Proof for parallelogram law of vector addition

Question

The Statement of Parallelogram law of vector addition is,

If two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of two vectors is given by the vector that is a diagonal passing through the point of contact of two vectors.

But how do we justify that the resultant is along the diagonal? Is it based on experimental evidence, or is it something that can be proved?

_{I know, the question might sound pretty obvious, but I'm new to this stuff. :)}

Two things. One, this is a mathematical question, so experimental evidence is not really a proof. Second, how do you define vector addition? — Javier, Aug 22 '15 at 17:14
Three things. First, you din't get me. Secondly, derivation of formula relating the two vectors and resultant can be derived from mathematics. Thirdly, vector addition is a definition based on experiments performed in real life. Vector addition is not a definition, it's a law. It's the generalization on the basis of which we can cloude that, when two particles are tied by two ropes making an angle theta w.r.t. to one another, the particle moves along the diagonal. Now, why the particle has to move along the diagonal, not any other direction? U got me now? — Anubhab Das, Aug 22 '15 at 17:40
Vectors are a purely mathematical construct. You seem to be asking about the superposition principle, which is the physical fact that forces add like vectors. Don't get it backwards! You seem to be implying that vectors add like they do because of the superposition principle, but it's the other way around: we represent forces with vectors because addition of forces is just like addition of vectors. In any case, I think you should make your question clearer, because apparently no one understood what you were asking. — Javier, Aug 22 '15 at 17:52
Yes exactly! How are we representing forces with vectors because addition of forces is just like addition of vectors? How can we justify that? — Anubhab Das, Aug 22 '15 at 17:56
This question then seems to mean "Why is the superposition principle valid for vectors (positions, velocities, forces)?" The mathematical proof should just be geometry. — jjack, Aug 22 '15 at 20:12

score 7 · Answer 1 · answered Aug 22 '15 at 18:14

First, an important remark. Vectors are mathematical objects, just like numbers, matrices, tensors, groups, manifolds, etc. These things are a part of physics because they happen to be a good model for the physical world, but they are not a part of it. Vectors are not something that exists in physical reality, whatever that might be. So let's separate the math and the physics.

You can define a vector as an ordered tuple of real numbers, or as an arrow in space; it doesn't make any difference because the definitions are equivalent. In the case of an ordered tuple of numbers, addition is defined component wise. If you define a vector as an arrow, then addition is defined either with the parallelogram or the triangle law. These notions of addition can be shown to be equivalent, but I don't think that's the point of your question.

On the other hand, we find that a good way to model the way things move is with forces. Suppose you're floating in empty space near a star. You will feel the force of gravity pulling you towards the star. We can model this force as a vector, where the magnitude and direction of the vector correspond to those of the force, and you can use your knowledge of this force (so far there's only one) to calculate your movement using Newton's law.

Now suppose the star magically disappears and is replaced by a different one, maybe with a different mass and a little bit to the side. Now the force (and its associated vector) points to the new star, and you can again use Newton's law. But what if both stars are present? So far all we know is that it's convenient to represent a force by a vector, but we have no idea what to do when there are various forces. It is found experimentally that when there are multiple forces acting on you, it's exactly as if there were a single force, whose vector is given by the vector sum of the individual forces' vectors.

Although this sounds obvious, it is a nontrivial statement. A priori, who says that multiple forces acting at once are equivalent to a single force? It happens to be true, and that is the reason we say that forces are vectors.

Of course, I've used forces as an example, but the superposition principle also applies to velocities, positions, electromagnetic fields, etc. Because it is found experimentally that the superposition principle is valid, we use vectors to model all those things.

score 3 · Answer 2 · edited Jun 12 '20 at 10:38

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Addition of Vectors basically found their origin from the Triangular Law of Vector Addition

The triangle law of vectors basically is a process that allows one to take two vectors, draw them proportional to each other, connect them head to tail, then draw the resultant vector as a result of the third side that is missing.

The Parallelogram law is just a furthermore explanation of Triangular law,

If two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.

The proof for the resultant vector in Parallelogram addition is as follows,

Consider a parallelogram $OABC$ as shown in the figure,

Let $P \;\&\; Q$ be two adjacent sides of parallelogram, and $R$ be the resultant vector obtained by addition of vectors $P \;\&\; Q$,

Now, drop a perpendicular from $C$ on $OA$ so that they meet at $A$.

From right angled triangle $\Delta OCD \;,\; OC^2=OD^2+DC^2$

$$[OD=OA+AD]$$ $$R^2=(OA+AD)^2+DC^2$$ $$R^2=OA^2+AD^2+2OA.AD+DC^2$$ From $\Delta$ADC, $AC^2=AD^2+DC^2$

and also $\cos\theta= \frac{AD}{AC}$

And hence $R^2= OA^2+AC^2+2 OA.AC \cos\theta$

And substituting A and B $$R^2=A^2+B^2+2 A.B \cos\theta$$

edited Jun 12 '20 at 10:38

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answered Aug 22 '15 at 16:18

axelonet

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You din't get my question. It was quite obvious anyways. :P I meant, you used the ||gm law to solve this, right? How'd you justify that? In fact, you got the ||gm law as a corollary of Triangle law. How would you justify the triangle law? In fact, if you prove triangle law in this manner, it would be a matter of circular reasoning. I am asking the absolute reason for validity of triangle law. – Anubhab Das Aug 22 '15 at 16:23
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Go read Euclid's elements. If that doesn't answer your question, then you're expecting too much. – Aug 22 '15 at 18:04
@AnubhabDas The fundamental reason that we treat physical quantity as obeying the mathematical properties of vectors is that doing so generates correct predictions. That's true of all science all the time. "Proofs" are a mathematicians tool; scientists use empirical tests. – dmckee --- ex-moderator kitten Aug 22 '15 at 19:22
@dmckee I guess, that does answer my question. – Anubhab Das Nov 30 '15 at 06:23

