Then, OA and OB are lines where DR′s are <l1,m1,n1> and <l2,m2,n2> respectively. Let OA=OB=r. Again OC denotes the angle bisector of ∠AOB. Then, C will be the mid-point of AB. ∴ Coordinates of C =(
l1r+l2r
2
,
m1r+m2r
2
,
n1r+n2r
2
) Now, DR′s of OC=
l1r+l2r
2
−0,
m1r+m2r
2
−0,
n1r+n2r
2
−0 =
r(l1+l2)
2
,
r(m1+m2)
2
,
r(n1+n2)
2
∴DR′s of OC=l1+l2:m1+m2:n1+n2 Similarly, DR's of obtuse ∠AOB will be l1−l2, m1−m2,n1−n2 So, DR′s of angle bisector will be l1±l2,m1±m2, n1±n2