Aditya Gupta
Last Activity: 5 Years ago
hello sridev the ans will be 0.
reason: we know that position vector of incentre of a triangle ABC is given by:
I= (aA + bB + cC)/(a + b + c) where A, B, C are position vectors of vertices A, B, C. also, a b c are side lengths.
now, IA= A – I so aIA= aA – aI.........(1)
IB= B – I so bIB= bB – bI.......(2)
IC= C – I so cIC= cC – cI..........(3)
adding these eqns
aIA + bIB + cIC= aA + bB + cC – (a + b + c)I. substitute value of I, we get
aIA + bIB + cIC= 0
kindly approve :)