# Why vector is defined as one straight line?

I'm starting my studies now, and wondering why a lot of things that at the beginning I don't know if the motivation is based on the intuition of concrete thinking or the logic of abstract thinking, and this is the case with the definition of vectors. I understood that the vectors are composed by direction and magnitude, which represent the difference between two points, which mathematically speaking makes intuitive sense to be a line. But why are they defined exactly as a line if not always when applied to physical concepts, for example, not necessarily what happens between these two is a line?

• May 23, 2021 at 5:36
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– rob
May 25, 2021 at 15:41

A vector is not a line. Nor is it "something that has magnitude and direction".

A vector is instead a type of mathematical object that can be used to represent things with magnitude and direction, or even just direction. That is, vectors come first, "magnitude and direction" come second. In fact, in the most general sense of vectors, vectors do not have "magnitude" in an absolute sense, and even their sense of "direction" is fairly weak.

What a vector "is" is an element of a vector space. That may seem really unhelpful, but one can say vector spaces are designed specifically to make working with quantities where we need to encode magnitude and direction information into a single object easy and useful.

A slightly less abstract way to look at it is vector spaces generalize the idea of a list of numbers:

$$\langle v_1, v_2, \cdots, v_n\rangle$$

where we can add those numbers in an elementwise fashion, as well as multiply them all by a single specific number (called a "scalar"). This lets us encode magnitude and direction in a neat way, because if we interpret that list of numbers as a list of Cartesian coordinates, we can say the direction represented is that of an arrow from the origin to the point given by the numbers we have so taken as coordinates (i.e. the point $$(v_1, v_2, \cdots, v_n)$$, where we have used different bracketing styles to distinguish the point from the vector), and the distance from the origin to the point is the magnitude. That's what you're seeing when you see "lines" or "arrows". But the vector itself is just this list of numbers, not the line or arrow, which is a pictorial representation of how the magnitude and direction information are encoded.

The relation between these two ideas is that when you go through the essential properties that "adding and multiplying in elementwise fashion" entail, you can prove that the only spaces that have all those properties can be considered "equivalent to" such a space of number lists, possibly infinitely long (though things get tricky in that case). (This is described in linear algebra texts as "there is one vector space of each dimension up to isomorphism", where "isomorphism" here means the equivalence in question.)

So when the vectors get used in physics, we are not using lines, we are using a purely abstract object which in some cases it is useful to interpret as describing a line, but in other circumstances, it may not be. And we may draw it using a little arrow or similar symbol when we want to convey or visualize the encoded magnitude and direction information.

• In conclusion, then: vectors are, ultimately, abstract objects able to carry information within a vector space that exists about certain properties. That vectors may or may not, not necessarily, be associate with magnitude and direction, or just direction, and sometimes it may be convenient to represent them as a straight line, and sometimes it can be the difference between two points, which in these last cases will not always be true, is that it? May 23, 2021 at 13:16
• As you said, it is possible for a vector to have only direction without magnitude, but could it be possible the other way in which it has magnitude and has no direction? May 23, 2021 at 13:16
• @Flávio Dorta I believe the answer to that is at least a moderate "no". The weak sense of direction that always exists for vectors is one can compare two vectors to say if they do or don't "have the same direction" by noting whether or not one can or cannot be multiplied by a scalar (that's the "elementwise multiplication by a given number" I mentioned) to obtain the other. However, to produce a strong sense of direction (i.e. we can write out bearings or something like that) and to produce a sense of magnitude, we must define additional structure on the vector space. May 23, 2021 at 18:42
• (That is, one must define a scheme by which vectors are given a specific magnitude, and one can do this in different ways, and same for giving a measurable, as opposed to just comparable, direction. The relevant concepts are called "normed" and "inner product" vector spaces. Basically, they are spaces where we have operations $||\cdot||$ and $\cdot$, or magnitude and dot product, defined in addition to the addition and scaling operations.) May 23, 2021 at 18:45
• Also, whether or not vectors are associated with magnitude and direction in an absolute and not merely comparative sense depends on whether or not you have defined the operations that map them to such - i.e. norm and inner product - on the type of vectors you are using. Then all such vectors will have such, because you can always use those operations to extract that information. In calculus and elementary physics, this is always the case, because you have norm (the thing denoted $|| \cdot ||$, e.g. $||\mathbf{v}||$) and inner product (i.e. $\cdot$, commonly called "dot product") operations. May 23, 2021 at 18:51

A vector is NOT a line. It has a magnitude, which is conveniently represented on paper as a line of a given length, but such a representation is purely for the convenience of the person who is working a particular physics problem. A vector also has a direction, which must be specified relative to some reference point or reference line, and that reference point or reference line is chosen by the person working a particular physics problem such that the problem can be properly defined in a convenient manner. While displacement is a vector quantity defined as ending position minus starting position, it is just a coincidence that this vector quantity happens to be the distance between two points, meaning that it is invalid to conclude that all vector quantities follow this pattern. As examples, note that force and acceleration are both vector quantities, neither of which is a line or the distance between two points on a line.

