The resultant of two vectors A and B is perpe | vedclass

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Question

(b) $\frac{B}{2} = \sqrt {{A^2} + {B^2} + 2AB\;\cos 	heta } $&hellip;$(i)$

 $	an 90^\circ = \frac{{B\sin 	heta }}{{A + B\cos 	heta }} \Rig

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(b) $\frac{B}{2} = \sqrt {{A^2} + {B^2} + 2AB\;\cos 	heta } $&hellip;$(i)$

 $	an 90^\circ = \frac{{B\sin 	heta }}{{A + B\cos 	heta }} \Rightarrow A + B\cos 	heta = 0$

$\cos 	heta = - \frac{A}{B}$

Hence, from $(i)$ $\frac{{{B^2}}}{4} = {A^2} + {B^2} - 2{A^2} \Rightarrow A = \sqrt 3 \frac{B}{2}$

$&rArr; \cos 	heta = - \frac{A}{B} = - \frac{{\sqrt 3 }}{2}$

$&rArr; 	heta = 150^\circ $

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 ....... $^o$

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