Given that →a=^i+^j+^k,−→b=−^i+^j−^k,→c=^i+^j−^k,The values of the following are i. (→a.→b)+(→b.→c)+(→c.→a) ii. (→a.→c)→c+(→c.→b)→a
A
i. 3 ii. 2^i
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B
i. 6 ii. −2^i
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C
i. 1 ii. −2^k
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D
i. 1 ii. 2^j
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Solution
The correct option is C i. 1 ii. −2^k i.(→a.→b)=(^i+^j+^k).(^i−^j+^k)=1−1+1=1 (→b.→c)=(^i−^j+^k).(^i+^j−^k)=1−1−1=−1 (→c.→a)=(^i+^j−^k).(^i+^j+^k)=1+1−1=1 ⇒(→a.→b)+(→b.→c)+(→c.→a)=1
ii. (→a.→c)→c=1(^i+^j−^k)=(^i+^j−^k) (→c.→b)→a=−1(^i+^j+^k)=−^i−^j−^k ⇒(→a.→c)→c+(→c.→b)→a=−2^k