• Distance actually is subjective term. One needs to mention "where" it measures that distance for it becoming meaningful. As such one can make force a distance quantity by subtracting points in some force vector field. May 22, 2021 at 21:54
• "A vector in Euclidean 3-space can be thought of as a straight arrow (or more formally a directed line segment) that reaches from one point, P, to another, Q" press.princeton.edu/books/hardcover/9780691159027/… May 22, 2021 at 22:28
• @AgniusVasiliauskas Actually distance should never be used for vectors in general, and length is a deceptive term (for it generally implies some type of distance unit). For vectors, it's always appropriate to talk about the magnitude of the vector, or possibly the modulus, although that term has different meanings in other mathematical arenas. May 25, 2021 at 14:30

Most introductory physics books do not attempt to rigorously define what a vector is, because a truly rigorous definition requires some heavy mathematical machinery that can distract from the physics that the books is trying to present.

Some important features of vectors relevant to your question are:

• A vector is not defined to be a line. However, a line is often a useful visual/geometric representation of a vector.
• Vectors are very frequently used as local linear approximations to some more complicated function. The classic geometric example is the tangent vector to a curve. If you zoom in to a complicated smooth curve at a point, then it will approximately look like a line (and this approximation will become better and better as you zoom in more and more). The best linear approximation at a point $$P$$ on the curve, lies on the same direction as the tangent vector at $$P$$.
• Vectors can be rescaled by a constant and maintain their direction (more technically: for any unit vector $$\hat{e}$$, the combination $$\vec{v}\cdot\hat{e}/|\vec{v}|$$ is invariant under rescaling $$\vec{v}$$ by a scalar). A line segment naturally has this property.
• Operations between vectors remain true in any coordinate system. This property is so important that it is really the reason vectors are defined the way that they are. In the 1800s, before vectors were defined, people had to write out each component of an equation explicitly in order to be precise. Newton's law $$\vec{F}=m \vec{a}$$ had to be written as 3 equations $$F_x=m a_x$$, $$F_y= m a_y$$, $$F_z=m a_z$$, which was extremely cumbersome, and really obscured crucial geometric facts, such as that the acceleration is in the same direction as the applied force. Representing the force and acceleration vectors as lines here is very useful, since the line along which the force is applied, is the same line along which an object accelerates.

If you do want to get into a more rigorous definition, you ultimately need to understand the notion of a vector space. But in physics, at least in introductory physics, the vector space in question is almost always $$R^n$$ (and usually $$n\leq 3$$ -- in introductory physics you are usually dealing with 3 dimensional space, or sometimes 1 or 2 dimensions). Then you can think of a vector as a list of numbers $$\vec{v}=\langle v_x,v_y,v_z\rangle$$, along with some operations defined on these sets of numbers:

• you can add vectors: $$\vec{v}_1+\vec{v}_2=\langle v_{1,x}+v_{2,x},v_{1,y}+v_{2,y}+v_{1,z}+v_{2,z}\rangle$$
• you can multiply a vector by a scalar: $$\alpha \vec{v} = \langle \alpha v_x, \alpha v_y, \alpha v_z\rangle$$
• There are certain other technical properties listed on a table on the wikipedia page about vector spaces.
• While this isn't true of all vector spaces, for $$R^n$$ there is a natural inner product that allows you to combine two vectors and create a scalar -- this is what gives vectors the idea of a "magnitude" (the inner product with itself) and a "direction" (the inner product between a vector and certain reference vectors with unit length). The inner product (or "dot product") is $$\vec{v}_1\cdot \vec{v}_2 = v_{1,x} v_{2,x} + v_{1,y} v_{2,y} + v_{1,z} v_{2,z}$$.
• Finally, specifically in 3 dimensions, one can define a cross product where two vectors can be combined to give another vector, $$\vec{v}_1 \times \vec{v}_2 = \langle v_{1,y} v_{2,y} - v_{1,z} v_{2,z}, v_{1,z} v_{2,z} - v_{1,x} v_{2,x}, v_{1,x} v_{2,x} - v_{1,y} v_{2,y} \rangle$$. This product is special to vectors in $$R^3$$, but very useful in physics since we actually do seem to live in a 3-dimensional space.

