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(B) The square of the scalar triple product of three vectors is given by the determinant of their Gram matrix:$[vec{a}, vec{b}, vec{c}]^2 = \begin{vma

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$\frac{2\pi}{3}$

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(B) The square of the scalar triple product of three vectors is given by the determinant of their Gram matrix:$[vec{a}, vec{b}, vec{c}]^2 = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{vmatrix}$Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,their dot products with themselves are $1$. Substituting the given values:$[vec{a}, vec{b}, vec{c}]^2 = \begin{vmatrix} 1 & \cos 	heta & \cos 	heta \ \cos 	heta & 1 & \cos 	heta \ \cos 	heta & \cos 	heta & 1 \end{vmatrix}$Evaluating the determinant:$[vec{a}, vec{b}, vec{c}]^2 = 1(1 - \cos^2 	heta) - \cos 	heta(\cos 	heta - \cos^2 	heta) + \cos 	heta(\cos^2 	heta - \cos 	heta)$$= 1 - \cos^2 	heta - \cos^2 	heta + \cos^3 	heta + \cos^3 	heta - \cos^2 	heta$$= 1 - 3\cos^2 	heta + 2\cos^3 	heta$$= (1 - \cos 	heta)^2 (1 + 2\cos 	heta)$Since the square of the scalar triple product must be non-negative,we have $(1 - \cos 	heta)^2 (1 + 2\cos 	heta) \ge 0$.Since $(1 - \cos 	heta)^2 \ge 0$ for all $	heta$,we must have $1 + 2\cos 	heta \ge 0$,which implies $\cos 	heta \ge -\frac{1}{2}$.Given $	heta \in [0, \pi]$,the condition $\cos 	heta \ge -\frac{1}{2}$ implies $	heta \le \frac{2\pi}{3}$.Thus,the maximum value of $	heta$ is $\frac{2\pi}{3}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \cos \theta$. Then the maximum value of $\theta$ is,where $\theta \in [0, \pi]$.

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