ABC is an equilateral triangle of side ‘a'. Then `bar(AB).bar(BC) +bar(BC).bar(CA)+ bar(CA).bar(AB) =`

If ABCDEF is a regular hexagon of side a, then bar(AB).bar(AF)=

In the figure 11.34, ABCD is a quadrilateral in which bar(AB) = bar(AD) and bar(BC) =bar(DC). Diagonals bar(AC) and BD intersect each other at O.Show that-

ABCDE is a regular hexagon. i) If bar(AB) + bar(AE)+ bar(BC) + bar(DC) + bar(ED) + bar(AC)=lambdabar(AC) then lambda=

If ABCDE is a pentagon, then bar(AB) + bar(AE) + bar(BC) + bar(DC) + bar(ED) =

If ABCD is a square, then bar(AB) + 2 bar(BC) + 3 bar(CD) + 4 bar(DA) =

DeltaABC be an equilateral triangle whose orthocentre is the origin 'O' . If bar (OA)=bar a.bar (OB)=bar b then bar (OC) is

DeltaABC be an equilateral triangle whose orthocentre is the origin 'O' . If bar (OA)=bar a.bar (OB)=bar b then bar (OC) is

([[bar(a),bar(b),bar(c)]])/([[bar(b),bar(a),bar(c)]]) =

If |bar(a)+bar(b)|<|bar(a)-bar(b)| then (bar(a),bar(b)) is