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A straight line L cuts the lines A B ,A ...

A straight line `L` cuts the lines `A B ,A Ca n dA D` of a parallelogram `A B C D` at points `B_1, C_1a n dD_1,` respectively. If `( vec A B)_1,lambda_1 vec A B ,( vec A D)_1=lambda_2 vec A Da n d( vec A C)_1=lambda_3 vec A C ,` then prove that `1/(lambda_3)=1/(lambda_1)+1/(lambda_2)` .

Text Solution

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Let `vec(AB) = veca, vec(AD) = vecb,` then `vec(AC) = veca+vecb`.
Given `vec(AB)_1 = lamda_1 veca, vec(AD)_1 = lamda_2 vecb, vec(AC)_1 = lamda_3 (veca+vecb) vec(B_1D_1) = vec(AD)_1 - vec(AB)_1= lamda_2 vecb-lamda_1veca`
Since vectors `vec(D_1C_1) and vec(B_1D_1)` are collinear, we have
`" "vec(D_1C_1)=kvec(B_1D_1)` for some `k in R`
`rArr " "vec(AC)_1 - vec(AD)_1= k vec(B_1D_1)`
`rArr " "lamda_3(veca+vecb) -lamda_2vecb=k(lamda_2 vecb-lamda_1veca)`

or `lamda_3 veca+ (lamda_3-lamda_2)vecb= klamda_2vecb - klamda_1veca`
Hence, `lamda_3=-klamda_1 and lamda_3-lamda_2=klamda_2`
`rArr k = -(lamda_3)/(lamda_1)= (lamda_3-lamda_2)/(lamda_2)`
or `" "lamda_1lamda_2= lamda_1lamda_3+lamda_2lamda_3`
or `" "(1)/(lamda_3)= (1)/(lamda_1) + (1)/(lamda_2)`
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