ABC is an equilateral triangle. Length of each side is 'a' and centroid is point O.Find
(i) `vec(AB)+vec(BC)+vec(CA)=?`
(ii) `vec(OA)+vec(OB)+vec(OC)=?`
(iii) If `|vec(AB)+vec(BC)+vec(AC)|=` na then n = ?

(iv) If `vec(AB) +vec(AC) =n vec(AO)` then n = ?

Explanation

(i) `vec(AB), vec(BC) & vec(CA)` form a closed triangle in the same order. Therefore their resultant is zero.
`therefore vec(AB) +vec(BC)+vec(CA)=vec0`
(ii) `vec(OA), vec(OB) and vec(OC)` are three vectors of equal magnitude which are separated by equal angles of 120°. Therefore their resultant is zero.
`therefore vec(OA)+vec(OA)+vec(OC)=vec0`
(iii) Here `vec(AB)+vec(BC)=vec(AC)`
`therefore (vec(AB)+vec(BC))+vec(AC)=vec(AC)+vec(AC)=2vec(AC)`
`therefore |vec(AB)+vec(BC)+vec(AC)|=|2vec(AC)|=2|vec(AC)|=2a`
`therefore n=2`
(iv) `vec(AB)+vec(AC)=?`
here `vec(AB)=vec(AO)+vec(OB)`
and `vec(AC)=vec(AO)+vec(OC)`
Adding the above two equations,
`therefore vec(AB)+vec(AC)=2vec(AO)+vec(OB)+vec(OC)" " `......(1)
but `vec(OA)+vec(OB)+vec(OC)=vec0`
[from (ii) part of question]
`therefore vec(OB)+vec(OC)= - vec(OA)= vec(AO)" "`....(2)
using equations (1) and (2)
`vec(AB)+vec(AC)=2vec(AO)+vec(AO)`
`vec(AB)+vec(AC)=3vec(AO) " " n=3`