In an octagon $A B C D E F G H$ of equal side, what is the sum of $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AF}}+\overrightarrow{\mathrm{AG}}+\overrightarrow{\mathrm{AH}}$, if, $\overrightarrow{\mathrm{AO}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$

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

(A) $16 \hat{i}+24 \hat{j}+32 \hat{k}$

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

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

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Vector

In an octagon

A B C D E F G H

of equal side, what is the sum of

AB + AC + AD + AE + AF + AG + AH

, if,

AO = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}

Answer: (D)

Route every vector through the centre,

A X = A O + O X

for each vertex

X

. Summing over the seven vertices from

B

H

\sum A X = 7 A O + X = B \sum H O X

In a regular octagon the eight position vectors drawn from the centre to the vertices cancel in pairs, so their total is zero and

\sum_{X = B}^{H} O X = - O A = A O

.

The required sum is therefore

8 A O = 8 (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 16 \hat{i} + 24 \hat{j} - 32 \hat{k}

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option