# 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

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 $$\vec{B}$$.