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Q. 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 }$
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Solution:

We Know
$\because \overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }+\overrightarrow{ OE }+\overrightarrow{ OF }+\overrightarrow{ OG }+\overrightarrow{ OH }=\overrightarrow{0}$
By triangle law of vector addition, we can write
$\overrightarrow{ AB }=\overrightarrow{ AO }+\overrightarrow{ OB } ; \overrightarrow{ AC }=\overrightarrow{ AO }+\overrightarrow{ OC }$
$\overrightarrow{ AD }=\overrightarrow{ AO }+\overrightarrow{ OD } ; \overrightarrow{ AE }=\overrightarrow{ AO }+\overrightarrow{ OE } $
$\overrightarrow{ AF }=\overrightarrow{ AO }+\overrightarrow{ OF } ; \overrightarrow{ AG }=\overrightarrow{ AO }+\overrightarrow{ OG }$
$\overrightarrow{ AH }=\overrightarrow{ AO }+\overrightarrow{ OH }$
Now
$ \overrightarrow{ AB }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ AF }+\overrightarrow{ AG }+\overrightarrow{ AH }$
$=(7 \overrightarrow{ AO })+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }+\overrightarrow{ OE }+\overrightarrow{ OF }+\overrightarrow{ OG }+\overrightarrow{ OH } $
$=(7 \overrightarrow{ AO })+\overrightarrow{0}-\overrightarrow{ OA } $
$=(7 \overrightarrow{ AO })+\overrightarrow{ AO } $
$= 8 \overrightarrow{ AO }=8(2 \hat{ i }+3 \hat{ j }-4 \hat{ k })$
$=16 \hat{ i }+24 \hat{ j }-32 \hat{ k } $