Question

Given that P + Q + R = 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. 90${}^{\circ }$, 45${}^{\circ }$, 45${}^{\circ }$
• C. 90${}^{\circ }$, 30${}^{\circ }$, 60${}^{\circ }$
  
• D. 60${}^{\circ }$, 45${}^{\circ }$, 135${}^{\circ }$

Answer 1

The correct option is B 90${}^{\circ }$, 45${}^{\circ }$, 45${}^{\circ }$

$\overline{P}+\overline{Q}=-\overline{R}$

$\therefore \overline{P},\overline{Q}and-\overline{R}formsidesoftriangle.$

$Fromcosinerule:$

$\cos \alpha =\frac{{\left|\overline{P}\right|}^2+{\left|\overline{R}\right|}^2-{\left|\overline{Q}\right|}^2}{2\left|\overline{P}\right|.\left|\overline{R}\right|}$

$If\left|\overline{P}\right|=\left|\overline{Q}\right|and;$

$\left|\overline{R}\right|=\sqrt{2}\left|\overline{P}\right|$

$then,\cos \alpha =\frac{2{\left|\overline{P}\right|}^2}{2{\left|\overline{P}\right|}^2.\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\cos \alpha =\frac{1}{\sqrt{2}}$

$⟹\alpha ={45}^{\circ }$

$and\sin \theta twoanglesareequal.$

$(\because twosidelengthsareequal)$

$\therefore Remaininganglesare{45}^{\circ },{90}^{\circ }.$

$If\left|\overline{R}\right|=\left|\overline{Q}\right|$

$and\left|\overline{P}\right|=\sqrt{2}\left|\overline{R}\right|$

$Thenalsowegetthesameresults$

$andif\left|\overline{P}\right|=\left|\overline{R}\right|$

$and\left|\overline{Q}\right|=\sqrt{2}\left|\overline{P}\right|$

$then\alpha ={90}^{\circ }andremaining2anglesare{45}^{\circ }each.$

$Now,\alpha ={45}^{\circ }\left(I\&IIcase\right)$

$but\alpha istheanglebetween\overline{P}\&-\overline{R}.$