ABCDEF is a regular hexagon. Find the vector `vec AB + vec AC + vec AD + vec AE + vec AF` in terms of the vector `vec AD`

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

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

In a regular hexagon ABCDEF,vec AE

ABCDEF be a regular hexagon in the xy-plane and vec AB=4hat i then vec CD=

Let ABCDEF be a regular hexagon and vec AB=vec a,vec BC=vec b,vec CD=vec c then vec AE is

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

ABCDEF is a regular hexagon, Fig. 2 (c ) .65. What is the value of (vec (AB) + vec (AC) + vec (AD) + vec (AE) + vec (AF) ? .

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 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,vec AB=a,vec BC=b and vec CD=c. Then ,vec AE=