The resultant of two vectors vec{A} and vec{B | vedclass

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(C) Let the resultant be $\vec{R} = \vec{A} + \vec{B}$.Given that $\vec{R} \perp \vec{A}$,the dot product $\vec{A} \cdot \vec{R} = 0$.$\vec{A} \cdot (

Vedclass · Accepted Answer

$150$

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(C) Let the resultant be $\vec{R} = \vec{A} + \vec{B}$.Given that $\vec{R} \perp \vec{A}$,the dot product $\vec{A} \cdot \vec{R} = 0$.$\vec{A} \cdot (\vec{A} + \vec{B}) = 0 \implies A^2 + AB \cos 	heta = 0 \implies AB \cos 	heta = -A^2$ ... $(i)$Given the magnitude $R = \frac{B}{2}$,we have $R^2 = \frac{B^2}{4}$.Using the law of vector addition,$R^2 = A^2 + B^2 + 2AB \cos 	heta$.Substituting $AB \cos 	heta = -A^2$ into the equation:$\frac{B^2}{4} = A^2 + B^2 + 2(-A^2) = B^2 - A^2$.Rearranging gives $A^2 = B^2 - \frac{B^2}{4} = \frac{3B^2}{4}$,so $A = \frac{\sqrt{3}}{2}B$.Substituting $A$ back into $(i)$:$B(\frac{\sqrt{3}}{2}B) \cos 	heta = -(\frac{\sqrt{3}}{2}B)^2$.$\frac{\sqrt{3}}{2} B^2 \cos 	heta = -\frac{3}{4} B^2$.$\cos 	heta = -\frac{3}{4} \cdot \frac{2}{\sqrt{3}} = -\frac{\sqrt{3}}{2}$.Thus,$	heta = 150^{\circ}$.

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to vector $\vec{A}$,and the resultant magnitude is equal to half of the magnitude of $\vec{B}$. Then,the angle between $\vec{A}$ and $\vec{B}$ is: (in $^{\circ}$)

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