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Question
If →a=^1+2^j+3^k,→b=2^i+3^j+^k,→c=3^i+^j+2^k and α→a+β→b+γ→c=−3(^i−^k)⋅ Then the triplet (α,β,γ) is
If →a=^1+2^j+3^k,→b=2^i+3^j+^k,→c=3^i+^j+2^k and α→a+β→b+γ→c=−3(^i−^k)⋅ Then the triplet (α,β,γ) is
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Solution
The correct option is A (2,−1,−1)
→a=^ı+2^ȷ+3ˆk→b=2^ı+3^ȷ+ˆk→c=3^ı+^ȷ+2ˆkα→a+β→b+γ→c=−3ˆi+3ˆk⇒α^i+2α^ȷ+3α^k+2β^i+3β^ȷ+β^k=−3^ı+3^k+3γ^ı+γ^ȷ+2Υ^k⇒^ı(α+2β+3γ)+^ȷ(2α+3β+γ)=−3^ı+3^k+^k(3α+β+2γ)Comparing both sidesα+2β+3γ=−3−(1)2α+3β+γ=0−(2)3α+β+2x=3−(3)α+2β=−3−3γ−(4)multiply equation (4) by 2 and subtracted by (2)2α+3β−(2α+4β)=−γ−(−6−6γ)⇒3β−4β=−γ+6+6γ⇒−β=5γ+6⇒β=−5γ−6−(5)put equation (5) in equation (4)α+2(−5γ−6)=−3−3γ⇒α−10γ−12=−3−3γ⇒α=7γ+9−(6)put value of β and α from equations(5) and (6) in equation (3)3(7x+9)+(−5y−6)+2γ=3⇒21x+27−5γ−6+2y=3⇒18γ=−18⇒x=−1β=−5γ−6=−5(−1)−6=−1α=7x+9=7(−1)+9=2∴α=2;β=−1;γ=−1(α,β,γ)≡(2,−1,−1)Answer: option (A)
→a=^ı+2^ȷ+3ˆk
→b=2^ı+3^ȷ+ˆk
→c=3^ı+^ȷ+2ˆk
α→a+β→b+γ→c=−3ˆi+3ˆk
⇒α^i+2α^ȷ+3α^k+2β^i+3β^ȷ+β^k=−3^ı+3^k+3γ^ı+γ^ȷ+2Υ^k
⇒^ı(α+2β+3γ)+^ȷ(2α+3β+γ)=−3^ı+3^k+^k(3α+β+2γ)
Comparing both sides
α+2β+3γ=−3−(1)
2α+3β+γ=0−(2)
3α+β+2x=3−(3)
α+2β=−3−3γ−(4)
multiply equation (4) by 2 and subtracted by (2)
2α+3β−(2α+4β)=−γ−(−6−6γ)
⇒3β−4β=−γ+6+6γ
⇒−β=5γ+6⇒β=−5γ−6−(5)
put equation (5) in equation (4)
α+2(−5γ−6)=−3−3γ
⇒α−10γ−12=−3−3γ
⇒α=7γ+9−(6)
put value of β and α from equations
(5) and (6) in equation (3)
3(7x+9)+(−5y−6)+2γ=3
⇒21x+27−5γ−6+2y=3
⇒18γ=−18⇒x=−1
β=−5γ−6=−5(−1)−6=−1
α=7x+9=7(−1)+9=2
∴α=2;β=−1;γ=−1(α,β,γ)≡(2,−1,−1)
Answer: option (A)
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