1. If  $l_1,m_1,n_1$and $l_2,m_2,n_2$ are the direction cosines of the two lines inclined to each other at an angle $\theta$ , then the direction cosines of the external angular bisector of the angle between these lines are $\frac{l_1-l_2}{2\sin \alpha },\frac{m_1-m_2}{2\sin \alpha },\frac{n_1-n_2}{2\sin \alpha }$ where$2\alpha =$

detailed solution

Correct option is A

Given that the direction cosines of two lines are l1,m1,n1 andl2,m2,n2Given  θ is the acute angle between the lines then cosθ=l1l2+m1m2+n1n2The direction ratios of angular bisector of the two lines whose direction cosines are l1,m1,n1 and l2,m2,n2 are l1−l2,m1−m2,n1−n2Consider the value of l1−l22+m1−m22+n1−n22=l12+m12+n12+l22+m22+n22+2l1l2+2m1m2+2n1n2=2−2cosθ=22sin2θ2=4sin2θ2It implies thatl1−l22+m1−m22+n1−n22=2sinθ2The direction cosines of angular bisector of given two lines are l1−l22sinθ2=m1−m22sinθ2=n1−n22sinθ2Comparing the above direction cosines with l1−l22 sinα,m1−m22 sinα,n1−n22 sinα, we have α=θ2.It implies that 2α=θ