If a unit $x$ velocity and a unit $y$ velocity are simultaneously applied to a point $p$ at the origin, the point $p$ will move at speed $ \sqrt{2} \enspace $ along the line $y=x$, to point $A = [0,1]$ in unit time.

How then does one clearly articulate how an orthonormal 2-dimensional vector basis can exist AT ALL without vanishing and being replaced by a single vector $ \vec{A} $ ?

That is, how does one clearly and succinctly articulate the difference in TYPE between actual acting vectors, and "basis vectors" -which merely 'record' the projection an actual vector has as a along the basis directions - and which do not in fact "$ \textbf act $" at all?

Is there a name for this difference?

  • $\begingroup$ I'm not sure what you're asking here. The particle in question has velocity $\vec v = \hat x + \hat y$ (in suitable units). What do you mean by an orthonormal basis "vanishing and being replaced?" $\endgroup$
    – J. Murray
    Commented Feb 5, 2023 at 2:09
  • $\begingroup$ What is an "acting vector"? What does it mean for a vector to "act"? $\endgroup$
    – d_b
    Commented Feb 5, 2023 at 2:14
  • $\begingroup$ I'm not sure it makes sense to "apply velocity to a point." If you have some point particle that is moving with velocity $\vec{A}$ in some frame of reference, then there's more than one way to describe it. One way is to define a coordinate system using a set of basis vectors in the same frame. That's handy because it affords you a numerical representation of $\vec{A},$ and it allows you to calculate answers to questions about the motion. But establishing a coordinate system does not in any way change the physical reality of the moving particle. $\endgroup$ Commented Feb 5, 2023 at 2:18
  • $\begingroup$ @solomon In the derivation of the relativistic lorentz transformations between two inertial frames, no matter particles are referred to, only the velocity of the frame ( x', y',z', t' ) relative to the frame ( x, y, z, t ). I guess I can ask it more succinctly: Two mutually orhthogonal unit vectors acting at a point p, produce a resultant, whereas the two orthogonal unit basis vectors do not, why? $\endgroup$
    – pete
    Commented Feb 5, 2023 at 3:45
  • 2
    $\begingroup$ Coordinate bases are not "real" (have no observable physical effects). They are just a helpful tool to calculate – a kind of graph paper upon which to make measurements. Vectors like velocity, acceleration, force, etc are physically real. You can tell because when you rotate or change a coord system, the basis vectors change, the physical vectors do not. $\endgroup$
    – RC_23
    Commented Feb 5, 2023 at 5:04

2 Answers 2


You use a vector when you have a situation where you can do one thing and then another, and the two things add to produce a result. You can get the same result by doing the "sum" thing instead of the two things. And there are always many different ways of doing two things that add up to the same sum.

There are many examples of this in physics. Here are some.

Walk to the east $1$ meter. Walk to the north $1$ meter. This is the same as walking northeast $\sqrt{2}$ meters.

Another is a tug-o-war between two equally strong people. One pulls his end of the rope to the left with a given force. The other pulls to the right with the same force. The two forces cancel. The effect is the same as if the force was $0$. In this case, we need to explicitly say the effect is to accelerate the rope. One force alone would accelerate the rope strongly. But for two, the acceleration is $0$.

Keep in mind what is real and what is mathematics. Moving around and pulling on a rope are real. Vectors are just a mathematical way of describing the movements and/or forces.

In reality, you can do get a result two different ways. These ways have two different mathematical descriptions. You use these vectors to describe one set of actions and those vectors to describe the other. The vectors don't "act" or "vanish" if they are not used. They are just math.

On the other hand, people get so used to vectors that they think of distances and forces as vectors. They treat math as if it was reality. They may well say the reason an object accelerates is a vector acts on it instead of saying a force acts on it. They may not care if the result comes from having two vectors act or just the sum of the two.

One reason vectors are useful is you don't have to care. The math may get easier if you just look at the sum. Or perhaps it is easier if you break a sum into pieces. They may use vectors for forces that didn't individually happen, but add up to the total that did happen. Here is a post where I use that idea to explain things. Toppling of a cylinder on a block.

So don't get confused by vocabulary.

Add on

Adding an answer to the (closed) follow on question Distinguishing different senses of 'vector'. @pete had asked much the same thing here. This addresses how a basis vector is different from any ordinary vector.

The short answer is - It isn't different. It is an ordinary vector chosen for a special use.

