Question

Given that $|\overrightarrow{\text{P}}|=|\overrightarrow{\text{Q}}|=|\overrightarrow{\text{R}}|$. If $\overrightarrow{\text{P}}+\overrightarrow{\text{Q}}=\overrightarrow{\text{R}}$ then the angle between $\overrightarrow{\text{P}}$ and $\overrightarrow{\text{R}}$ is ${\theta }_1$, and if $\overrightarrow{\text{P}}+\overrightarrow{\text{Q}}+\overrightarrow{\text{R}}=0$ then the angle between $\overrightarrow{\text{P}}$ and $\overrightarrow{\text{R}}$ is ${\theta }_2$. Then

• A. ${\theta }_2={120}^{\circ }$
• B. ${\theta }_1={\theta }_2={60}^{\circ }$
• C. $2{\theta }_1={\theta }_2$
• D. ${\theta }_1={60}^{\circ }$

Answer 1

Answer: (A), (C), (D)

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

Condition $\left(1\right):$

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

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

Taking magnitude of the vector followed by squaring,

${\text{R}}^2+{\text{P}}^2-2\text{RP}\cos {\theta }_1={\text{Q}}^2$

$2-2\cos {\theta }_1=1(\because \text{P}=\text{Q}=\text{R})$

$\cos {\theta }_1=\frac{1}{2}\Rightarrow {\theta }_1={60}^{\circ }$

Condition $\left(2\right):$

$\overrightarrow{\text{P}}+\overrightarrow{\text{Q}}+\overrightarrow{\text{R}}=0$

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

Taking magnitude of the vector followed by squaring,

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

$\cos {\theta }_2=-\frac{1}{2}\Rightarrow {\theta }_2={120}^{\circ }$

$\therefore 2{\theta }_1={\theta }_2={120}^{\circ }$

Therefore, options $\left(\text{B}\right),\left(\text{C}\right)\&\left(\text{D}\right)$ are the correct answer.