Question

Three forces $\overrightarrow{P}$ , $\overrightarrow{Q}$ and $\overrightarrow{R}$ are acting at a point in a plane. The angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ , $\overrightarrow{Q}$ and $\overrightarrow{R}$ are ${150}^{\circ }$ and ${120}^{\circ }$ respectively. Then for the equilibrium, forces $\overrightarrow{P}$ , $\overrightarrow{Q}$ and $\overrightarrow{R}$ are in the ratio :-

• A. $1:2:3$
• B. $1:2:\sqrt{3}$
• C. $3:2:1$
• D. $\sqrt{3}:2:1$

Answer 1

The correct option is D $\sqrt{3}:2:1$

For equilibrium, $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$

Let, $P=xQ$ and $R=yQ$

Taking dot product with $\overrightarrow{Q}:$

$\Rightarrow \overrightarrow{P}\cdot \overrightarrow{Q}+Q^2+\overrightarrow{R}\cdot \overrightarrow{Q}=0\Rightarrow PQ\cos \left(\frac{5\pi }{6}\right)+Q^2+RQ\cos \left(\frac{2\pi }{3}\right)=0\Rightarrow \left(\frac{-\sqrt{3}}{2}\right)PQ+Q^2+\left(\frac{-1}{2}\right)RQ=0\Rightarrow 2Q^2=\sqrt{3}PQ+RQ\Rightarrow 2=\sqrt{3}x+y.....\left(1\right)$

$\overrightarrow{P}\cdot \overrightarrow{R}+\overrightarrow{R}\cdot \overrightarrow{Q}+R^2=0\Rightarrow \frac{R^2}{2y}-\frac{x}{y}\cos \theta =R^2$

Now, $\theta =\pi -(\pi -\frac{5\pi }{6})-(\pi -\frac{2\pi }{3})=\frac{\pi }{2}$

Because these three vectors should form a triangle when arranged in a cyclic order.

$\Rightarrow y=\frac{1}{2}.....\left(2\right)$

From $\left(1\right)$ and $\left(2\right):$

$\Rightarrow x=\frac{\sqrt{3}}{2}$

Hence, option D is correct.