[This is a very silly question. Will be a great help if someone could answer this question]

How is this law proven? I am not asking about the proof itself rather asking about the preliminary assumptions. That is how does two vectors be represented as two adjacent sides of a parallelogram and how does their resultant correspond to the diagonal? I could understand that this is applicable for displacement since it coincides with intuition. But how is it applicable for velocity or force? All these seems quite wierd to me.

In case of displacement, say a person is moving in such a manner that his path is representing two adjacent sides of a parallelogram. For example in parallelogram ABCD, the person is moving from A to B and then B to C. Well this concludes that the person has been displaced from A to C (along a straight line connecting A and C). But what if two forces are acting on a person at A directing towards B and D? In this case we cannot put forward the idea we used in case of displacement. Then how do we get to know that AC is the resultant of those two forces?

  • 1
    $\begingroup$ You can try to learn the derivation of triangle law of vector addition, parallelogram law is essentially a restatement of triangle law $\endgroup$
    – Draculin
    Jun 10, 2021 at 10:08
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    $\begingroup$ It's worth noting that some people think force should actually be thought of as a covector. This makes sense since we have $f = \frac{dp}{dt}$, and momentum is naturally a covector. This distinction almost never matters in practice because we usually have a way of identifying vectors and covectors. $\endgroup$ Jun 11, 2021 at 15:37

4 Answers 4


Your first question

How is this law proven? I am not asking about the proof itself rather asking about the preliminary assumptions.

The parallelogram law, mathematically speaking, is a result of an affine space. In simple terms, an affine space is a space that is flat. For example, the surface of a table and standard 3D space are affine spaces. The surface of a sphere, on the other hand, is not an affine space.

We first need to understand what a vector space is. A vector space is a set whose elements must obey certain rules. The elements are called vectors only if all the rules are satisfied.

An affine space is a set of points and a vector space, together with an addition operation that adds vectors (from the vector space) to points.

Let $P,Q$ be points and $\mathbf{u},\mathbf{v}$ be vectors. The addition is defined to satisfy two properties:

  1. $(P+\mathbf{u})+ \mathbf{v} = P+(\mathbf{u}+ \mathbf{v})$
  2. There is a unique vector $\overrightarrow{PQ}$ that goes from $P$ to $Q$.

An aside

We can add vectors to vectors (from vector addition in the vector space), as well as vectors to points (with the definition above), but it would make no sense to add points to points without defining a privileged point known as the origin.

Simply put, this is just saying that vectors are the displacements from point to point. From this definition, the following identities (which hopefully look familiar) can be proven without much difficulty:

  1. $\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$
  2. $P + \mathbf{0} = P$
  3. If $\overrightarrow{PQ} = \mathbf{0}$, then $P = Q$
  4. $\overrightarrow{PQ} = -\overrightarrow{QP}$

where $\mathbf{0}$ is the zero vector. We can then prove the parallelogram law: $$\text{If} \;\; \overrightarrow{P_1Q_1} = \overrightarrow{P_2Q_2} \\ \overrightarrow{P_1Q_1} + \overrightarrow{Q_1P_2} = \overrightarrow{P_2Q_2} + \overrightarrow{Q_1P_2} \\ \overrightarrow{P_1P_2} = \overrightarrow{Q_1P_2} + \overrightarrow{P_2Q_2} \\ \therefore \overrightarrow{P_1P_2} = \overrightarrow{Q_1Q_2}$$ where we have used the first identity as well as the commutativity of vector addition.

Your second question

I could understand that this is applicable for displacement since it coincides with intuition. But how is it applicable for velocity or force?

From the above we have shown that displacements are vectors in an affine space. Therefore we need to prove that velocity, acceleration and force are vectors. Velocity is defined to be the rate of change of displacement. Let $\mathbf{r}(t)$ be the displacement from a fixed point. The velocity is then $$\frac{\text{d} \mathbf{r}}{\text{d}t} = \lim \limits_{\Delta t \to 0} \frac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t}$$ The numerator $\mathbf{r}(t+\Delta t)- \mathbf{r}(t)$ is the difference of two vectors which is also a vector. The numerator multiplied by $1/\Delta t$ is also a vector by scalar multiplication. Therefore we have proven that velocity is also a vector. By the same logic, acceleration is also a vector and so on.


The parallelogram rule is simply the way in which you add two quantities that are vectors. So the real question is ... how do we know that displacement, velocity and force are all vectors ?

For displacements it is, as you say, almost intuitive. Displacements in Euclidean space are Euclidean vectors, and so you can add then like vectors - if you walk $x$ miles in one direction and then $y$ miles in another direction, your final displacement $\vec x + \vec y$ is the diagonal of the parallelogram formed by the two vectors $\vec x$ and $\vec y$.

