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(C) Given $\vec{v} 	imes \vec{b} = \vec{c}$. Taking the cross product with $\vec{b}$ on both sides: $\vec{b} 	imes (\vec{v} 	imes \vec{b}) = \vec{b

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$\vec{v} = \frac{1}{4}(\vec{b} + \vec{b} 	imes \vec{c})$

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(C) Given $\vec{v} 	imes \vec{b} = \vec{c}$. Taking the cross product with $\vec{b}$ on both sides: $\vec{b} 	imes (\vec{v} 	imes \vec{b}) = \vec{b} 	imes \vec{c}$ Using the vector triple product identity $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$: $(\vec{b} \cdot \vec{b})\vec{v} - (\vec{b} \cdot \vec{v})\vec{b} = \vec{b} 	imes \vec{c}$ Since $|\vec{b}| = 2$,$|\vec{b}|^2 = 4$. Thus,$4\vec{v} - (\vec{b} \cdot \vec{v})\vec{b} = \vec{b} 	imes \vec{c}$. From $\vec{v} 	imes \vec{b} = \vec{c}$,we have $|\vec{v} 	imes \vec{b}| = |\vec{c}|$. $|\vec{v}||\vec{b}| \sin 	heta = \sqrt{3}$,where $	heta$ is the angle between $\vec{v}$ and $\vec{b}$. Since $\vec{v}$ is a unit vector,$|\vec{v}| = 1$. $1 	imes 2 	imes \sin 	heta = \sqrt{3} \implies \sin 	heta = \frac{\sqrt{3}}{2} \implies \cos 	heta = \frac{1}{2}$. Now,$\vec{b} \cdot \vec{v} = |\vec{b}||\vec{v}| \cos 	heta = 2 	imes 1 	imes \frac{1}{2} = 1$. Substituting this into the equation: $4\vec{v} - (1)\vec{b} = \vec{b} 	imes \vec{c}$ $4\vec{v} = \vec{b} + \vec{b} 	imes \vec{c}$ $\vec{v} = \frac{1}{4}(\vec{b} + \vec{b} 	imes \vec{c})$.

Let $\vec{v}$ be a unit vector which follows the equation $\vec{v} \times \vec{b} = \vec{c}$. Also,$|\vec{b}| = 2$ and $|\vec{c}| = \sqrt{3}$. Then,which of the following is true?

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