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Question

Match the elements of List I with their behaviour in List II:

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Solution

(A) L1:x12=y23=z34=λ
so A(a)=(1,2,3) and b=2^i+3^j+4^k
and L2:x13=y34=z55=μ
so C(c)=(1,3,5) and d=3^i+4^j+5^k
we find [ca b d]=∣ ∣012234345∣ ∣=0
so the lines are intersecting in a point

(B) L1:x12=y23=z34=λ
and L2:x32=y53=z74=μ
observe that D.Rs are identical also for λ=1 and μ=0 we get (x,y,z)=(3,5,7) so lines are coincident (superimposition)

(C) L1:x25=y+34=5z2=λ
and L2:x75=y+14=z22=μ
observe that D.Rs are identical
If the lines are coincident then 5λ+2=5μ+7 ...(i)
4λ3=4μ1 ...(ii)
and 2λ+5=2μ+2 ....(iii)
Now from (i) and (ii) we get λμ=3
and from (i) and (ii) we get 3(λμ)=2
so the lines do not intersect lines are parallel and distinct

(D) L1:x32=y+24=z45=λ
and L2:x33=y22=z75=μ
so a=(3,2,4);b=(2,4,5);c=(3,2,7);d=(3,2,5)
Now [ca b d]=∣ ∣043245325∣ ∣=2024=4
since[ca b d] 0
skew lines

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