# What is the proof of parallelogram law of vector additon?

I have a question. How force can be defined as the magnitude of a ray? For instance, when we deal with two forces having magnitude 4 N and 5 N working 60 degree apart we solve the resultant by simple geometry taking the 4 &5 N as 4 and 5 unit length.Normally,we consider this 2 force as the magnitude of two adjacent sides of a parallelogram and then using very simple geometry we find the length of a diagonal of that parallelogram.(and we call it the resultant force which is directly a length.....5 N--converts into--5 metet or cm.weired!!!!)How it can be done??? • It is just the matter of definition. – Shreyansh Pathak May 18 '19 at 8:49
• It you want to draw a picture, use graph paper and a compass. Make off 5 squares for 5 N - one square for each Newton - use that as your ruler. Then construct the angle and the 4 N force, And finally measure the resultant force with the compass by comparing it to the 5 N line, – Cinaed Simson May 18 '19 at 9:29

## 4 Answers

There is no “proof” of how vectors add. Vectors are defined to add component-wise, which produces the parallelogram result.

That velocities, accelerations, forces, etc. in the real world can be described by mathematical vectors is based on observational evidence of physical systems. We model certain physical quantities as vectors, we make predictions based on this model (such as how two different forces “add” in Newtonian mechanics), and we observe that our model works.

As the saying goes, “The proof is in the pudding!” Nature doesn’t have to work this way.

Well, the parallelogram is not in a space where length is measured in meters if you want to see it that way. It is in a space where lengths are measured in Newtons. And the geometry is still Euclidean (i.e., the Pythagorean "theorem" is followed) so you can do all the geometry that you used to do with a space in which lengths were measured in meters.

Consider two vector, One along positive x axis.$$\vec{A}= a_1 \hat{i}$$ $$\vec{B}= b_1 \hat{i}+b_2\hat{j}$$$$\vec{A}+\vec{B}=(a_1+ b_1 )\hat{i}+b_2\hat{j}$$ $$| {\vec{A}+\vec{B}}|^2=(a_1+ b_1 )^2+b_2^2=a_1^2 +b_1^2+b_2^2+2a_1b_1$$$$| {\vec{A}+\vec{B}}|^2=|\vec{A}| ^2+|\vec{B}|^2+2\vec{A}.\vec{B}$$$$| {\vec{A}+\vec{B}}|^2=|\vec{A}| ^2+|\vec{B}|^2+2|\vec{A}||\vec{B}|cos\alpha$$ $$\alpha$$ is the angle between $$\vec{A}$$ and $$\vec{B}$$.The magnitude of the resultant vector is same as that of the square of length of the diagonal of a parallelogram with side $$\vec{A}$$ and $$\vec{B}$$.

The force is not defined as the magnitude of a ray, the "ray" only represents the force wherein its length represents the magnitude or amount of force and its direction is same as that of the force. Force is not same as length, they don't even have same dimensions.