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 $\overrightarrow{\mathrm{C}}$ is

Let $\mathrm{C} = \mathrm{A} + \mathrm{B}$ be the original resultant vector. When the magnitude of $\mathrm{B}$ is doubled, the new resultant vector becomes $\mathrm{C}' = \mathrm{A} + 2\mathrm{B}$, which is perpendicular to $\mathrm{A}$. For perpendicular vectors, $\mathrm{C}' \cdot \mathrm{A} = 0$.

Expanding the dot product gives $(\mathrm{A} + 2\mathrm{B}) \cdot \mathrm{A} = 0$, which simplifies to $A^2 + 2(\mathrm{A} \cdot \mathrm{B}) = 0$. Solving yields $\mathrm{A} \cdot \mathrm{B} = -\frac{A^2}{2}$.

The magnitude of $\mathrm{C}$ is found from $|\mathrm{C}|^2 = (\mathrm{A} + \mathrm{B}) \cdot (\mathrm{A} + \mathrm{B}) = A^2 + B^2 + 2(\mathrm{A} \cdot \mathrm{B})$. Substituting the dot product gives $|\mathrm{C}|^2 = A^2 + B^2 - A^2 = B^2$. Thus, $|\mathrm{C}| = B$.