Given this more abstract and complete definition (actually only the first three bullet points are strictly part of the definition, the second two are extra but important properties enjoyed by the vectors we are most interested in introductory physics), the reason we represent vectors with lines can be explained as follows. The line we draw for a given vector $$\vec{v}$$ is a line connecting the origin of coordinates to the point $$\langle v_x, v_y, v_z\rangle$$. This geometric point of view allows us to convert some of the algebraic operations into geometric ones. For example, vector addition can be represented geometrically via the "tail-to-tip" method; the dot product can be represented as the "projection" of one vector onto the other.

The geometric point of view also makes it clear that relationships between vectors do not depend on the specific coordinates used to represent vectors. This is something that is not immediately clear from the definition, but nonetheless true (and really, this property is the reason why vectors are defined the way they are). For example, suppose we know that $$\vec{a} + \vec{b} = \vec{c}$$. To make it concrete suppose $$\vec{a}=\vec{b}=\vec{c}/2=\langle 1,0,0 \rangle$$. Now we can do a coordinate transformation by rotating the coordinates about the $$z$$ axis by an angle $$\theta$$. Then the components of the vectors change: $$\vec{a}=\vec{b}=\vec{c}/2=\langle \cos \theta, -\sin \theta, 0 \rangle$$. However, the relationship $$\vec{a}+\vec{b}=\vec{c}$$ remains, for any value of $$\theta$$. The fact that relationships between vectors hold in any coordinate system, is really the reason vectors are defined to begin with.

To summarize: vectors are rigorously defined as algebraic objects, but their usefulness comes from the way they allow us to connect algebra and geometry. The geometric intuition is really the most important thing in introductory physics, and this is why introductory books focus on this geometric intuition without showing you the full machinery of the rigorous definition (which -- I would argue -- is ultimately is there to justify and support the intuitive, geometric operations). The reason vectors are geometrically represented as lines, is because this makes it easy to visualize the most important operations we can do with vectors: addition, multiplication by a scalar, and the dot and cross products. It also lets us visualize some of the most important applications of vectors, such as using the tangent vector to a curve as a local linear approximation to that curve.

• As a curiosity: in the evolution of mathematics, were these definitions defined discovered by observing their usefulness in nature or definitions invented to test their usefulness that were in fact useful? May 24, 2021 at 12:54
• @FlávioDorta In general in mathematics, a good definition is usually a synthesis of many different results and allows one to precisely define the exact set of properties needed to prove interesting theorems. So even though a definition logically comes first, a good definition is almost always discovered toward the end of a long process! In the specific case of vectors, the history is very much entangled with physics. The history section of the wikipedia article on vectors is a good starting point. May 24, 2021 at 13:55

Vectors are modeled after “displacements from a point in a plane” (or more generally “[flat] space”) that satisfy the “parallelogram rule” for addition. That’s their definition. https://en.wikipedia.org/wiki/Vector_space#First_example:_arrows_in_the_plane

Vectors are also useful in describing forces at a point applied to an object.

Furthermore, vectors are useful in describing “linear approximations” to quantities that vary smoothly. Thus, a velocity vector at a point helps describe the “linear approximation” of part of a curved path. Is this along the lines of your concern?

(In passing [since I don’t think this is your main concern], not every directed quantity is naturally described by an arrow. [In the spirit of Schouten’s Tensor Analysis for Physicists] In some cases, a flat oriented area or an oriented tube may be more appropriate .)

UPDATE: here is a photo from Schouten's Tensor Analysis for Physicists,
from my answer to What is the way to represent a Lorentz tensor field?

In three dimensions, there are eight type of directed quantities.
With the structure implied by a metric and an orientation, these can be reduced to an "arrow". However, invoking such symmetry may hide its arguably more physical representation. There are two ways to approach vectors.

The first way:

1. Start with the idea of a line between two points.
2. Notice it has some neat geometric properties (such as magnitudes and directions)
3. Determine ways to calculate values for these properties.
4. Notice that there are some things that have some of these properties, but aren't lines between two points.
5. Find clever ways to think of these things as lines between two points.
6. Find more cases where these properties seem really useful.
7. Create more obtuse metaphors to treat them as lines between points.
8. Repeat 6 and 7 until your brain hurts

This is a very natural way. And, arguably, it's how we invented the concept of vectors. We started with things that were useful and generalized them until it didn't make sense to do it anymore.

