If α→+β→+γ→=aδ→ and β→+γ→+δ→=bα→,α→ and δ→ are non-collinear, then α→+β→+γ→+δ→ equal
aα→
bδ→
0→
(a+b)γ→
Given α→+β→+γ→=aδ→----i
β→+γ→+δ→=bα→----ii
From(i), α→+β→+γ→+δ→=(a+1)δ→---iii
From (ii), α→+β→+γ→+δ→=(b+1)α→----iv
From (iii) and (iv), we get
(a+1)δ→=(b+1)α→-----v
Since α→ is not parallel to δ→,
From (v),a+1=0andb+ 1=0
From (iii), α→+β→+γ→+δ→=0→