If |vec A + vec B| = |vec A| + |vec B|,then t | vedclass

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

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$0$

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(C) The magnitude of the resultant of two vectors $\vec A$ and $\vec B$ is given by the formula:$|\vec R| = |\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$,where $	heta$ is the angle between the two vectors.Given the condition $|\vec A + \vec B| = |\vec A| + |\vec B|$,we substitute the magnitude formula:$\sqrt{A^2 + B^2 + 2AB \cos 	heta} = A + B$Squaring both sides:$A^2 + B^2 + 2AB \cos 	heta = (A + B)^2$$A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 + 2AB$$2AB \cos 	heta = 2AB$$\cos 	heta = 1$Therefore,$	heta = 0^o$.

If $|\vec A + \vec B| = |\vec A| + |\vec B|$,then the angle between $\vec A$ and $\vec B$ will be ....... $^o$.

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