Now that mathematicians have done this, there's another way to approach vectors:

1. Look at the properties of vectors
2. Realize how nicely these properties capture the relationships of lines between two points.
3. Realize how these properties support far more problems, even ones where it makes no sense to think of them as lines.

Lets thank the mathematicians for doing it the hard way first, and take the easy road!

Vectors are elements from a vector space. We pay attention to the vector space because, while a single vector is just a vector, vector spaces describe how you can operate on vectors, just like how a number is just a number, but real numbers describe the arithmetic you can do. In particular, a vector space can guarantee the following (in this, $$u$$, $$v$$, and $$w$$ are arbitrary vectors in the vector space, and $$a$$ and $$b$$ are scalars e.g. real numbers)

• Addition of two vectors is:
• (Associative) $$u+(v+w)=(u+v)+w$$
• (Commutative) $$u+v=v+u$$
• (Has an Identity) There is some vector, called $$0$$ which has the property $$v+0=v$$ for all $$v$$.
• Multiplication with a scalar which is
• (Compatible) $$a(bv)=ab(v)$$
• (Has an Identity) $$1v=v$$ (where $$1$$ is the multiplicative identity for the scalar)
• Combinations:
• (Distributive over addition) $$a(u+v)=au+av$$
• (Distributive over multiplication) $$(a+b)v=av+bv$$

Now, what we have found is that a surprising amount of the time, we come across things that have these properties. When we do, we thank whatever deity we believe in, because once we phrase our thing as a vector in a vector space, hundreds of years of mathematicians' work can suddenly spring to our aid, proving all sorts of things that we would otherwise have to spend our time proving, ourselves.

Now it is rather easy to take spatial vectors, which are the vectors you are thinking of, with points and lines, and demonstrate that they indeed have these properties. After all, vectors are a generalization of this concept. But we find that the vast majority of their value is not in their points and lines, but in the things we can prove with the above rules.

In fact, spatial vectors have other properties. They have a norm operator, $$||v||$$ which returns the magnitude of the vector, and a dot product $$v\cdot u$$, the latter of which permits us to define the angle between two vectors using $$v\cdot u=||v||\cdot||u||\cdot\theta$$. So now we have the idea of magnitude and direction you were thinking of. But we are actually able to do a great deal with vectors without these operations, so we permit vectors that don't have these operators.

Now as you move forward in math and science, you may come across other things that are treated as vectors which are not spatial vectors. For example, all of quantum mechanics eventually finds its way down to something called a "Hilbert space," which is an infinite dimensional vector space. If you thought going from 2d to 3d to 4d was a trip, try going to infinity (and beyond!). But, despite the infinite headaches, you can use those vector rules above to make sense of the quantum madness. Or perhaps you might be interested in sympletcic vector spaces. These oddly named creatures show up in Lagrangian Mechanics. As a computer engineer, I use them because they help me make sure that when my computer makes rounding errors in a physics simulation, those rounding errors don't violate rules like the conservation of energy. It turns out you can do a lot without that pesky rule, so its good to abuse that vector space's properties to not claim you can do the impossible.

The list goes on and on, of course. Mathematicians have found the concept of a vector enormously useful over the years. But they're quite a lot more generalized than the points and lines that they originated from.

In many areas of physics, vectors are used to denote some kind of difference in position/velocity in some euclidean space. Displacement, for example, is the final minus initial position. Velocity is derivative of displacement, which is also a vector. Intuitively, if velocity at a point $$x$$ is $$v$$, and if it continues with that velocity from that point, then its position after unit time would be $$x+v$$, so the intuition of vector as a difference makes sense (even though you still need the formal definition of derivative to truly understand velocity).

A less intuitive use of vectors in physics is in cross-products, such as angular momentum. For that kind of thing, note that the direction of the cross-product is actually truly arbitrary even if you want it to be parallel to the axis of rotation. This is because we arbitrarily choose by convention to use a right-handed coordinate system for $$ℝ^3$$, and a right-hand formula for cross-product, to get a right-handed (i.e. counter-clockwise) rotation about the axis for positive angular momentum. That relation remains correct if you switch any two of them to left-handed, so you can actually choose arbitrarily from four possible choices for these three.

