Answer:
Let
a particle be displaced from location P to Q. Fig. 2(c).53. So the displacement
vector
\[=\overset{\to
}{\mathop{PQ}}\,=\overset{\to }{\mathop{r}}\,\]
With
respect to origin O, let, \[(\overset{\to }{\mathop{OP}}\,)=\overset{\to
}{\mathop{{{r}_{1}}}}\,\] and \[(\overset{\to }{\mathop{OQ}}\,)=\overset{\to
}{\mathop{{{r}_{2}}}}\,\].
With
respect to origin \[O'\], Let
\[(\overset{\to
}{\mathop{O'P}}\,)=\overset{\to }{\mathop{{{r}_{1}}'}}\,\] and \[(\overset{\to
}{\mathop{O'Q}}\,)=\overset{\to }{\mathop{{{r}_{2}}'}}\,\]
Using
triangle law of vectors addition, we have
\[\overset{\to
}{\mathop{{{r}_{1}}}}\,+\overset{\to }{\mathop{r}}\,=\overset{\to
}{\mathop{{{r}_{2}}}}\,\]
or
\[\overset{\to }{\mathop{{{r}_{{}}}}}\,=\overset{\to
}{\mathop{{{r}_{2}}}}\,-\overset{\to }{\mathop{{{r}_{1}}}}\,\]
Also,
\[\overset{\to }{\mathop{{{r}_{{}}}}}\,+\overset{\to
}{\mathop{{{r}_{1}}}}\,'=\overset{\to }{\mathop{{{r}_{2}}}}\,'\]
or
\[\overset{\to }{\mathop{r}}\,=\overset{\to
}{\mathop{{{r}_{2}}'}}\,-\overset{\to }{\mathop{{{r}_{1}}'}}\,\]
so,
\[\overset{\to }{\mathop{r}}\,=\overset{\to
}{\mathop{{{r}_{2}}}}\,-\overset{\to }{\mathop{{{r}_{1}}}}\,=\overset{\to
}{\mathop{{{r}_{2}}'}}\,-\overset{\to }{\mathop{{{r}_{1}}'}}\,\]
This
shows that the displacement vector \[\overset{\to }{\mathop{r}}\,\] is
independent of the choice of origin.
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