If α→+β→+γ→=kδ→ and β→+γ→+δ→=mα→,α→ and δ→ are non-collinear, then α→+β→+γ→+δ→ equal
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a
kα→
b
mδ→
c
0→
d
(k+m)γ→
answer is C.
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Detailed Solution
Given α→+β→+γ→=kδ→----iβ→+γ→+δ→=mα→----iiFrom(i), α→+β→+γ→+δ→=(k+1)δ→---iiiFrom (ii), α→+β→+γ→+δ→=(m+1)α→----ivFrom (iii) and (iv), we get (k+1)δ→=(m+1)α→-----vSince α→ is not parallel to δ→, From (v),k+1=0and m+ 1=0 From (iii), α→+β→+γ→+δ→=0→