But for general vector spaces, it is a bad idea to always try to think of a vector space in terms of arrows. For example, every finite field $$K$$ is a vector space over the finite field $$\mathbb{F}_p$$ with $$p$$ elements, where $$p$$ is the characteristic of $$K$$. This can be used to show that $$K$$ is generated by adjoining a root of some polynomial over $$\mathbb{F}_p$$ with degree equal to the dimension of $$K$$ over $$\mathbb{F}_p$$. Finite fields are very useful in coding theory and cryptography, which has relevance to quantum encryption.

• Cross products are basically wedge or exterior products, with the note that the bivectors in $\mathbb{R}^3$ can be hodge-star mapped (using the right or left hand rule) back to vectors in $\mathbb{R}^3$. They are the space of oriented parallelograms. Vectors "live in" "3 choose 1" of $\mathbb{R}^3$, wedge products in "3 choose 2" of $\mathbb{R}^3$, and cross product maps the "3 choose 2" (bivectors) back to "3 choose 1".
– Yakk
May 25, 2021 at 14:36

A vector has magnitude (size) and direction, but no particular fixed start or end point like a line.

If you drew a line 5cm long going East (with the end of it due East of the starting point) - and so did your next door neighbour, you might say that they were two different lines. However the two would count as the same vector as they have the same magnitude and direction.

• So, is the vector always movable in space in such a way that a vector of direction and magnitude forms a family or can I just say that they are just the same types in different positions? May 22, 2021 at 12:19
• Anyway, I still don't understand why the vector must necessarily be represented by a line. May 22, 2021 at 12:20
• The line form you might have seen of a vector, is just a way to visualise it. the vector could be the magnitude and direction of a velocity, or of a force, or of an electric field, or any physical quantity that has both magnitude and direction. May 22, 2021 at 12:23
• Saying that vectors have no particular start is the best way to introduce misconceptions that will be difficult to eradicate at a later time. What you are referring to are extensions of the concept of vector, sometimes called free vectors. see for instance en.wikipedia.org/wiki/Affine_space May 22, 2021 at 12:42
• @FlávioDorta: A vector has no start point. It is like a difference. You would not say that $3-2$ starts somewhere; it is simply $1$. Similarly $(3,5)-(2,9) = (1,-4)$ does not start anywhere. May 24, 2021 at 4:39

Except for displacement vectors (which define the operation of vector addition), a vector generally represents the magnitude and direction of some physical entity at a point.

Smart question. Basic answer is that it is so more or less due to a nature of Euclidean geometry. If you draw displacement vector on curved surfaces, it's no longer a line. So strictly speaking a car displacement vector drawn on Earth surface is an arc, not a line. Of course, you can recalculate displacement of a car $$\vec r_2 - \vec r_1$$ using Earth center as an origin of Euclidean 3D geometry. So it depends on what type of geometry you are using for vectors.

• I did not get to the point of having studied what I am going to ask, but beforehand: from what I understood then, the displacement vector is Euclidean when it comes to Newton's mechanics, for example, I don't know if it is the case with other mechanics like that of Lagrange that I read superficially, but if there are any mechanics other than Euclidean, in this case the displacement vector is curved? I ask this because, from what I researched about displacement, it seemed to me universally represented by the vector in a straight line. May 22, 2021 at 23:11
• And is it meaningless to associate the word trajectory with a displacement vector? May 22, 2021 at 23:11
• @FlávioDorta It's because classical mechanics deals with Euclidean geometry. Check for example Riemannian Geometry, which Einstein has used in general relativity for describing curved space-time. Riemannian Geometry locally falls back to Euclidean geometry. Like if you take a very short segment on an arc, it will be almost a line. May 23, 2021 at 10:42

You write:

I don't know if the motivation is based on the intuition of concrete thinking or the logic of abstract thinking.

I think it is based on the abstract thought that you can stop time. When you do that it is easy to define vectors. These can change in time, of course, but, on their own, they don't contain information about past or future events. Only if they would contain this information they would be curved (or not, depending on the process). But in that case, you would have to make clear (besides the changing of a direction of a quantity) the time-varying "strength" (magnitude) of the quantity. Say by varying the thickness of the curved vector (zero-magnitude would be difficult to show though...).
For this kind of vector you have to choose two points that are stopped in time (say the start of a pulling force and the end), instead of one. Or two fixed points in space, while time goes on (say the floating of current in a wire).

But don't let this stop you from using the accepted definition! I just gave you an idea of how it could be done in your way.

