Let ABC be a triangle, AD, BE and CF be the angular bisectors of its interior angles. These bisectors are concurrent at a point I called incentre of the triangle. We know from geometry that 
.
If BC = α , CA = β and AB = γ and with reference to same origin let 
, 
, 
be position vectors of A, B and C respectively. Then-
(i) The position vector of I must be
Text Solution
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Ans.
(i)
Sol. 
= 
= 
⇒ Position vector of D = 
In Δ BDA, BI is bisector of Δ BDA also
⇒ 
= 
= 
= 
Thus position vector of I must be 
= 
(ii)
Sol. 
. 
= | 
| 
| cos 
= r cosec 
r cosec 
= –r2 cosec 
cosec 
sin 
(iii)
Sol . We get | 
× 
|
as r2 cosec 
cosec 
cos 
Using | 
× 
| = ab sin θ
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