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 $|\vec{a}| = |\vec{b}| = |\vec{c}| = |\vec{d}| = 1$.The given condition i

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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 $|\vec{a}| = |\vec{b}| = |\vec{c}| = |\vec{d}| = 1$.The given condition is $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = 1$.Using the property of the dot product,$|(\vec{a} 	imes \vec{b})| |(\vec{c} 	imes \vec{d})| \cos \phi = 1$,where $\phi$ is the angle between the vectors $(\vec{a} 	imes \vec{b})$ and $(\vec{c} 	imes \vec{d})$.Since $|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin 	heta_1 = \sin 	heta_1 \le 1$ and $|\vec{c} 	imes \vec{d}| = |\vec{c}| |\vec{d}| \sin 	heta_2 = \sin 	heta_2 \le 1$,the product $\sin 	heta_1 \sin 	heta_2 \cos \phi = 1$ is only possible if $\sin 	heta_1 = 1$,$\sin 	heta_2 = 1$,and $\cos \phi = 1$.This implies $	heta_1 = \frac{\pi}{2}$,$	heta_2 = \frac{\pi}{2}$,and $\phi = 0$.Since $\phi = 0$,the vectors $(\vec{a} 	imes \vec{b})$ and $(\vec{c} 	imes \vec{d})$ are parallel,which implies that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar.Given $\vec{a} \cdot \vec{c} = \frac{1}{2}$,the angle between $\vec{a}$ and $\vec{c}$ is $\frac{\pi}{3}$.Since all vectors are coplanar and unit vectors,the geometric configuration forces $\vec{b}$ and $\vec{d}$ to be non-parallel.

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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