IIT JEE , AIEEE Forum - Mathematics , Vectors

karthik_abiram

Q] In a parallogram ABCD ,/AB/ =a , /AD/=b and /AC/=c ,then

DB.AB {vectors} has the value of

1]{3a²+b²-c²}/2

2]{a²+3b²-c²}/2

3]{a²-b² +3c²}/2

4]{a²+3b²+c²}/2

plzz post a detailed solution.plzz help

rajat

is it 1.

i'll post the soln. if its correct

karthik_abiram

yea.plzz post it.

thank u

rajat

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

do rate me !!

karthik_abiram

hey rajat,

The question is correct.......u made a small mistake

i.e. DB= a-b and not b-a

and so there will be no "-"

but ur method is correct .........thank you