In an octagon $ABCDEFGH$ of equal side, what is the sum of $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} + \overrightarrow{AG} + \overrightarrow{AH}$, if $\overrightarrow{AO} = 2\hat i + 3\hat j - 4\hat k$?
Answer: (A) $16\hat i + 24\hat j - 32\hat k$
Write each vector through the centre: $\overrightarrow{AX} = \overrightarrow{AO} + \overrightarrow{OX}$. Summing over the seven vertices $X = B, \dots, H$:
$$\sum \overrightarrow{AX} = 7\overrightarrow{AO} + \left(\sum_{\text{all 8}}\overrightarrow{OX} - \overrightarrow{OA}\right)$$
For a regular octagon $\sum_{\text{all 8}}\overrightarrow{OX} = \vec 0$, and $-\overrightarrow{OA} = \overrightarrow{AO}$. So the sum is $8\overrightarrow{AO} = 16\hat i + 24\hat j - 32\hat k$.
Solution by Sreeraj P, M.Sc Physics