Question

Given that $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=\overrightarrow{0}$. Two out of the three vectors are equal in magnitude. The magnitude of the third vector is $\sqrt{2}$ times that of the other two. 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

Answer: (A)

${90}^{\circ },{135}^{\circ },{135}^{\circ }$

Let magnitude of vectors $P$ and $Q$ be $A$ i.e $|\overrightarrow{P}|=|\overrightarrow{Q}|=A$

$\therefore$ $|\overrightarrow{R}|=\sqrt{2}A$

Let the angle between vectors $P$ and $Q$ be $\theta$

Using $|\overrightarrow{R}{|}^2=|\overrightarrow{P}{|}^2+|\overrightarrow{Q}{|}^2+2|\overrightarrow{P}||\overrightarrow{Q}|cos\theta$

$(\sqrt{2}A{)}^2=A^2+A^2+2A^{2}cos\theta$ $⟹cos\theta =0$ OR $\theta ={90}^o$

Also as $|\overrightarrow{P}|=|\overrightarrow{R}|cos{\theta }_1$

$\therefore$ $A=\sqrt{2}Acos{\theta }_1$ $⟹cos{\theta }_1=\frac{1}{\sqrt{2}}$ OR ${\theta }_1={45}^o$

Thus the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$ $={180}^o-{45}^o={135}^o$

Similarly $|\overrightarrow{Q}|=|\overrightarrow{R}|cos{\theta }_2$

$\therefore$ $A=\sqrt{2}Acos{\theta }_2$ $⟹cos{\theta }_2=\frac{1}{\sqrt{2}}$ OR ${\theta }_2={45}^o$

Thus the angle between $\overrightarrow{R}$ and $\overrightarrow{Q}$ $={180}^o-{45}^o={135}^o$