The resultant of two vectors vec{A} and vec{B | vedclass

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(B) Let the angle between $\vec{A}$ and $\vec{B}$ be $	heta$. The resultant $\vec{C} = \vec{A} + \vec{B}$.When the magnitude of $\vec{B}$ is doubled,

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$B$

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(B) Let the angle between $\vec{A}$ and $\vec{B}$ be $	heta$. The resultant $\vec{C} = \vec{A} + \vec{B}$.When the magnitude of $\vec{B}$ is doubled,the new resultant is $\vec{C}' = \vec{A} + 2\vec{B}$.Given that $\vec{C}'$ is perpendicular to $\vec{A}$,their dot product is zero: $\vec{A} \cdot (\vec{A} + 2\vec{B}) = 0$.$A^2 + 2(\vec{A} \cdot \vec{B}) = 0 \implies A^2 + 2AB \cos 	heta = 0$.Thus,$2AB \cos 	heta = -A^2$.The magnitude of the original resultant $\vec{C}$ is given by $C^2 = A^2 + B^2 + 2AB \cos 	heta$.Substituting $2AB \cos 	heta = -A^2$ into the equation:$C^2 = A^2 + B^2 - A^2 = B^2$.Therefore,the magnitude of $\vec{C}$ is $B$.

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$. If the magnitude of $\vec{B}$ is doubled,the new resultant vector becomes perpendicular to $\vec{A}$. Then,the magnitude of $\vec{C}$ is:

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