Question

If $\sum _{i=1}^n{\overrightarrow{a}}_i=0$ where $\left|{\overrightarrow{a}}_i\right|=1,\forall i,$ then the value of $\sum _{1\le i<j\le n}^{}\sum _{}^{}{\overrightarrow{a}}_i.{\overrightarrow{a}}_j$ is equal to

• A. $-\frac{n}{2}$
• B. $-n$
• C. $\frac{n}{2}$
• D. $n$

Answer 1

The correct option is A $-\frac{n}{2}$

$\left(\sum _{i=1}^n{a}_i\right).\left(\sum _{i=1}^n{a}_i\right)=\sum _{i=1}^n{\left|a_i\right|}^2+2\sum _{1\le i<j\le n}^{}\sum _{}^{}{a}_i.a_j$

$\Rightarrow 0=n+2\sum _{1\le i<j\le n}^{}\sum _{}^{}{a}_i.a_j$

$\Rightarrow \sum _{1\le i<j\le n}^{}\sum _{}^{}{a}_i.a_j=-\frac{n}{2}$