If `f` is the centre of a circle inscribed in a triangle `ABC`, then `|vec(BC)|vec(IA)+|vec(CA)|vec(IB)+|vec(AB)|vec(IC)` is

If f is the centre of a circle inscribed in a triangle ABC, then |vec BC|vec IA+|vec CA|vec IB+|vec AB|vec IC is

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 (vecOA)+vec(OB)+vec(OC))/3 (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

In a right angled triangle ABC, the hypotenuse AB =p, then vec(AB).vec(AC) + vec(BC).vec(BA)+vec(CA).vec(CB) is equal to:

In a right angled triangle hypotenuse AC= p, then vec(AB). vec(AC ) + vec(BC) .vec(BA) + vec(CA). vec(CB) equal to ?

If D is the mid-point of the side BC of a triangle ABC, prove that vec AB+vec AC=2vec AD .

ABCDE is a pentagon prove that vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EA)=vec0

If in a right-angled triangle ABC, hypotenuse AC=p, then what is vec(AB).vec(AC)+vec(BC).vec(BA)+vec(CA).vec(CB) equal to ?

If ABCDE is a pentagon, then vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) is equal to

If G is the centroid of a triangle ABC, prove that vec GA+vec GB+vec GC=vec 0

Orthocenter of an equilateral triangle ABC is the origin O. If vec(OA)=veca, vec(OB)=vecb, vec(OC)=vecc , then vec(AB)+2vec(BC)+3vec(CA)=