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(C) The component of vector $\vec{A}$ along $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.The component of vector $\vec{B}$ along $\

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$2: 1$

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(C) The component of vector $\vec{A}$ along $\vec{B}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.The component of vector $\vec{B}$ along $\vec{A}$ is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$.According to the problem, the component of $\vec{A}$ along $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$:$\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 2 \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|} ight)$.Assuming $\vec{A} \cdot \vec{B} 
eq 0$, we can divide both sides by $(\vec{A} \cdot \vec{B})$:$\frac{1}{|\vec{B}|} = \frac{2}{|\vec{A}|}$.Rearranging the terms to find the ratio of magnitudes $\frac{|\vec{A}|}{|\vec{B}|}$:$\frac{|\vec{A}|}{|\vec{B}|} = 2$.Thus, the ratio is $2: 1$.

Column-$I$	Column-$II$
$(a) \|\vec{A} \times \vec{B}\| = 0$	$(i) \theta = 30^{\circ}$
$(b) \|\vec{A} \times \vec{B}\| = 8$	$(ii) \theta = 45^{\circ}$
$(c) \|\vec{A} \times \vec{B}\| = 4$	$(iii) \theta = 90^{\circ}$
$(d) \|\vec{A} \times \vec{B}\| = 4\sqrt{2}$	$(iv) \theta = 0^{\circ}$

If the component of the vector $\vec{A}$ along the vector $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$, then the ratio of magnitudes of vectors $\vec{A}$ and $\vec{B}$ is

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