In a regular hexagon `ABCDEF, bar(AB) + bar(AC)+bar(AD)+ bar(AE) + bar(AF)=k bar(AD)` then k is equal to

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

ABCDEF is a regular hexagon. Show that : vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=6vec(AO) . Where O is the centre of the hexagon.

In the figure bar(AC), bar(AD), bar(BD), bar(BE) and (CE) are diagonals. Is bar(BC) a diagonal. Why or why not? If you try to join adjacent vertices, will the result be a diagonal? Name all the sides, adjacent vertices of the figure ABCDE.

Prove that : z= bar(z) iff z is real.

If ABCDEF is a regular hexagon, prove that AD+EB+FC=4AB .

Consider the regular hexagon ABCDEF with centre at O (origin). Q. Five forces AB,AC,AD,AE,AF act at the vertex A of a regular hexagon ABCDEF. Then, their resultant is

M and N are mid-point of the diagnols AC and BD respectivley of quadrilateral ABCD, then bar(AB) + bar(AD)+bar(CB)+bar(CD) =

Given four non zero vectors bar a,bar b,bar c and bar d . The vectors bar a,bar b and bar c are coplanar but not collinear pair by pairand vector bar d is not coplanar with vectors bar a,bar b and bar c and hat (bar a bar b) = hat (bar b bar c) = pi/3,(bar d bar b)=beta ,If (bar d bar c)=cos^-1(mcos beta+ncos alpha) then m-n is :

If (3+i)(z+bar(z))-(2+i)(z-bar(z))+14i=0 , where i=sqrt(-1) , then z bar(z) is equal to