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

In a regular hexagon ABCDEF, vec(AE)

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

Let O be the centre of the regular hexagon ABCDEF then find vec(OA)+vec(OB)+vec(OD)+vec(OC)+vec(OE)+vec(OF)

In a parallelogram ABCD. Prove that vec(AC)+ vec (BD) = 2 vec(BC)

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

vec(AC) and vec(BD) are the diagonals of a parallelogram ABCD. Prove that (i) vec(AC) + vec(BD) - 2 vec(BC) (ii) vec(AC) - vec(BD) - 2vec(AB)

Let O be the centre of a regular pentagon ABCDE and vec(OA) = veca , then vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA) is equals:

Let O be the centre of a regular pentagon ABCDE and vec(OA) = veca , then vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA) is equals:

If A,B,C,D are any four points in space prove that vec(AB)xxvec(CD)+vec(BC)xvec(AD)+vec(CA)xxvec(BD)=2vec(AB)xxvec(CA)

Let B_1,C_1 and D_1 are points on AB,AC and AD of the parallelogram ABCD, such that vec(AB_1)=k_1vec(AC,) vec(AC_1)=k_2vec(AC) and vec(AD_1)=k_2 vec(AD,) where k_1,k_2 and k_3 are scalar.