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

In Fig. ABCDEF is a ragular hexagon. Prove that vec(AB) +vec(AC) +vec(AD) +vec(AE) +vec(AF) = 6 vec(AO) .

In a regular hexagon ABCDEF, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

ABCDE is a pentagon prove that vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EA)=vec0

ABCDEF is a regular hexagon. Show that : vec(OA)+vec(OB)+vec(OC)+vec(OD)+vec(OE)+vec(OF)=vec(0)

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

ABCDEF is a regular hexagon with point O as centre. The value of vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF) is

In a regular hexagon ABCDEF,vec AE

Assertion ABCDEF is a regular hexagon and vec(AB)=veca,vec(BC)=vecb and vec(CD)=vecc, then vec(EA) is equal to -(vecb+vecc) , Reason: vec(AE)=vec(BD)=vec(BC)+vec(CD) (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

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 ABCDE is a pentagon, then vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) is equal to