Question

# If (l1,m1,n1) and (l2,m2,n2,) are d.c.'s of ¯¯¯¯¯¯¯¯¯¯OA, ¯¯¯¯¯¯¯¯OB such that ∠AOB=θ where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle ∠AOB are

A
l1+l22sinθ2,m1+m22sinθ2,n1+n22sinθ2
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B
l1+l22cosθ2,m1+m22cosθ2,n1+n22cosθ2
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C
l1l22sinθ2,m1m22sinθ2,n1n22sinθ2
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D
l1l22cosθ2,m1m22cosθ2,n1n22cosθ2
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Solution

## The correct option is B l1+l22cosθ2,m1+m22cosθ2,n1+n22cosθ2Let OA and OB be two lines with d.c’s l1,m1,n1 and l2,m2,n2,. Let OA = OB = 1. Then, the coordinates of A and B are (l1,m1,n1) and (l2,m2,n2), respectively. Let OC be the bisector of ∠AOB .Then, C is the mid point of AB and so its coordinates are (l1+l22,m1+m22,n1+n22). ∴ d.r's of OC are l1+l22,m1+m22,n1+n22 We have, OC=√(l1+l22)2+(m1+m22)2+(n1+n22)2=12√(l21+m21+n21)+(l22+m22+n22)+(l1l2+m1m2+n1n2)=12√2+2cosθ[Q cos θ=l1l2+m1m2+n1n2]=12√2(1+cosθ)=cos(θ2) ∴ d.c's of OC are l1+l22(OC),m1+m22(OC),n1+n22(OC)

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