actually i am getting the negative value of "1" .

it should BD.AB

ok

now draw the parallelogram .

take A as the oorigin . now AD = position vector of D and AB = postion vevtor of B.

now DB = b - a (all in vector form )

therefore DB.AB = (b - a ).a  = b.a - a^2............................1

now let the angle between a,b (vectors) be theta .

now angle ADC = 180 - thgeta.

from triangle ADC,cos(180-theta) = a^2 + b^2 - c^2/2ab

=>  - abcos(thteta) = a^2 + b^2 - c^2/2 ...............................2

now b.a = abcos(theta)

from 2 put value of b.a in 1

u'll get it

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