis set. – The Photon Apr 3 at 17:01
• You raise an important point about what dimension we assign to $|v|$. Normally, scalars are dimensionless, so that dividing by them we retain the dimension of the vector. However, it doesn't seem correct to call speed, a scalar quantity, a dimensionless scalar. And if we give it the same dimensions as the vector, then the unit vector will be dimensionless ...! – Mozibur Ullah Apr 3 at 18:19
• @MoziburUllah I’m struggling to understand this statement. Since when are most scalars dimensionless?? – J. Murray Apr 4 at 4:28
• Counter-example: Halliday, Resnick, Walker, Fundamentals of Physics Extended, 5th ed.: "A unit vector is a vector that has a magnitude of exactly 1 and points in a particular direction. It lacks both dimension and unit. Its sole purpose is to point, that is, to specify a direction." (emphasis mine.) – ilkkachu Apr 4 at 15:03
• @ThePhoton - Just the opposite. Vector spaces arise naturally in mathematics (in physics, they arise naturally in the mathematical model being used to describe the physical phenomena). For example, given some set $A$, the collection of functions from $A$ into $\Bbb R$ is a vector space. If there is some condition on the functions which is compatible with pointwise addition and multiplication by constants, then the set of functions satisfying the condition is also a vector space. Basis vectors are something we impose on the vector space so that we can compare it to $\Bbb R^n$ or $\Bbb C^n$. – Paul Sinclair Apr 5 at 17:27

A unit vector has magnitude $$1$$ - as in, the dimensionless number $$1$$. Not $$1\ \mathrm{cm}$$ or $$1\ \mathrm{kg}$$ or $$1\ \mathrm{N}$$ or $$1\ \mathrm{J}$$. It's also not hard to show that for any vector $$\vec A$$, the dimensions of $$\vec A$$ and $$\vert \vec A \vert$$ are the same.

• Just for anyone googling here, in the humble field of computer graphics (so, you're a programmer at Pixar or an animator at Fortnite, or whatever....) a "unit vector" has a different meaning, which may cause confusion for some readers. – Fattie Apr 5 at 16:05
• @Fattie Are you sure? I'm fairly certain that, in computer graphics, a unit vector (or a normalized vector) is a vector of length 1. E.g. surface normals and tangents are often unit vectors; directions are generally represented by unit vectors. This notion is pretty universal across disciplines - linear algebra, geometry, physics, engineering... Maybe you're confusing the term with the notion of basis vectors? Or are you referring to the use of homogenized coordinates [x, y, z, w], where the w component is disregarded when it comes to calculating magnitude (as it's more of a "flag" value)? – Filip Milovanović Apr 5 at 18:18
• @Fattie - but in computer graphics, unit vectors aren't a priory associated with meters or any other unit. It's not one meter, it's just one. An orthonormal standard basis is assumed; this defines a sort of a global coordinate system and scale - what's sometimes called "world units". Then there's usually some global parameter that sets the scaling factor, determining how many of these "world units" fit in a meter or in a foot or whatever, but this is mostly for the artists, for consistency. After scaling, the implementation can proceed without making any reference to units. 1/2 – Filip Milovanović Apr 5 at 19:55
• @Fattie - In other words, what I'm saying is that in computer graphics, vectors and associated operations are based on linear algebra and projective geometry, so on the implementation side, it's all actually more in line with the mathematical (dimensionless) notion then physics is. 2/2 – Filip Milovanović Apr 5 at 19:55
• @Fattie A unit vector is any vector whose magnitude is one. It's the same definition in physics, mathematics, engineering and computer science, including in computer graphics. I say this as someone whose degree is in Math/CS and who has worked for many decades as a software developer including lots of CG. Here's the course notes from a CMU course on Computer Graphics if you need confirmation (page 5): cs.cmu.edu/afs/cs/academic/class/15462-s09/www/lec/04/lec04.pdf. Software publishers often invent their own terms and redefine existing ones, but that's on them. – RBarryYoung Apr 6 at 16:47

I think that your confusion lies in the word "unit". In the definition of unit vector, "A unit vector is every vector whose magnitude is 1 unit," the word does not really refer to units like meter and second. It rather means just '1' and could be skipped.

A true unit vector has no physical dimension (like force) but only a direction.

If $$\vec{F}$$ is a physical vector (e.g. a force), then $$|\vec{F}|$$ is its size and has the same physical dimension (e.g. force), and $$\hat{u} = \vec{F}/|\vec{F}|$$ is a unit vector without physical dimension.

• Fair enough. But "1 unit" in scientific context usually (or at least sometimes) means "the unit of measure in the system being used" (like 1m for length, when using SI), rather than a dimensionless 1. This means that "whose magnitude is 1 unit" is prone to being misunderstood (as the existence of this question demonstrates), while "whose magnitude is 1" is not. I would then say that "unit" in that phrase not only could be omitted but should be omitted, and its presence is simply an error. – Marc van Leeuwen Apr 7 at 3:11

$$(1,0,0)={(2,0,0)\over2}$$. According to your naming scheme, the left hand side is not a unit vector but the right hand side is. This seems to be a problem.

For that matter, whenever $$(x,y,z)$$ has length $$1$$, then $$(x,y,z)={(x,y,z)\over 1}$$ both is and is not a unit vector. Oops!

