Explain the parallelogram method for vector addition. Also explain that this is comparable to triangle method.
$\vec{A}$ and $\vec{B}$ are to be added as shown in figure $(a).$
Select a point $\mathrm{O}$ as shown in figure $(b)$.
Represent $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ such that their lengths and directions remain unchanged and their tails remain at $\mathrm{O}$.
Draw a parallelogram $\square^{\mathrm{m}}$ OPSQ in which $\vec{A}$ and $\vec{B}$ are adjacent sides of it. Draw a diagonal OS from $\mathrm{O}$.
Vector $\overrightarrow{\mathrm{OS}}$ represent resultant vector of addition of $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$.
$\overrightarrow{\mathrm{OS}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}} \quad \therefore \overrightarrow{\mathrm{R}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$
Triangle method for vector addition is shown in figure $(c)$.
If is clear that both methods give equal resultant vector. Hence, both methods are comparable to each other.
Here, magnitude of resultant vector $\overrightarrow{\mathrm{R}},|\overrightarrow{\mathrm{R}}| \leq|\overrightarrow{\mathrm{A}}|+|\overrightarrow{\mathrm{B}}|$
Two forces of $12 \,N$ and $8 \,N$ act upon a body. The resultant force on the body has maximum value of........$N$
Assertion $A$ : If $A, B, C, D$ are four points on a semi-circular arc with centre at $'O'$ such that $|\overrightarrow{{AB}}|=|\overrightarrow{{BC}}|=|\overrightarrow{{CD}}|$, then $\overrightarrow{{AB}}+\overrightarrow{{AC}}+\overrightarrow{{AD}}=4 \overrightarrow{{AO}}+\overrightarrow{{OB}}+\overrightarrow{{OC}}$
Reason $R$ : Polygon law of vector addition yields $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{A D}=2 \overrightarrow{A O}$
In the light of the above statements, choose the most appropriate answer from the options given below
Two forces ${F_1} = 1\,N$ and ${F_2} = 2\,N$ act along the lines $x = 0$ and $y = 0$ respectively. Then the resultant of forces would be
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