215
Question
If →a=ˆi+2^j+3^k,→b=2^i+3^j+^k,→c=3^i+^j+2^k are vectors satisfying
α→a+β→b+γ→c=−3(^i−^k), then the ordered triplet (α,β,γ) is
If →a=ˆi+2^j+3^k,→b=2^i+3^j+^k,→c=3^i+^j+2^k are vectors satisfying
α→a+β→b+γ→c=−3(^i−^k), then the ordered triplet (α,β,γ) is
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Solution
The correct option is A (2,−1,−1)
Given : →a=ˆi+2^j+3^k,→b=2^i+3^j+^k,→c=3^i+^j+2^kConsider, α→a+β→b+γ→c=(α^i+2α^j+3α^k)+(2β^i+3β^j+β^k)+(3γ^i+γ^j+2γ^k)
=(α+2β+3γ)^i+(2α+3β+γ)^j+(3α+β+2γ)^k
=−3^i+3^k ..... [Given]By comparing coefficients, we get α+2β+3γ=−3 ...... (i)
2α+3β+γ=0 ...... (ii)
3α+β+2γ=3 ...... (iii)Solving (i) and (ii) we get−5α−7β=−3 .... (iv)Solving (ii) and (iii) we get−α−5β=3 ...... (v)Then solving (iv) and (v) we getβ=−1 and α=2Substituting α and β in (i) we getγ=−1Hence from the above expressions, we get (α,β,γ)=(2,−1,−1).
Given : →a=ˆi+2^j+3^k,→b=2^i+3^j+^k,→c=3^i+^j+2^k
Consider, α→a+β→b+γ→c
=(α^i+2α^j+3α^k)+(2β^i+3β^j+β^k)+(3γ^i+γ^j+2γ^k)
=(α+2β+3γ)^i+(2α+3β+γ)^j+(3α+β+2γ)^k
=−3^i+3^k ..... [Given]
=(α+2β+3γ)^i+(2α+3β+γ)^j+(3α+β+2γ)^k
=−3^i+3^k ..... [Given]
By comparing coefficients, we get
α+2β+3γ=−3 ...... (i)
2α+3β+γ=0 ...... (ii)
3α+β+2γ=3 ...... (iii)
2α+3β+γ=0 ...... (ii)
3α+β+2γ=3 ...... (iii)
Solving (i) and (ii) we get
−5α−7β=−3 .... (iv)
Solving (ii) and (iii) we get
−α−5β=3 ...... (v)
Then solving (iv) and (v) we get
β=−1 and α=2
Substituting α and β in (i) we get
γ=−1
Hence from the above expressions, we get
(α,β,γ)=(2,−1,−1).
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