Question

Three vectors $\overrightarrow{P},\overrightarrow{Q}$ and $\overrightarrow{R}$ are such that $|\overrightarrow{P}|=|\overrightarrow{Q}|,|\overrightarrow{R}|=\sqrt{2}|\overrightarrow{P}|$, and $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$. The angles between $\overrightarrow{P}$ and $\overrightarrow{Q},\overrightarrow{Q}$ and $\overrightarrow{R}$, and $\overrightarrow{P}$ and $\overrightarrow{R}$ are

• A. ${90}^{\circ },{135}^{\circ },{135}^{\circ }$
• B. ${90}^{\circ },{45}^{\circ },{45}^{\circ }$
• C. ${45}^{\circ },{90}^{\circ },{90}^{\circ }$
• D. ${45}^{\circ },{135}^{\circ },{135}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ },{135}^{\circ },{135}^{\circ }$

Vectors $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$
Given: $|\overrightarrow{P}|=|\overrightarrow{Q}|\text{and}|\overrightarrow{R}|=\sqrt{2}|\overrightarrow{P}|$
Let $\varphi$ be the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}.$
So, using vector addition $\overrightarrow{P}+\overrightarrow{Q}=-\overrightarrow{R}$
$|\overrightarrow{P}+\overrightarrow{Q}{|}^2=|-\overrightarrow{R}{|}^2=|\overrightarrow{R}{|}^2$
$|\overrightarrow{P}{|}^2+|\overrightarrow{Q}{|}^2+2|\overrightarrow{P}||\overrightarrow{Q}|\times \cos \varphi =2|\overrightarrow{P}{|}^2$
Hence, $\cos \varphi =0\Rightarrow \varphi =\frac{\pi }{2}={90}^{\circ }$

As $\overrightarrow{P}$ and $\overrightarrow{Q}$ vectors of equal magnitude, $-\overrightarrow{R}$ will be at $\frac{\pi }{4}$ angle from both $\overrightarrow{P}$ and $\overrightarrow{Q}$. So $\overrightarrow{R}$ will be at ${180}^{\circ }$ from - $\overrightarrow{R}.$
Angle between $\overrightarrow{Q}$ and $\overrightarrow{R}$ $={180}^{\circ }-{45}^{\circ }={135}^{\circ }$
Angle between vectors $\overrightarrow{R}$ and $\overrightarrow{P}={135}^{\circ }$