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If `r_(1),r_(2) and r_(3)` are the position vectors of three collinear points and scalars l and m exists such that `r_(3)=lr_(1)+mr_(2)`, then show that l+m=1.

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Let A,B and C be the three points whose position vectors referred to O are `r_(1),r_(2)` and `r_(3)` respectively.
`AB=OB-OA=r_(2)r_(1)`.
`BC=OC-OB=r_(3)-r_(2)`
Now, if A,B and C are collinear points, then AB and AC are the same line and BC=`lamda(AC)`
`implies(r_(3)-r_(2))=lamda(r_(2)-r_(1))`
`implies r_(3)=-lamdar_(1)+(lamda+1)r_(2)`
`therefore r_(3)=-lamdar_(1)+mr_(2)`
where, `l=-lamda and m=lamda+1`
`implies l+m=-lamda+(lamda+1)=1`
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