Question

Given that $P=Q=R$. If $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$ then the angle between $\overrightarrow{P}$ and $\overrightarrow{R}$ is ${\theta }_1$. If $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$ then the angle between $\overrightarrow{P}$ and $\overrightarrow{R}$ is ${\theta }_2$. The relation between ${\theta }_1$ and ${\theta }_2$ is

• A. ${\theta }_1={\theta }_2$
• B. ${\theta }_1=\frac{{\theta }_2}{2}$
• C. ${\theta }_1=2{\theta }_2$
• D. ${\theta }_1=4{\theta }_2$

Answer 1

The correct option is B ${\theta }_1=\frac{{\theta }_2}{2}$

$\text{Step 1: Using first condition to find}{\theta }_1$

When $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$

$\Rightarrow$ $\overrightarrow{P}-\overrightarrow{R}=-\overrightarrow{Q}$

Squaring on both the sides

$(\overrightarrow{P}-\overrightarrow{R})\cdot (\overrightarrow{P}-\overrightarrow{R})=(-\overrightarrow{Q})\cdot (-\overrightarrow{Q})$

$\Rightarrow$ $P^2+R^2-2\overrightarrow{P}\cdot \overrightarrow{R}=Q^2$

$\Rightarrow$ $P^2+R^2-2PR\cos {\theta }_1=Q^2$

$\Rightarrow$ $P^2+P^2-2P^2\cos {\theta }_1=P^2$ $[\because P=Q=R$]

$\Rightarrow$ $\cos {\theta }_1=\frac{1}{2}$

$\Rightarrow$ ${\theta }_1={60}^{\circ }$

$\text{Step 2: Using second condition to find}{\theta }_2$

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

$\Rightarrow$ $\overrightarrow{P}+\overrightarrow{R}=-\overrightarrow{Q}$

Squaring on both sides

$(\overrightarrow{P}+\overrightarrow{R}).(\overrightarrow{P}+\overrightarrow{R})=(-\overrightarrow{Q})\cdot \left(\overrightarrow{-Q}\right)$

$\Rightarrow$ $P^2+R^2+2PR\cos {\theta }_2=Q^2$

$\Rightarrow$ $P^2+P^2+2P^2\cos {\theta }_2=P^2$ $[\because P=Q=R$]

$\Rightarrow$ $\cos {\theta }_2=-\frac{1}{2}$

$\Rightarrow$ ${\theta }_2={120}^{\circ }$

$\therefore {\theta }_1=\frac{{\theta }_2}{2}$

Hence option $B$ is correct.