If ABCDEF is a regular hexagon then `vec(AD)+vec(EB)+vec(FC)` equals :

If ABCDEF be a regular hexagon then (vec(AD)+vec(EB)+vec(FC)) is equal to -

If ABCDEF is a regular hexagon,them vec AD+vec EB+vec FC equals 2vec AB b.vec 0 c.3vec AB d.4vec AB

If A B C D E F is a regular hexagon, them vec A D+ vec E B+ vec F C equals 2 vec A B b. vec0 c. 3 vec A B d. 4 vec A B

If ABCDEF is a regular hexagon with vec(AB) = vec(a) ,vec(BC ) = vec(b) then vec(CE) equals :

If ABCDEF is a regular hexagon , then A vec D + E vec B + F vec C equals

If ABCDEF is a regualr hexagon, then vec(AC) + vec(AD) + vec(EA) + vec(FA)=

If ABCDEF is a regular hexagon with vec(AB) = veca and vec(BC)= vecb, then vec(CE) equals

If ABCDEF is a regular hexagon, prove that vec(AC)+vec(AD)+vec(EA)+vec(FA)=3vec(AB)

If ABCDEF is a regular hexagon, prove that vec(AC)+vec(AD)+vec(EA)+vec(FA)=3vec(AB)