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Solution:

Let the particles collide at the point A (Fig.) whose position vector is  \( {{\vec{r}}_{3}} \) (say). If t be the time taken by each particle to reach at point A, from triangle law of vector addition:

 \( {{\vec{r}}_{3}}={{\vec{r}}_{1}}+{{\vec{v}}_{1}}t={{\vec{r}}_{2}}+{{\vec{v}}_{2}}t \)

So,  \( {{\vec{r}}_{1}}-{{\vec{r}}_{2}}=({{\vec{v}}_{2}}-{{\vec{v}}_{1}})t\,\,\,\,\,\,\,\,\,\,(1) \)

Therefore,  \( t=\frac{\left| {{{\vec{r}}}_{1}}-{{{\vec{r}}}_{2}} \right|}{\left| {{{\vec{v}}}_{2}}-{{{\vec{v}}}_{1}} \right|}\,\,\,\,\,\,\,\,\,(2) \)

From Eqs. (1) and (2)

 \( {{\vec{r}}_{1}}={{\vec{r}}_{2}}-({{\vec{v}}_{2}}-{{\vec{v}}_{1}})\frac{\left| {{{\vec{r}}}_{1}}-{{{\vec{r}}}_{2}} \right|}{\left| {{{\vec{v}}}_{2}}-{{{\vec{v}}}_{1}} \right|} \)

Or,  \( \frac{{{{\vec{r}}}_{1}}-{{{\vec{r}}}_{2}}}{\left| {{{\vec{r}}}_{1}}-{{{\vec{r}}}_{2}} \right|}=\frac{{{{\vec{v}}}_{2}}-{{{\vec{v}}}_{1}}}{\left| {{{\vec{v}}}_{2}}-{{{\vec{v}}}_{1}} \right|} \), which is the sought relationship.