Question

A point $I$ is the centre of a circle inscribed in a triangle $ABC$, then the vector sum $|\overrightarrow{BC}|\overrightarrow{IA}+|\overrightarrow{CA}|\overrightarrow{IB}+|\overrightarrow{AB}|\overrightarrow{IC}$ is

• A. zero
• B. $\frac{\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}}{3}$
• C. $3(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC})$
• D. None of these

Answer 1

Answer: (A)

If $|\overrightarrow{BC}|=a;|\overrightarrow{CA}|=b;|\overrightarrow{AB}|=c$ then $\frac{aIA+\overrightarrow{IB}+\overrightarrow{cIC}}{a+b+c}$
is the position vector of I with respect to I \& this is equal to zero.
$[\because$ p.v. of incentre of a triangle is $\frac{a\alpha +b\vec{\beta }+c\vec{\gamma }}{a+b+c}$, where
$\vec{\alpha },\vec{\beta }$ and $\vec{\gamma }$ are p.v. of vertices $A,B$ and $C$ of a triangle ABC respectively].