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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
α+γ=3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
α+β+γ=3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
β+γ=2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cα+β+γ=3 2→a−3→b+4→c=→d=α→p+β→q+γ→r substituting 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 get α−β+γ=2α+β−γ=−3 and −α+β+γ=4 ∴α−β+γ+α+β−γ−α+β+γ=2−3+4 ⇒α+β+γ=3