Question

In a triangle ABC, I is the incentre. If BC = a, CA = b,

AB = g, then a$\overset{\rightarrow}{IA}$+ b$\overset{\rightarrow}{IB}$ + g$\overset{\rightarrow}{IC}$ is equal to-

• A. zero vector
• B. (a + b + g)$\overset{\rightarrow}{IA}$
• C. (a + b + g)$\overset{\rightarrow}{IB}$
• D. (a + b + g)$\overset{\rightarrow}{IC}$

Answer 1

Answer: (A)

zero vector

Answer 2

If the incentre I be chosen as the origin and $\overline{a}$,$\overline{b}$,$\overline{c}$ be the position vectors of A,B,C then the position vector of

I = $\frac{\alpha\overline{a} + \beta\overline{b} + \gamma\overline{c}}{\alpha + \beta + \gamma}$

But position vector of I is zero, since it is chosen as the origin.

\ $\frac{\alpha\overline{a} + \beta\overline{b} + \gamma\overline{c}}{\alpha + \beta + \gamma}$ = $\overset{\rightarrow}{0}$ ̃ a$\overline{a}$ + b$\overline{b}$ + g$\overline{c}$ = $\overset{\rightarrow}{0}$