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Question

(C) The magnitude of the resultant vector $C$ of two vectors $A$ and $B$ with an angle $	heta$ between them is given by $C = \sqrt{A^2 + B^2 + 2AB \c

Vedclass · Accepted Answer

$C > |A - B|$

Vedclass · Answer

(C) The magnitude of the resultant vector $C$ of two vectors $A$ and $B$ with an angle $	heta$ between them is given by $C = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $	heta = 120^{\circ}$,we have $\cos 120^{\circ} = -0.5$.Thus,$C = \sqrt{A^2 + B^2 - AB}$.Now,consider the magnitude of the difference vector $D = |A - B| = \sqrt{A^2 + B^2 - 2AB \cos 	heta} = \sqrt{A^2 + B^2 - 2AB \cos 120^{\circ}} = \sqrt{A^2 + B^2 + AB}$.Comparing $C = \sqrt{A^2 + B^2 - AB}$ and $D = \sqrt{A^2 + B^2 + AB}$,it is clear that $C Wait,re-evaluating the options provided: If $A=B$,then $C = A$ and $|A-B| = 0$,so $C > |A-B|$. Generally,for $	heta > 90^{\circ}$,the resultant magnitude is smaller than the sum but larger than the difference of the magnitudes for specific vector configurations. Given the standard interpretation of this vector problem,the correct relation is $C > |A - B|$.

If the angle between two vectors $A$ and $B$ is $120^{\circ}$,then their resultant $C$ will be:

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