Explain commutative law for vector addition.
Consider the vector $\vec{A}$ and $\vec{B}$. According to Parallelogram of vector addition we get the figure.
Here, $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{RQ}} ; \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{OR}}=\overrightarrow{\mathrm{PQ}}$
Draw a Parallelogram,
From $\Delta \mathrm{OPQ} \quad \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PQ}}$
$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{OQ}}$
From $\Delta$ $ORQ$ $\vec{B}+\vec{A}=\overrightarrow{O R}+\overrightarrow{R Q}$
$=\overrightarrow{\mathrm{PQ}}+\overrightarrow{\mathrm{OP}}[\because \overrightarrow{\mathrm{OR}}=\overrightarrow{\mathrm{PQ}} \text { and } \overrightarrow{\mathrm{RQ}}=\overrightarrow{\mathrm{OP}}]$
$\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{OQ}} \ldots \text { (ii) }$
From $(i)$ and $(ii)$ we get,
$\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{A}}$
Following sets of three forces act on a body. Whose resultant cannot be zero
The vectors $\vec{A}$ and $\vec{B}$ are such that
$|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$
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Two vectors $\vec A$ and $\vec B$ have equal magnitudes. The magnitude of $(\vec A + \vec B)$ is $‘n’$ times the magnitude of $(\vec A - \vec B)$. The angle between $ \vec A$ and $\vec B$ is
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
If $\vec{P}+\vec{Q}=\vec{P}-\vec{Q}$, then