Question

The resultant of two vectors A and B perpendicular to the vector A and its magnitude is half of the magnitudeof vector B....find out the angle between A and B???

Answer 1

the angle between A Nd B would be 120 degree

Answer 2

The angle between “A and B” 150°.

Solution:

Formula of “Magnitude of resultant” of ‘A and B’ is,  

$R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}$

According to question,  

Magnitude of resultant = Half of magnitude of B.  

Hence, we can say that  

$\sqrt{A^{2}+B^{2}+2 A \cdot B \cos \theta}=\frac{B}{2}$

Taking square both sides,

$A^{2}+B^{2}+2 A . B \cos \theta=\frac{B^{2}}{4}$

$A^{2}+2 A B \cos \theta+\frac{3 B^{2}}{4}=0 \ldots \ldots \ldots \ldots(1)$

A and R is perpendicular upon each other.  Hence, we can say that  

e.g., $\mathrm{A}(\mathrm{A}+\mathrm{B})=0$

$A . A+A . B=0$

$A^{2}+A . B \cos \theta=0$

$\cos \theta=-\frac{A}{B}$, put it in equation (1)  

$A^{2}+2 A . B\left(-\frac{A}{B}\right)+\frac{3 B^{2}}{4}=0$

$A^{2}-2 A^{2}+\frac{3 B^{2}}{4}=0$

$A=\frac{\sqrt{3 B}}{2}$

Now, $\cos \theta=-\frac{A}{B}=-\frac{\sqrt{3 B}}{2 B}=-\frac{\sqrt{3}}{2}$

$\cos \theta=\cos 150^{\circ}$

$\bold{\Rightarrow \theta=150^{\circ}}$