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Question
If →a=→i+2→j+3→k,→b=2→i+3→j+→k,→c=3→i+→j+2→k and α→a+β→b+γ→c=−3(ˆi−ˆk), then the order triplet (α,β,γ) is
If →a=→i+2→j+3→k,→b=2→i+3→j+→k,→c=3→i+→j+2→k and α→a+β→b+γ→c=−3(ˆi−ˆk), then the order triplet (α,β,γ) is
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Solution
The correct option is C (2,−1,−1)
Solution -→a=^i+2^j+3^k→b=2^i+3^l+^k→c=3^i+^j+2^kα→a+β→b+γ→c=−3^i+3^k
α+2β+3γ=−3___(1)2α+3β+γ=0____(2)3α+β+2γ=3______(3)Solve for α,β,γα=2β=−1γ=−1(2,−1,−1) A is correct
Solution -
→a=^i+2^j+3^k
→b=2^i+3^l+^k
→c=3^i+^j+2^k
α→a+β→b+γ→c=−3^i+3^k
α+2β+3γ=−3___(1)
2α+3β+γ=0____(2)
3α+β+2γ=3______(3)
Solve for α,β,γ
α=2
β=−1
γ=−1
(2,−1,−1)
A is correct
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