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.
$\left[\because\right.$ 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].