• I really liked this view of you about vectors being stopped in time, this is really an abstract thought, different from what I thought when intuiting about the straight line as less space between two points as I even thought to formulate my doubt... but anyway, what you meant by your idea would be like a curved vector that has been extended by the effect of time on its action in a given space, that would not be a form of displacement or perhaps a scalable variable in terms of the duration of that time vector? May 23, 2021 at 13:44
• @FlávioDorta If you displace an object, the curved vector I have in mind would show how (the thickness shows with what velocity) you have displaced an object. The total displacement would be just the distance between the endpoints of the displacement. If you would subtract two different displacement vectors you will get another displacement vector that contains information about a different way to go from the beginning position to the final position. Multiply it by two, and you get the same form of the curved line, and the begin and endpoint will be two times as far from each other. May 23, 2021 at 14:10

The given answers are all correct (as far as I know), but I feel that there's a disconnect between the basic nature of your question and the very advanced content of these answers. So here's my layman's attempt at explaining it in simple terms:

Vectors are defined as the "travel" from point A to point B.

You'd think this requires us to define two points: the start (A) and the end (B). However, vectors have chosen to use the convention whereby the start is always presumed to be the origin point:

• (0,0) in a 2D plane
• (0,0,0) in a 3D space
• and so on

Therefore, all we have to write down is the end point (B), which is what we call "the vector".

From a data perspective, the vector is really just a point coordinate. However, due to the "start from the origin" convention used with vectors, vectors actually represent the travel from the origin to the specified point.

I understood that the vectors are composed by direction and magnitude, which represent the difference between two points

It's the other way around. Vectors are composed of the two points (well, as discussed, the implicit origin and the specifically chosen point), and the direction/magnitude of the vector is a consequence from that specifically chosen point.

You might be conflating/thinking of polar (angle and distance) and cartesian (classic x,y) coordinates instead of vectors here.

While there's a lot of similarity between these two kinds of coordinates (angle, distance, x,y) and a vector (direction, magnitude, and defined point B) in terms of information they contain, vectors express something that is conceptually very different.

Coordinates represent a fixed state, i.e. a snapshot. London is [here]. Paris is [here].

Vectors represent a transformation, i.e. the difference between two snapshots. The plane flies [from London to Paris].

## But why are they straight?

Well, simply put, because that is the easiest and most straightforward (pun intended) way to think of the travel from one point to another.

If I ask you what the distance between London and Paris is, you think of the straight-line distance, don't you? You don't give me the travel distance going from Paris to Berlin to Beijing to Los Angeles to New York to London, do you?

And even when you now think about this Paris-Berlin-Beijing-LA-NY-London route, I'll bet that you are still thinking of the individually straight-line distances between each city.

That sort of proves the point that straight lines are the easiest way to think about things.
There are infinitely many "curvy" ways to travel between two points, but there is only one way to travel in a straight line. Therefore, straight line travel is taken as the implicit standard.

## But what if I want to travel in a curve?

An important point to consider is that when talking about vectors, you're only talking about the start and end point, with an instantaneous travel time. You don't really consider the inbetween locations. All you are interested in is the precise beginning of the journey, and the precise end of it.

To keep it simple: think of it like teleportation.

Why is that? Well, let's say there's an interesting point C that you pass by when traveling from A to B.
Instead of having to somehow define the interesting point in the middle of the vector, it's much easier to simply break down the vector into two separate vectors: A-to-C and C-to-B.

This makes vectors a lot simpler to consider, because you can wholly ignore anything that's inbetween A and B. And if there is something you don't want to ignore (C), then you break down your vectors so they go from one interesting point (A or C) to the next (C or B, respectively).

This is not unlike planning a GPS route as individual routes between your stops, as opposed to a single route with many stops. Both make sense in their own way to a human. But from a purely mathematical perspective, breaking a vector down into individual routes makes it a whole lot easier to work with, which is why it's the preferred approach.

So if you want to model a curve, you model it as a combination of vectors, like so: The more vectors you use, the more accurate your curve will be. This is a balance between accuracy and effort required that you must strike.

Tangentially: computer graphics cannot accurately render curves. They use a LOT of straight lines, so many that it becomes indiscernable to the human eye that it's technically not a curve. Example: Every black line you see in all images are straight. But you really can't see that anymore in the images on the right, because your mind assumes that it must be a perfect sphere since it sort of looks like it.

Why is this the case? Well, simply put, because straight lines are significantly easier to deal with than curves, computationally speaking. The difference in computational complexity is so dramatic that a computer is more capable of handling the computation involving tens of thousands of defined polygons as opposed to the computation for a single defined sphere.