Q. If $\sum_{i = 1}^{n}{\left| \overrightarrow{\mathbf{a}_{i}} \right| = \overrightarrow{\mathbf{0}}}$, where $\left| \overrightarrow{\mathbf{a}_{i}} \right| = 1\forall i$, then the value of $\sum_{1 \leq i <}^{}{\sum_{j \leq n}^{}{\overrightarrow{\mathbf{a}_{i}} \bullet {\overrightarrow{\mathbf{a}}}_{j}}}$ is
Explanation
(d)
$\because\ \sum_{i = 1}^{n}{{\overrightarrow{\mathbf{a}}}_{i} = \overrightarrow{\mathbf{0}}}$
$\therefore\left( \sum_{i = 1}^{n}{\overrightarrow{\mathbf{a}}}_{i}\ \right)\ \left( \ \sum_{i = 1}^{n}{\overrightarrow{\mathbf{a}}}_{j}\ \right) = \sum_{i = 1}^{n}{|{\overrightarrow{\mathbf{a}}}_{i}}\left. \ \right|^{2} + 2\sum_{1 \leq i <}^{}\ \sum_{j \leq n}^{}{\overrightarrow{\mathbf{a}}}_{i} \bullet \ {\overrightarrow{\mathbf{a}}}_{j}$
$\Rightarrow 0 = n + 2\ \sum_{1 \leq i <}^{}\ \sum_{j \leq n}^{}{\overrightarrow{\mathbf{a}}}_{i} \bullet \ {\overrightarrow{\mathbf{a}}}_{j}$
$\therefore\ \sum_{1 \leq i <}^{}\ \sum_{j \leq n}^{}{\overrightarrow{\mathbf{a}}}_{i} \bullet \ {\overrightarrow{\mathbf{a}}}_{j} = - \frac{n}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.