Five forces ` vec A B` , ` vec A C` , ` vec A D` , ` vec A E` and ` vec A F` act at the vertex of a regular hexagon `A B C D E Fdot` Prove that the resultant is `6 vec A O` , where `O` is the centre of heaagon.

Five forces vec AB,vec AC,vec AD,vec AE and vec AF act at the vertex of a regular hexagon ABCDEF . Prove that the resultant is 6vec AO, where O is the centre of heaagon.

Two forces vec A B and vec A D are acting at vertex A of a quadrilateral ABCD and two forces vec C B and vec C D at C prove that their resultant is given by 4 vec E F , where E and F are the midpoints of AC and BD, respectively.

Given a regular hexagon ABCDEF with centre o,show that vec OB-vec OA=vec OC-vec OD

Let O be the centre of a regular hexagon ABCDEF. Find the sum of the vectors vec OA,vec OB,vec OC,vec OD,vec OE and vec OF .

In a regular hexagon ABCDEF,vec AB=a,vec BC=b and vec CD=c. Then ,vec AE=

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

[vec a, vec b + vec c, vec d] = [vec a, vec b, vec d] + [vec a, vec c, vec d]

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

If vec a , vec b , vec c and vec d are the position vectors of points A, B, C, D such that no three of them are collinear and vec a + vec c = vec b + vec d , then ABCD is a