સમાન બાજુ ધરાવતાં અષ્ટકોણ ABCDEFGH માટે overr | vedclass

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Question

We know,

$\because \overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }+\overrightarrow{ OE }+\overrightarrow{ O

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$\overrightarrow{ AB }=\overrightarrow{ AO }+\overrightarrow{ OB } ; \overrightarrow{ AC }=\overrightarrow{ AO }+\overrightarrow{ OC }$

$\overrightarrow{ AD }=\overrightarrow{ AO }+\overrightarrow{ OD } ; \overrightarrow{ AE }=\overrightarrow{ A O }+\overrightarrow{ OE }$

$\overrightarrow{ AF }=\overrightarrow{ AO }+\overrightarrow{ OF } \quad ; \overrightarrow{ AG }=\overrightarrow{ AO }+\overrightarrow{ OG }$

$\overrightarrow{ AH }=\overrightarrow{ A O }+\overrightarrow{ OH }$

Now

$\overrightarrow{ AB }+\overrightarrow{ AC }+\overrightarrow{ AD }+\overrightarrow{ AE }+\overrightarrow{ AF }+\overrightarrow{ AG }+\overrightarrow{ AH }$

$=(7 \overrightarrow{ AO })+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }+\overrightarrow{ OE }+\overrightarrow{ OF }+\overrightarrow{ OG }+\overrightarrow{ OH }$

$=(7 \overrightarrow{ AO })+\overrightarrow{0}-\overrightarrow{ OA }$

$=(7 \overrightarrow{ AO })+\overrightarrow{ AO }$

$=8 \overrightarrow{ A O }=8(2 \hat{ i }+3 \hat{ j }-4 \hat{ k })$

$=16 \hat{i}+24 \hat{j}-32 \hat{k}$

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