If |vec{A} + vec{B}| = |vec{A}| + |vec{B}|,th | vedclass

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(A) For two vectors $\vec{A}$ and $\vec{B}$,let the angle between them be $	heta$. The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}

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(A) For two vectors $\vec{A}$ and $\vec{B}$,let the angle between them be $	heta$. The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by:$|\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta}$Given the condition $|\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}|$,we square both sides:$|\vec{A} + \vec{B}|^2 = (|\vec{A}| + |\vec{B}|)^2$$|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|$Subtracting $|\vec{A}|^2 + |\vec{B}|^2$ from both sides:$2|\vec{A}||\vec{B}| \cos 	heta = 2|\vec{A}||\vec{B}|$Assuming $\vec{A}$ and $\vec{B}$ are non-zero vectors,we divide by $2|\vec{A}||\vec{B}|$:$\cos 	heta = 1$Since $\cos 	heta = 1$,the angle $	heta = 0^o$.

If $|\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}|$,the angle between $\vec{A}$ and $\vec{B}$ is ....... $^o$.

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