CGPET 2022 Solved Paper

Let O be the origin


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