• Yes, obviously the problem of units and dimensions disappears if you reduce everything to unit-less quantities. That's hardly an answer to a question specifically about units. – ilkkachu Apr 4 at 14:58
• $(1,0,0)$ can always be written as $(1,0,0)/1$. But it will only be a unit vector if it has no physical units attached to it. Otherwise it will be only a vector of length 1, that is not a unit vector by definition. – Girardi Apr 5 at 23:46

In many physics textbooks it is given the following definition of unit vector: "A unit vector is every vector whose magnitude is 1 unit".

Yes, that is the definition of unit vector. I cannot remember seeing any other definitions.

Is correct my definition of unit vector?

You did not actually give a definition, unless you are referring to this formula:

$$\hat{u} = \frac{\vec{A}}{|\vec{A}|}$$

But this formula is equivalent to the definition above, as shown below.

On one hand, it is quite common to use a notation for unit vectors (for instance a hat, $$\hat{u}$$) different to the one used for vectors in general (usually, an arrow, $$\vec{A}$$).

Yes, if we know that a vector is a unit vector, we can use a different notation to convey this extra information. But we do not have to.

We could have a vector $$\vec{A}$$ in an exercise or problem and find at the end that it has a magnitude of 1 unit. ¿Would $$\vec{A}$$ be a unit vector?

Yes, it would be a unit vector. We might not say that it is a unit vector; this does not matter.

I think that unit vectors arise as a consecuence of normalizing another vector $$\vec{A}$$, that is, by dividing it by its magnitude,

$$\hat{u} = \frac{\vec{A}}{|\vec{A}|}$$

so that we get a vector $$\hat{u}$$ with just vector $$\vec{A}$$'s direction information.

According to units, at this point we can consider two paths:

1. Suppose that $$|\vec{A}|$$ has the same units as $$\vec{A}$$. Then $$\hat{u}$$ is a dimensionless quantity.

2. Suppose that $$|\vec{A}|$$ is dimensionless. Then $$\hat{u}$$ has the same units as $$\vec{A}$$.

I think the first option is the one that is usually used.

It is not a choice. $$|\vec{A}|$$ does have the same dimension as $$\vec{A}$$, by definition. $$|\vec{A}|$$ can only be dimensionless if $$\vec{A}$$ itself is dimensionless.

You actually say this yourself: $$\hat{u}$$ has “just vector $$\vec{A}$$'s direction information”. This means that $$\hat{u}$$ is dimensionless.

• But isn't the "every vector whose magnitude is 1 unit" what they're asking about? If $\vec{A}$ is one meter long and points that way, then $|\vec{A}|$ is one meter, and the resulting $\hat{u} = \frac{\vec{A}}{|\vec{A}|}$ has a length of just one. Not one meter, just a unit-less one. Which contradicts the first definition. – ilkkachu Apr 4 at 14:57
• @ilkkachu A “unit-less one” is 1 unit, so there is no contradiction. This issue has already been addressed in md2perpe’s answer. My answer addresses other, equally important, important issues with the question. – Brian Drake Apr 5 at 15:02
• @BrianDrake Saying "a 'unit-less one' is one unit" rather than "a 'unit-less one' is the number 1" is utterly confusing (or just wrong if you prefer that). Why bring up the word "unit" (which could very well mean "unit of measure") again when trying to express its absence? And if you cannot remember ever seeing a definition of unit vector that does not mention "unit" in its definiens, then I think you just did not look (or remember) well. – Marc van Leeuwen Apr 7 at 3:18

Just to expand on J. Murrays answer: Every unit vector is a vector, but not every vector is a unit vector. Hence, if you use the "hat" notation to indicate unit vectors its fine to write $$\hat u = \vec u = \vec e_u = \hat e_u = \ldots$$. So, if we consider a force of magnitude 1N pointing in the $$x$$-direction, we write $$\vec F = F \cdot \hat x = 1N \;\hat x = 1N \;\vec e_x$$, where $$\vec e_u$$ is the unit vector, not $$\vec F$$.

Definition of unit vector:

$$\hat{u}=\dfrac{\vec{A}}{|\vec{A}|}$$

So a unit vector is a mathematical object that is different from a vector of length 1 [UNIT], be it meters, Newtons, whatever. This is because by definition the unit vector has no physical units, since its given by the ratio of the vector by its norm (and vector and norm always have the same physical units).

Thus, you can always write any vector in terms of any unit vector without any trouble having to figure out the physical units of the quantities they represent.

For instance, say you have a force $$\vec{F}=[3\hat{x}+4\hat{y}]$$ Newtons. And you want a particle to displace with a velocity of 10 m/s in the same direction as $$\vec{F}$$. You could represent it in terms of the base unit vectors, $$\hat{x}$$ and $$\hat{y}$$, by doing the scalar product, BUT YOU DON'T NEED TO DO IT.

Just write,

$$\vec{v}=10 \dfrac{\vec{F}}{|\vec{F}|} [{m/s}]$$

This is possible, again, because by definition the unit vector has no physical unit attached to it. So it is very important to separate it from vectors of length 1 [unit].

The only remark here is that if you want to derive $$\vec{v}$$ written in terms of the unit vector $$\vec{F}/|\vec{F}|$$, and $$\vec{F}$$ is not constant, you have to derive $$\vec{F}$$ as well. This is particularly important when representing your problem in a rotating frame of reference.