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(B) Given that $\vec{p}, \vec{q}, \vec{r}$ are unit vectors,so $|\vec{p}| = |\vec{q}| = |\vec{r}| = 1$. Since they are equally inclined at an angle $\

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$2\cos 	heta \sin \left( \frac{	heta}{2} ight)$

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(B) Given that $\vec{p}, \vec{q}, \vec{r}$ are unit vectors,so $|\vec{p}| = |\vec{q}| = |\vec{r}| = 1$. Since they are equally inclined at an angle $	heta$,we have $\vec{p} \cdot \vec{q} = \vec{q} \cdot \vec{r} = \vec{r} \cdot \vec{p} = \cos 	heta$. Using the vector triple product formula,$\vec{p} 	imes (\vec{q} 	imes \vec{r}) = (\vec{p} \cdot \vec{r})\vec{q} - (\vec{p} \cdot \vec{q})\vec{r} = \cos 	heta \vec{q} - \cos 	heta \vec{r} = \cos 	heta (\vec{q} - \vec{r})$. Taking the magnitude,$|\vec{p} 	imes (\vec{q} 	imes \vec{r})| = |\cos 	heta| |\vec{q} - \vec{r}|$. Since $	heta$ is an acute angle,$\cos 	heta > 0$. $|\vec{q} - \vec{r}| = \sqrt{|\vec{q}|^2 + |\vec{r}|^2 - 2(\vec{q} \cdot \vec{r})} = \sqrt{1 + 1 - 2 \cos 	heta} = \sqrt{2(1 - \cos 	heta)} = \sqrt{4 \sin^2 \left( \frac{	heta}{2} ight)} = 2 \sin \left( \frac{	heta}{2} ight)$. Thus,$|\vec{p} 	imes (\vec{q} 	imes \vec{r})| = \cos 	heta \cdot 2 \sin \left( \frac{	heta}{2} ight) = 2 \cos 	heta \sin \left( \frac{	heta}{2} ight)$.

Let $\vec{p}, \vec{q},$ and $\vec{r}$ be three non-coplanar unit vectors equally inclined to each other at an acute angle $\theta$. The value of $|\vec{p} \times (\vec{q} \times \vec{r})|$ is:

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