Question

The resultant of two vectors A and B is perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is

• A. ${120}^{\circ }$
  
  
• B. ${150}^{\circ }$
  
  
• C. ${135}^{\circ }$
• D. None of these

Answer 1

The correct option is B

${150}^{\circ }$

$\frac{B}{2}=\sqrt{A^2+\overrightarrow{B}+2ABcos\theta }...\left(i\right)\therefore tan{90}^{\circ }=\frac{Bsin\theta }{A+Bcos\theta }\Rightarrow A+Bcos\theta =0\therefore cos\theta =-\frac{A}{B}$
Hence, from (i) $\frac{B^2}{4}=A^2+B^2-2A^2\Rightarrow A=\sqrt{3}\frac{B}{2}$
$\Rightarrow cos\theta =\frac{A}{B}=-\frac{\sqrt{3}}{2}\therefore \theta ={150}^{\circ }$