Assertion: If I is the incentre of /\ABC...

Assertion: If I is the incentre of `/_\ABC, then`|vec(BC)| vec(IA) +|vec(CA)| vec(IB) +|vec(AB)| vec(IC) =0` Reason: If O is the origin, then the position vector of centroid of `/_\ABC` is (vec(OA)+vec(OB)+vec(OC))/3`

A

Both A and R are correct and R is the correct explanation of A

B

Both A and R are correct but R is not the correct explanation of A

C

A is correct but R is incorrect

D

R is correct but A is incorrect

Verified by Experts

The correct Answer is:

B

We know that,
`OI=(|CB|OA+|CA|OB+|AB|OC)/(|BC|+|CA|+|AB|)`
and `OG=(OA+OB+OC)/(3)`

Similar Questions

Explore conceptually related problems

If the position vector of a point A is vec a + 2 vec b and vec a divides AB in the ratio 2:3 , then the position vector of B, is

If G is the centroid of Delta ABC and G' is the centroid of Delta A' B' C' " then " vec(A A)' + vec(B B)' + vec(C C)' =

If vec(a) is a unit vector such that (vec(x) - vec(a)).(vec(x) + vec(a)) = 8 find |vec(x)| .

If A, B and C are the vertices of a triangle ABC, then what is the value of vec(AB) + vec(BC) + vec(CA) ?

ABCDE is a pentagon. Prove that the resultant of forces vec (AB), vec(AE), vec(BC), vec(DC), vec(ED) and vec(AC) is 3vec(AC) .

If theta is the angle between the vectors vec a and vec b then (|vec a xx vec b|)/(|vec a . vec b|) =

Prove that 3vec(OD)+vec(DA)+vec(DB)+vec(DC) is equal to vec(OA)+vec(OB)+vec(OC) .