Question

Three forces $\mathrm{P},\mathrm{Q},\mathrm{R}$ act along the sides $\mathrm{BC},\mathrm{CA},\mathrm{AB}$ of triangle $\mathrm{ABC}$, taken in order. If their resultant passes through the incentre of$∆\mathrm{ABC}$, then

• A. $\mathrm{P}+\mathrm{Q}+\mathrm{R}=0$
• B. $\left(\raisebox{1ex}{$\mathrm{P}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right.\right)+\left(\raisebox{1ex}{$\mathrm{Q}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{b}$}\right.\right)+\left(\raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{c}$}\right.\right)$
• C. $\mathrm{aP}+\mathrm{bQ}+\mathrm{cR}=0$
• D. None of these

Answer 1

The correct option is A

$\mathrm{P}+\mathrm{Q}+\mathrm{R}=0$

The explanation for the correct answer.

Find the required condition?

Let $\mathrm{I}$ be the incentre of the triangle and $\mathrm{r}$ be the radius of the circle.

Moment $=$Force $\times$Distance

Moment of the resultant about $\mathrm{I}=$Sum of the moments of the given forces about $\mathrm{I}$

Moment $=\mathrm{P}\times \mathrm{ID}+\mathrm{Q}\times \mathrm{IE}+\mathrm{R}\times \mathrm{IF}$

$=\mathrm{P}\times \mathrm{r}+\mathrm{Q}\times \mathrm{r}+\mathrm{R}\times \mathrm{r}$

Since at the center sum of all moments is zero

$$\begin{gathered} \therefore \mathrm{Pr}+\mathrm{Qr}+\mathrm{Rr}=0 \\ \Rightarrow \mathrm{P}+\mathrm{Q}+\mathrm{R}=0 \end{gathered}$$

Hence option (a) is the correct answer.