Returning to the first example vector space above, I can specify any point on a plane with two numbers, $x$ and $y$, by saying how many meters east and north of me it is.

$$\vec v = (x, y) \tag{1}$$

I can specify that same point in much the same way by adding two vectors. One, which I will call $\vec v_1$, is x meters long and points to the east. The other, $\vec v_2$, is y meters long and points north. So

$$\vec v = \vec v_1 + \vec v_2$$

The numbers $x$ and $y$ are buried inside $\vec v_1$ and $\vec v_2$. We can make the connection clear by choosing two vectors to be basis vectors. The first, called $\vec e_1$, is $1$ meter long and points east. The other, called $\vec e_2$, is $1$ meter long and points north.

Now we have $$\vec v_1 = x \space \vec e_1$$ $$\vec v_2 = y \space \vec e_2$$ and $$\vec v = x \space \vec e_1 + y \space \vec e_2 \tag{2}$$

So the basis vectors $\vec e_1$ and $\vec e_2$ are just helpers to help us use numbers to write down a vector. If someone writes down equation $(1)$, you can mentally expand it to equation $(2)$.

Note 1

Basis vectors can be chosen many ways. Any two vectors that are not parallel and have a positive length can be used. Each choice creates a way to use two numbers to represent vectors.

The choice above is the most convenient way, and is overwhelmingly the most popular choice. It is called the "usual basis". Often it is not mentioned, and you are expected to know it is being used.

If you wanted, you could choose a vector $2$ meters long to the south and another $1$ meter long to the northeast as a basis. You could figure out what numbers you would need to represent $\vec v$. They would be different from $x$ and $y$.

This kind of thing comes up all the time in physics. It is an important topic.

Note 2

This has talked about points in a plane. Often we need points in space.

It works the same way. But you need $3$ numbers to reach any point in space. These are distances east, north, and up. Likewise, you need to add $3$ vectors. You need a basis with $3$ vectors in it.

The number of vectors you need in the basis is the dimension of the space.


Question not really clear.

However, a basis $R^2$ must be defined by 2 linearly indepenant vectors. The span of these vectors $ \vec{S} = \lambda \vec{V}_{1} + \lambda V_{2}$ spans a plane. A vector contained within this span, what you call "$\vec{A}$" spans a one dimensional line and hence would not be an equivelant basis , Span{S} ≠ Span{A}.

I don't understand what your question really is, if you want to take a vector and express it in the form of a specific basis or just take the vector defined as it self (much like the cartesian unit vectors), then it doesn't really matter, you get the same results.

A basis is just used to define vectors in a specific way that makes computation easier. Whether or not you want to think about this vector acting itself , or think about it as the sum of two acting basis vectors is irrelevant.

  • $\begingroup$ I guess I can ask it more succinctly: Two mutually orhthogonal unit vectors acting at a point p, produce a resultant, whereas the two orthogonal unit basis vectors do not, why? $\endgroup$
    – pete
    Commented Feb 5, 2023 at 3:38
  • $\begingroup$ Right I think I understand your question. We define a coordinate system, x,y. We express a position vector at a specific point in space as xi + yj. You now ask in the context of adding velocities , if the coexisting velocities existing in a point in space combine, why don't the axial unit vectors do the same? This is because we are talking about 2 very different mathematical ideas. Position vectors, are defined as the vector connecting the origin to a point in space. A point in space can be defined by the sum of basis vectors these basis vectors aren't defined as existing- $\endgroup$ Commented Feb 5, 2023 at 15:09
  • $\begingroup$ At the point that the resultant vector lands, the idea of a coordinate system is not the operation of adding unit vectors coexisting in the same place ( ie for position vectors the unit vectors can be thought of as only existing at the origin) the only reason they combine is because we define a point as the ADDITION of vectors , if we do so, these basis vectors combine to give a resultant vector the position vector. In general we don't define 2 vectors existing on the same point as adding them. $\endgroup$ Commented Feb 5, 2023 at 15:11
  • $\begingroup$ What you are describing in your question. Is the existence of a vector field. (Adding velocities is abit ill-defined depending on what you mean so I'll stick with forcefields). A vector field is defined as having a vectors attached to each point in space. $F = f_{x}(x,y)\hat i+ f_{y}(x,y)\hat j$ here there is a vector attached to each point in space that we DEFINE as the addition of 2 vectors. In general we don't just add 2 random vectors coexisting in the same point. And thus the idea of a coordinate system wouldn't just be the sum of I and J, since we define a coordinate system as $\endgroup$ Commented Feb 5, 2023 at 15:14
  • $\begingroup$ The SPAN of these vectors, which is very different than the addition of the 2 individual basis vectors $\endgroup$ Commented Feb 5, 2023 at 15:15

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