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

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

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

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

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

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

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

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

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.

In a regular hexagon ABCDEF, vec(AE)

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