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(C) The magnitude of the resultant vector $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula:$R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$T

Vedclass · Accepted Answer

$180$

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(C) The magnitude of the resultant vector $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula:$R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$To minimize $R$,the value of $\cos 	heta$ must be as small as possible.The minimum value of $\cos 	heta$ is $-1$,which occurs when $	heta = 180^{\circ}$.Substituting $\cos 	heta = -1$ into the formula:$R_{\min} = \sqrt{A^2 + B^2 - 2AB} = \sqrt{(A - B)^2} = |A - B|$Thus,the angle $	heta$ must be $180^{\circ}$ for the resultant to be minimum.

What should be the angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ so that the magnitude of the resultant vector is minimum (in $^{\circ}$)?

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