Question

If $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$ and out of these, two vectors are equal in magnitude and the third vector has magnitude $\sqrt{2}$ times that of any of these two vectors, then angles among the three vectors are

• A. ${45}^o,{75}^o,{75}^o$
• B. ${45}^o,{90}^o,{135}^o$
• C. ${90}^o,{135}^o,{180}^o$
• D. ${90}^o,{135}^o,{135}^o$

Answer 1

The correct option is D ${90}^o,{135}^o,{135}^o$

Let $P=Q=x$ and $R=\sqrt{2}x$. Using $\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}=0$, we get $\overrightarrow{P}+\overrightarrow{Q}=-\overrightarrow{R}$
Then $P^2+Q^2+2PQ\cos \theta =R^2$
i.e $x^2+x^2+2x^2\cos \theta =2x^2$
i.e $\cos \theta ={90}^{\circ }$
Again, $\overrightarrow{Q}+\overrightarrow{R}=-\overrightarrow{P}$
Then, $Q^2+R^2+2QR\cos \alpha =P^2$
or
$x^2+2x^2+2\sqrt{2}{x}^{2}cos\alpha =x^2$

i.e $\cos \alpha =-1/\sqrt{2}$ or $\varphi ={135}^{\circ }$
Third angle $={360}^0-({135}^0+{90}^0)={360}^0-{225}^0={135}^{\circ }$