A) \[{{a}_{1}}+{{a}_{2}}+....+{{a}_{n}}=n\]
B) \[{{a}_{1}}+{{a}_{2}}+....+{{a}_{n}}=1\]
C) \[{{a}_{1}}+{{a}_{2}}+....+{{a}_{n}}=0\]
D) \[{{a}_{1}}={{a}_{2}}={{a}_{3}}=....={{a}_{n}}=0\]
Correct Answer: C
Solution :
[c] Given \[{{a}_{1}}{{\vec{r}}_{1}}+{{a}_{2}}{{\vec{r}}_{2}}+.....+{{a}_{n}}{{\vec{r}}_{n}}=0\] Now \[\vec{a}+{{\vec{r}}_{1}}'={{\vec{r}}_{1}}\] and so on Hence, \[{{a}_{1}}(\vec{a}+{{\vec{r}}_{1}})+{{a}_{2}}(\vec{a}+{{\vec{r}}_{2}})+....+{{a}_{n}}(\vec{a}+{{\vec{r}}_{n}}')=0\] \[{{a}_{1}}{{\vec{r}}_{1}}'+{{a}_{2}}{{\vec{r}}_{2}}'+....+{{a}_{n}}{{\vec{r}}_{n}}'+\vec{a}({{a}_{1}}+{{a}_{2}}+....+{{a}_{n}})=0\] Hence, \[{{a}_{1}}{{\vec{r}}_{1}}'+{{a}_{2}}{{\vec{r}}_{2}}'+....+{{a}_{n}}{{\vec{r}}_{n}}'=0\] if \[{{a}_{1}}+{{a}_{2}}\] \[+....+{{a}_{n}}=0\].You need to login to perform this action.
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