Question

Given that $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$. Two out of the three vectors are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of $1^{\text{st}}$ and $2^{\text{nd}}$ vector. Which of the following can be the angles between these vectors?

• A. ${90}^{\circ },{135}^{\circ },{135}^{\circ }$
• B. ${45}^{\circ },{45}^{\circ },{90}^{\circ }$
• C. ${30}^{\circ },{60}^{\circ },{90}^{\circ }$
• D. ${45}^{\circ },{90}^{\circ },{135}^{\circ }$

Answer 1

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

Let the magnitudes of $\overrightarrow{P},\overrightarrow{Q}$ and $\overrightarrow{R}$ be $x,x$ and $\sqrt{2}x$ respetively.

For $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$ , the resultant of any two vectors must be equal to and opposite to the third vector
Let us take resultant of $\overrightarrow{P}\text{and}\overrightarrow{Q}$ opposite to $\overrightarrow{R}$
Then, magnitude of resultant:
$R=\sqrt{x^2+x^2+2\times x\times x\cos \theta }$
$\sqrt{2}\times x=\sqrt{2}\times x\times \sqrt{\cos \theta +1}$
Squaring both sides:
$\cos \theta =0$
$\therefore \theta ={90}^{\circ }$
Angle between $\overrightarrow{P}$ and$\overrightarrow{Q}$ is ${90}^{\circ }$


As magnitudes of $\overrightarrow{P}$ and $\overrightarrow{Q}$ are equal, the remaining two angles of vector triangle shown in figure will be ${45}^{\circ }$ each.

$\because$ The angle is measured when vectors are tail-to-tail, hence $\overrightarrow{R}$ will be making an angle $({180}^{\circ }-{45}^{\circ })={135}^{\circ }$ with both $\overrightarrow{P}$ and $\overrightarrow{Q}$.

$\therefore$ Angles between vectors($\overrightarrow{P},\overrightarrow{Q}$), ($\overrightarrow{Q},\overrightarrow{R}$), ($\overrightarrow{R},\overrightarrow{P}$) are ${90}^{\circ },{135}^{\circ },{135}^{\circ }$