Question

In an octagon ABCDEFGH of equal side, what is the sum of $\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF}+\overline{AG}+\overline{AH}$ if, $\overline{AO}=2\hat{i}+3\hat{j}-4\hat{k}$ ?

• A. $-16\hat{i}-24\hat{j}+32\hat{k}$
• B. $16\hat{i}+24\hat{j}-32\hat{k}$
• C. $16\hat{i}+24\hat{j}+32\hat{k}$
• D. $16\hat{i}-24\hat{j}$

Answer 1

Answer: (B)

$16\hat{i}+24\hat{j}-32\hat{k}$

Answer 2

Let O be the center of the regular octagon ABCDEFGH. The sum of vectors from the center O to the vertices is zero: $\sum_{i=A}^H \overline{Oi} = \vec{0}$.

We want to calculate $S = \sum_{i=B}^H \overline{Ai}$.

Using the property $\overline{Ai} = \overline{Oi} - \overline{OA}$, we have $S = \sum_{i=B}^H (\overline{Oi} - \overline{OA}) = \sum_{i=B}^H \overline{Oi} - \sum_{i=B}^H \overline{OA}$.

$\sum_{i=B}^H \overline{Oi} = (\sum_{i=A}^H \overline{Oi}) - \overline{OA} = \vec{0} - \overline{OA} = -\overline{OA}$.

$\sum_{i=B}^H \overline{OA} = 7 \overline{OA}$.

So, $S = -\overline{OA} - 7\overline{OA} = -8\overline{OA}$.

Given $\overline{AO} = 2\hat{i}+3\hat{j}-4\hat{k}$, we have $\overline{OA} = -\overline{AO} = -(2\hat{i}+3\hat{j}-4\hat{k}) = -2\hat{i}-3\hat{j}+4\hat{k}$.

$S = -8(-2\hat{i}-3\hat{j}+4\hat{k}) = 16\hat{i}+24\hat{j}-32\hat{k}$.

Alternatively, using the property that for a regular n-sided polygon with center O, the sum of vectors from a vertex A to all other vertices is $n \overline{AO}$.

For an octagon (n=8), the sum of vectors from A to all other vertices (B, C, D, E, F, G, H) is $8 \overline{AO}$.

Given $\overline{AO} = 2\hat{i}+3\hat{j}-4\hat{k}$.

The sum is $8(2\hat{i}+3\hat{j}-4\hat{k}) = 16\hat{i}+24\hat{j}-32\hat{k}$.