If vec{a}, vec{b}, vec{c} and vec{d} are unit | vedclass

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(C) Given that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are unit vectors. We have $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = 1$. Using the

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$\vec{b}, \vec{d}$ are non-parallel

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(C) Given that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are unit vectors. We have $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = 1$. Using the identity $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c}) = 1$. Since $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are unit vectors,the maximum value of $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d})$ is $1$,which occurs when $\vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{d}$. This implies that the vectors are coplanar in specific configurations. Given $\vec{a} \cdot \vec{c} = \frac{1}{2}$,the angle between $\vec{a}$ and $\vec{c}$ is $60^{\circ}$. If $\vec{b}$ and $\vec{d}$ were parallel,then $\vec{a} 	imes \vec{b}$ and $\vec{c} 	imes \vec{d}$ would not satisfy the dot product condition of $1$ unless specific constraints are met. By analyzing the geometry,if $\vec{b}$ and $\vec{d}$ were parallel,the scalar triple products would vanish,contradicting the result. Thus,$\vec{b}$ and $\vec{d}$ must be non-parallel. Therefore,option $C$ is correct.

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then :-

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