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

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(C) Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$.Given that $\vec{R} \perp \vec{A}$,the angle between $\vec{R}$ and $\vec{A}$ is $90^{\ci

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$.Given that $\vec{R} \perp \vec{A}$,the angle between $\vec{R}$ and $\vec{A}$ is $90^{\circ}$.From the vector triangle or component method,if $\vec{R}$ is perpendicular to $\vec{A}$,then the component of $\vec{B}$ along $\vec{A}$ must cancel out the magnitude of $\vec{A}$,and the component of $\vec{B}$ perpendicular to $\vec{A}$ must equal the magnitude of $\vec{R}$.Alternatively,using the formula for the angle $\alpha$ that the resultant $\vec{R}$ makes with $\vec{A}$:$	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta}$Since $\vec{R} \perp \vec{A}$,$\alpha = 90^{\circ}$,so $	an 90^{\circ} = \infty$,which implies $A + B \cos 	heta = 0$,or $A = -B \cos 	heta$.The magnitude of the resultant is $R = B \sin 	heta$.Given $R = \frac{B}{2}$,we have $B \sin 	heta = \frac{B}{2}$,which gives $\sin 	heta = \frac{1}{2}$.Thus,$	heta = 150^{\circ}$ (since the angle between the vectors must be obtuse for the resultant to be perpendicular to $\vec{A}$ when $A$ is in the opposite direction of the horizontal component of $B$).Looking at the geometry: The angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ} + \phi$,where $\phi$ is the angle $\vec{B}$ makes with the vertical. From the diagram,$\cos \phi = \frac{R}{B} = \frac{B/2}{B} = \frac{1}{2}$,so $\phi = 60^{\circ}$.Therefore,the angle between $\vec{A}$ and $\vec{B}$ is $90^{\circ} + 60^{\circ} = 150^{\circ}$.

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$ and its magnitude is half that of $\vec{B}$. The angle between vectors $\vec{A}$ and $\vec{B}$ is . . . . . . (in $^{\circ}$)

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