Question

Given that $P=Q=R.$ If $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$ then the angle between $\overrightarrow{P}\text{ }and\text{ }\overrightarrow{Q}$ 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) ${{\theta }_{1}}=\dfrac{{{\theta }_{2}}}{2}$
C) ${{\theta }_{1}}=2{{\theta }_{2}}$
D) ${{\theta }_{1}}=4{{\theta }_{2}}$

Answer 1

Hint: We have been provided with a magnitude of vector $P=Q=R$. In this question we have given two conditions using the first condition and given the equation i.e. $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$. Since angle ${{\theta }_{1}}$ is the angle between $\overrightarrow{P}and\overrightarrow{R}$ so use parallelogram laws of vector addition apply $P=Q=R.$ condition and we get value of ${{\theta }_{1}}$ similarly in second condition use parallelogram law of vector addition and given condition to calculate ${{\theta }_{2}}$ the angle between $\overrightarrow{P}and\overrightarrow{R}$.

Complete step by step solution:
In this question we have been provided with vector P, Q, R. given that $P=Q=R$ we have first condition over here that if $\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$ then angle between $\overrightarrow{P}and\overrightarrow{R}$ is ${{\theta }_{1}}$. And we have second condition which is, if $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$ then angle between $\overrightarrow{P}and\overrightarrow{R}$ is ${{\theta }_{2}}$ now we need to calculate the relation between ${{\theta }_{1}}$ and ${{\theta }_{2}}$.
Given that, magnitude of vector is $\left| \overrightarrow{P} \right|=\left| \overrightarrow{Q} \right|=\left| \overrightarrow{R} \right|$ let, use first condition i.e. $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$ since ${{\theta }_{1}}$ is the angle between $\overrightarrow{P}and\overrightarrow{R}$ so let’s take $\overrightarrow{R}$ on left side and vector $\overrightarrow{Q}$ on another side.
$\overrightarrow{=>P}-\overrightarrow{R}=-\overrightarrow{Q}$
Take dot product we get,
$=>\left( \overrightarrow{P}-\overrightarrow{R} \right)\left( \overrightarrow{P}-\overrightarrow{R} \right)=\left( -\overrightarrow{Q} \right)\left( -\overrightarrow{Q} \right)$
According to parallelogram law of vector addition, we get,
$=>{{\overrightarrow{P}}^{2}}+{{\overrightarrow{R}}^{2}}-2\overrightarrow{P}\overrightarrow{R}\cos {{\theta }_{1}}={{\overrightarrow{Q}}^{2}}$
It is given that $P=Q=R$ therefore,
$\begin{align}
  & =>{{Q}^{2}}+{{Q}^{2}}-2QQ\cos {{\theta }_{1}}={{Q}^{2}} \\
 & =>{{Q}^{2}}(2-2\cos {{\theta }_{1}})={{Q}^{2}} \\
 & =>2-2\cos {{\theta }_{1}}=1 \\
 & =>1-\cos {{\theta }_{1}}=\dfrac{1}{2} \\
 & =>\cos {{\theta }_{1}}=\dfrac{1}{2} \\
 & =>\theta ={{60}^{o}}.......(1) \\
\end{align}$
Hence, the angle between P and R, for first condition is ${{\theta }_{1}}={{60}^{o}}$
Now use second condition which is given as, when $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$ then angle between P and Q is ${{\theta }_{2}}$ so,
$\overrightarrow{=>P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$,
can be written as,
$=>\overrightarrow{P}+\overrightarrow{R}=-\overrightarrow{Q}$
Take dot product we get,
$=>\left( \overrightarrow{P}+\overrightarrow{R} \right)\left( \overrightarrow{P}+\overrightarrow{R} \right)=\left( -\overrightarrow{Q} \right)\left( -\overrightarrow{Q} \right)$
Again use parallelogram law of vector addition, we get,
$=>{{\overrightarrow{p}}^{2}}+{{\overrightarrow{R}}^{2}}+2\overrightarrow{P}\overrightarrow{R}\cos {{\theta }_{2}}={{\overrightarrow{Q}}^{2}}$
(Since ${{\theta }_{2}}$ is the angle between$\overrightarrow{P}and\overrightarrow{R}$)
It is given that $P=Q=R$ therefore,
$\begin{align}
  & =>{{Q}^{2}}+{{Q}^{2}}+2QQ\cos {{\theta }_{2}}={{Q}^{2}} \\
 & =>{{Q}^{2}}(2+2\cos {{\theta }_{2}})={{Q}^{2}} \\
 & =>2(1+\cos {{\theta }_{2}})=1 \\
 & =>1+\cos {{\theta }_{2}}=\dfrac{1}{2} \\
 & =>\cos {{\theta }_{2}}=-\dfrac{1}{2} \\
 & =>{{\theta }_{2}}={{120}^{o}}.....(2) \\
\end{align}$
So, if you compare equation (1) and equation (2) then we can say
$=>{{\theta }_{2}}=2{{\theta }_{1}}$

Therefore, option (c) is the correct option.


Note: According to parallelogram law of vector addition, if two vectors of the same type is starting from the same point, are represented in magnitude and direction by two adjacent sides of a parallelogram then their resultant vector is given in magnitude and direction by the diagonal of the parallelogram starting from the same point the diagonal of the parallelogram made by two vectors as adjacent sides is not passing through common point of two vectors.