If vec{r} is a unit vector satisfying vec{r} | vedclass

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(D) Given, $\vec{r} 	imes \vec{a}=\vec{b}$.Taking the cross product with $\vec{a}$ on both sides: $(\vec{r} 	imes \vec{a}) 	imes \vec{a}=\vec{b}

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$\frac{1}{4}[\vec{a}-(\vec{b} 	imes \vec{a})]$

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(D) Given, $\vec{r} 	imes \vec{a}=\vec{b}$.Taking the cross product with $\vec{a}$ on both sides: $(\vec{r} 	imes \vec{a}) 	imes \vec{a}=\vec{b} 	imes \vec{a}$.Using the vector triple product formula $(\vec{A} 	imes \vec{B}) 	imes \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$, we get:$(\vec{r} \cdot \vec{a})\vec{a} - (\vec{a} \cdot \vec{a})\vec{r} = \vec{b} 	imes \vec{a}$.Since $|\vec{a}|=2$, we have $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 4$.Thus, $4\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} - (\vec{b} 	imes \vec{a})$.Now, $|\vec{r} 	imes \vec{a}| = |\vec{b}| \Rightarrow |\vec{r}||\vec{a}|\sin 	heta = \sqrt{3}$.$1 	imes 2 \sin 	heta = \sqrt{3} \Rightarrow \sin 	heta = \frac{\sqrt{3}}{2} \Rightarrow 	heta = \frac{\pi}{3}$.Then, $\vec{r} \cdot \vec{a} = |\vec{r}||\vec{a}|\cos 	heta = 1 	imes 2 	imes \cos \frac{\pi}{3} = 2 	imes \frac{1}{2} = 1$.Substituting $\vec{r} \cdot \vec{a} = 1$ into the equation $4\vec{r} = (\vec{r} \cdot \vec{a})\vec{a} - (\vec{b} 	imes \vec{a})$:$4\vec{r} = \vec{a} - (\vec{b} 	imes \vec{a}) \Rightarrow \vec{r} = \frac{1}{4}[\vec{a} - (\vec{b} 	imes \vec{a})]$.

If $\vec{r}$ is a unit vector satisfying $\vec{r} \times \vec{a}=\vec{b}$, $|\vec{a}|=2$ and $|\vec{b}|=\sqrt{3}$, then one such $\vec{r}=$

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