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(B) Let the resultant vector be $R$. Given that $R \perp A$,the angle between $R$ and $A$ is $90^\circ$. From the vector addition formula,the angle $\

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(B) Let the resultant vector be $R$. Given that $R \perp A$,the angle between $R$ and $A$ is $90^\circ$. From the vector addition formula,the angle $\alpha$ that the resultant $R$ makes with vector $A$ is given by $	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta}$. Since $\alpha = 90^\circ$,$	an 90^\circ$ is undefined,which implies the denominator must be zero: $A + B \cos 	heta = 0$,so $\cos 	heta = -\frac{A}{B}$. The magnitude of the resultant is $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$. Given $R = \frac{B}{2}$,we have $\frac{B^2}{4} = A^2 + B^2 + 2AB \cos 	heta$. Substituting $\cos 	heta = -\frac{A}{B}$ into the equation: $\frac{B^2}{4} = A^2 + B^2 + 2AB(-\frac{A}{B}) = A^2 + B^2 - 2A^2 = B^2 - A^2$. Rearranging gives $A^2 = B^2 - \frac{B^2}{4} = \frac{3B^2}{4}$,so $A = \frac{\sqrt{3}}{2}B$. Finally,$\cos 	heta = -\frac{A}{B} = -\frac{\sqrt{3}}{2}$. Therefore,$	heta = 150^\circ$.

Column $I$	Column $II$
$(A) \|A+B\|$	$(p) \frac{\sqrt{3}}{2} x^2$
$(B) \|A-B\|$	$(q) x$
$(C) A \cdot B$	$(r) \sqrt{3} x$
$(D) \|A \times B\|$	$(s) \frac{x^2}{2}$

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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