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Q.

L1 and L2 are two lines whose vector equations areL1:r→=λ((cos⁡θ+3)i^+(2sin⁡θ)j^+(cos⁡θ−3)k^)L2:r→=μ(ai^+bj^+ck^)  where λ and μ pare scalars and a is the acute angle between L1and L2.If the angle α is independent of θ then the value of α is

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a

π6

b

π4

c

π3

d

π2

answer is A.

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Detailed Solution

Both the lines pass through the origin. Line L1, is parallel to the vector  V1→V→1=(cos⁡θ+3)i^+(2sin⁡θ)j^+(cos⁡θ−3)k^Line L1, is parallel to the vector  V2→V→2=ai^+bj^+ck^∴ cos⁡α=V→1⋅V→2V→1V→2=a(cos⁡θ+3)+(b2)sin⁡θ+c(cos⁡θ−3)a2+b2+c2(cos⁡θ+3)2+2sin2⁡θ+(cos⁡θ−3)2=(a+c)cos⁡θ+b2sin⁡θ+(a−c)3a2+b2+c22+6For cos a to be independent of θ we get     a+c=0 and b=0∴     cos⁡α=2a3a222=32 or  α=π6
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L1 and L2 are two lines whose vector equations areL1:r→=λ((cos⁡θ+3)i^+(2sin⁡θ)j^+(cos⁡θ−3)k^)L2:r→=μ(ai^+bj^+ck^)  where λ and μ pare scalars and a is the acute angle between L1and L2.If the angle α is independent of θ then the value of α is