Let vec{C} = vec{A} + vec{B},then: | vedclass

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(B) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by $|\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$,where $	heta$ is

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It is possible to have $|\vec{C}| < |\vec{A}|$ and $|\vec{C}| < |\vec{B}|$.

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(B) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by $|\vec{C}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$,where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.The range of the magnitude of the resultant vector is $|A - B| \le |\vec{C}| \le A + B$.If the angle $	heta$ between $\vec{A}$ and $\vec{B}$ is obtuse (i.e.,$90^\circ < 	heta \le 180^\circ$),the resultant vector $\vec{C}$ can have a magnitude smaller than either $|\vec{A}|$ or $|\vec{B}|$. For example,if $\vec{A}$ and $\vec{B}$ are vectors of equal magnitude $A=B$ and the angle between them is $120^\circ$,then $|\vec{C}| = \sqrt{A^2 + A^2 + 2A^2 \cos(120^\circ)} = \sqrt{2A^2 - A^2} = A$. If the angle is greater than $120^\circ$,$|\vec{C}|$ will be less than $A$ and $B$. Thus,option $(b)$ is correct.

Let $\vec{C} = \vec{A} + \vec{B}$,then:

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