- JEE Mains 2021
- Hard
- MCQ
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 & Solution
(B) \(16 \hat{i}+24 \hat{j}-32 \hat{k}\)
Step-by-step Solution
Detailed explanation
\(\because \overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }+\overrightarrow{ OE }+\overrightarrow{ OF }+\overrightarrow{ OG }+\overrightarrow{ OH }=\overrightarrow{0}\)
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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