Velocity is the rate of change of displacement. And we can prove mathematically that the rate of change of one vector is another vector. So velocities are vectors, and can be added using the parallelogram rule.

Force is not so obvious, but we know from Newton's second law that force is proportional to acceleration, which is the rate of change of velocity. Since velocity is a vector then so is acceleration, and since acceleration is a vector then so is force. In the same way, we can conclude that momentum is also a vector because it is proportional to velocity.

  • $\begingroup$ sir, in case of displacement say there is a parallelogram ABCD and a person moves from A to B and then to C, then we can easily understand that the person has been displaced from A to C and as per the definition of displacement, the straight line connecting A and C should be the displacement of the person. But when it comes to force, two forces are acting from A to B and A to D then how does AC represent their resultant? $\endgroup$
    – MSKB
    Jun 10, 2021 at 10:21
  • $\begingroup$ I could not understand the last part of the previous comment $\endgroup$
    – MSKB
    Jun 10, 2021 at 10:23
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    $\begingroup$ @MohammadSakibShahriar Displacement is a vector and the rate of change of a vector is another vector. Velocity is the rate of change of displacement (a vector) therefore velocity is a vector. Acceleration is the rate of change of velocity (a vector) therefore acceleration is a vector. Force is mass (a scalar) times acceleration (a vector) so force is also a vector. Momentum is mass (a scalar) times velocity (a vector) so momentum is also a vector. $\endgroup$
    – gandalf61
    Jun 10, 2021 at 11:01
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    $\begingroup$ I think it is always good to also think when these things fail (especially for obvious things), and this is on curved surfaces. Assume you are close to the norhpol (like 89 degree). Walking 100 km east and then 3000km south will give a dramatically different location than first going south then east. $\endgroup$
    – lalala
    Jun 10, 2021 at 21:57
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    $\begingroup$ I think it's worth noting that some people think force should be considered a co-vector. This consistent with the law $f = \frac{d p}{dt}$, since $p$ is naturally a co-vector. $\endgroup$ Jun 11, 2021 at 14:15

It is not so much a problem of physics as a problem of choosing a given formalism to represent a physical problem.

The choice of choosing the metaphor of mathematical vectors to represent a problem whose entities have a [fixed] number parameters is quite straightforward if one is familiar with the mathematical concept of vectors (they nicely map to each other).



Operations on physical entities which have typically 3 dimensions in classical mechanics are made simple if one chooses to represent them with vectors from a 3 dimensional space. It allows one to use all the standard operations available for vectors (a ready-made toolkit if you like) and to apply them to : position, speed, momentum, angular momentum. Amongst these operations are: addition, scalar product, measure of distances, calculation of surfaces, angular momentum, etc.

In other words there is a mapping between:

  • A problem whose variables have N parameters, N fixed.
  • A vector space with N dimensions.
  • Geometry in a N-dimensional space (the parallelograms you mention).

Let's tackle your question in single steps, and take the "force" example. Then I understand your question as "If two forces are applied to an object at the same time, why does the parallelogram method give the correct resulting force?"

Vectors are a mathematical concept. Often, we use 2- or 3-dimensional vectors, being composed from 2 or 3 numbers, e.g. representing x, y, and z components of points or distances or forces.

First, there is the question "If I represent vectors as arrows on paper, does the parallelogram method give a diagonal arrow that correctly represents the mathematical addition of the two input vectors?" Mathematics answers that as being true (I don't have a proof for that at hand, but I'm quite sure that one exists).

Then, physics found it useful to use vectors as representations of forces. So, forces can be added according to the laws established by math, e.g. numerically or by drawing the parallelogram.

Finally, there's the question if the vector addition of two forces has any physical meaning. Especially, can an object that experiences two combined forces be replaced by the same object experiencing one single force (being the vector addition of the individual forces), without changing the resulting movement of the object? That's a physics question, and is has been observed to be true many, many times - so often that we believe this to be universally true.

To sum it up:

  • Mathematics can prove that the parallelogram method correctly adds two vectors.
  • Physics found out (by observation) that vector addition is a proper method for describing the combined effect of individual forces.
  • $\begingroup$ So does that mean addition of forces is something empirical rather than proven from scratch? $\endgroup$
    – MSKB
    Jun 10, 2021 at 11:07
  • $\begingroup$ All physics is empirical, essentially. $\endgroup$ Jun 10, 2021 at 11:18
  • $\begingroup$ Yeah thats true but some equations are based on only observations and some are derived from other equations. So I wanted to know in which category does this law belong to? $\endgroup$
    – MSKB
    Jun 10, 2021 at 11:35

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