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Question

# Let →a,→b,→c be non-coplanar vectors such that →p=→a+→b−→c,→q=→b+→c−→a,→r=→c+→a−→b,→d=2→a−3→b+4→c. If →d=α→p+β→q+γ→r, then

A
α=γ
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B
α+γ=3
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C
α+β+γ=3
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D
β+γ=2
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Solution

## The correct option is C α+β+γ=32→a−3→b+4→c=→d=α→p+β→q+γ→rsubstituting for →p=→a+→b−→c,→q=→b+→c−→a,→r=→c+→a−→b we get⇒→d=α(→a+→b−→c)+β(→b+→c−→a)+γ(→c+→a−→b)By removing the coefficients of →a,→b and →c we get=(α−β+γ)→(a)+(α+β−γ)→(b)+(−α+β+γ)→c Comparing the left hand side with right hand side, we getα−β+γ=2 α+β−γ=−3 and −α+β+γ=4∴α−β+γ+α+β−γ−α+β+γ=2−3+4⇒α+β+γ=3

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