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

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Let $\mathbf{a}\;,\; \mathbf{b}$ be two vectors & let these constitute the sides of a parallelogram such that they are co-initial from one of the vertices of a parallelogram. So, considering assumption that the law be true(this is what we'll judge now whether by taking the assumption for the validity of the law, we can prove something which is earlier ascertained by elementary geometry), the two diagonals are $\mathbf{a + b}\;,\;\mathbf{b -a}$. Sum of the squares of the diagonals is $\left| \mathbf{a + b} \right|^2 + \left| \mathbf{b - a} \right|^2$. Now, from Euclidean Geometry, we get from Parallelogram law which states that,

The sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals.$^1$

So, the sum of the squares of the diagonals i.e. $\left| \mathbf{a + b} \right|^2 + \left| \mathbf{b - a} \right|^2 $ must be equal to the sum of the squares of the sides $2(|\mathbf{a}|^2 + |\mathbf{b}|^2)$.

This can be proved by inner product or for this case dot product.

$$\begin{align} \left| \mathbf{a + b} \right|^2 + \left| \mathbf{b - a} \right|^2 = (\mathbf{a + b}) \cdot (\mathbf{a + b}) + (\mathbf{b - a}) \cdot (\mathbf{b - a}) \end{align} \implies \left| \mathbf{a + b} \right|^2 + \left| \mathbf{b - a} \right|^2 = 2\mathbf{a}\cdot \mathbf{a} +2\mathbf{b}\cdot \mathbf{b} =2(|\mathbf{a}|^2 + |\mathbf{b}|^2)$$ which makes our assumption that the law is true absolutely true.

^{$^1$ Courtsey: Wikipedia-Parallelogram law; If you want to deduce the parallelogram law in elementary geometry, then check Is the parallelogram law a theorem or an axiom?.}

edited Apr 13 '17 at 12:20

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answered Aug 22 '15 at 16:28

Yeah, I got your point. But, I've one major confusion. U assumed a statement to be true and then by manipulation, you reached a more general statement which is true, thus making your assumption true. Is this valid? – Anubhab Das Aug 22 '15 at 17:46
@Anubhab Das: Which assumption are you talking of? – Aug 22 '15 at 17:48
At first, you assumed ||gm law to be true right? – Anubhab Das Aug 22 '15 at 17:49
@Anubhab Das: See what I've done: I've assumed the law to be valid & if the assumption is true that the law is valid, then it must come to the same conclusion, that has been proved in other way i.e. by using elementary geometry. If the law wasn't valid, then our assumption which would be incorrect then, would bring us to such a conclusion which would be incorrect. So, proving that the assumption of the validity of the law brings forth to a correct conclusion means our assumption is correct. – Aug 22 '15 at 17:56
Yeah, I got your point. :) Thanks! Actually it's tough to visualize physical quantities as lengths and how laws of geometry holds for them, as well too! – Anubhab Das Aug 22 '15 at 17:59
@Anubhab Das: Feel free to ask, if you've any problem. I've removed your downvote since the question was good. Many students of your age would cram it but would never ask how it came. So, good luck :) – Aug 22 '15 at 18:02
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"If the law wasn't valid, then our assumption which would be incorrect then, would bring us to such a conclusion which would be incorrect. So, proving that the assumption of the validity of the law brings forth to a correct conclusion means our assumption is correct." That is entirely wrong. – Aloizio Macedo Sep 01 '18 at 04:44
Perhaps anyone can help me in this. https://math.stackexchange.com/q/3082098/474166 – prashant sharma Feb 05 '19 at 10:41
You basically proved the statement "if the law is true then the sum of the squares $\cdots$ of the two diagonals". You cannot conclude the truth value of the assumption from this. – Jamāl Mar 14 '20 at 17:27

score 0 · Answer 4 · answered Aug 22 '15 at 15:41

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If you hold a mirror to the two vectors, you get a mirror image of both. Turn the mirror image upside down, and the resultant vector of the upside down mirror image is the same magnitude as the original. Combine the upside-down mirror image with the original and you have a parallelogram. This is a proof of magnitude, only.

answered Aug 22 '15 at 15:41

That's just one of the case. How'd you generalize for all directional physical quantities? – Anubhab Das Aug 22 '15 at 16:23

score 0 · Answer 5 · answered Sep 28 '19 at 18:28

This was already answered, but I figured I'd break it down into more intermediate steps for those who, like me, want to see all of the work.

First, note that:

$||\vec{x}|| = \sqrt{x_1^2 + ... + x_n^2}$

$||\vec{x}||^2 = x_1^2 + ... + x_n^2 = \vec{x} \cdot \vec{x}$

Now, let $\vec{a}$ and $\vec{b}$ be two vectors of size $n$ (i.e., they are both $\in R^n $.

Substituting symbolically using $\vec{a} + \vec{b} = \vec{x}$ in the above, we find that:

$||\vec{a} + \vec{b}||^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = \vec{a}\cdot\vec{a} + \vec{a}\cdot \vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot \vec{b} = ||\vec{a}||^2 + 2(\vec{a}\cdot\vec{b}) + ||\vec{b}||^2$

Repeat these steps using the same reasoning, this time for $\vec{a} - \vec{b}$, and you'll find that the $2(\vec{a}\cdot\vec{b})$ term will disappear, while the other terms will double.

Proof for parallelogram law of vector addition

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