If $\vec{a}$, $\vec{b}$, $\vec{c}$, and $\vec{d}$ are unit vectors such that $$\Large(\normalsize\vec{a}\times\vec{b}\Large)\normalsize\cdot\Large(\normalsize\vec{c}\times\vec{d}\Large)\normalsize=1$$ and $$\vec{a}\cdot\vec{c} = \frac{1}{2},$$ then
(A) $\vec{a}$, $\vec{b}$, $\vec{c}$ are non-coplanar
(B) $\vec{b}$, $\vec{c}$, $\vec{d}$ are non-coplanar
(C) $\vec{b}$, $\vec{d}$ are non-parallel
(C) $\vec{a}$, $\vec{d}$ are parallel and $\vec{b}$, $\vec{c}$ are parallel
My approach is as follow $\left( {\overrightarrow a \times \overrightarrow b } \right).\left( {\overrightarrow c \times \overrightarrow d } \right) = 1$
$\left( {\frac{{\overrightarrow a \times \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right).\left( {\frac{{\overrightarrow c \times \overrightarrow d }}{{\left| {\overrightarrow c } \right|\left| {\overrightarrow d } \right|}}} \right) = 1 \Rightarrow \sin \alpha .\sin \beta = 1$. hence $\alpha ,\beta = \frac{\pi }{2}$
Similarly the angle between $\overrightarrow a \& \overrightarrow c $ is $60^{\circ}$ but how will I derieve the conditions from it.
Since $( \vec{a}\times \vec{b} ) \cdot (\vec{c} \times \vec{d} ) = 1 $ and $ \vec{a}, \vec{b}, \vec{c},\vec{d} $ are unit vectors, then we must have
$ \vec{a} \perp \vec{b} $
$ \vec{c} \perp \vec{d} $
$ \vec{c} , \vec{d} $ is a rotation of $\vec{a}, \vec{b} $ about the vector $\vec{a} \times \vec{b}$
This means that choices $(A)$ and $(B)$ are false.
We now have $\vec{a} \cdot \vec{c} = \frac{1}{2} $
Therefore, the only correct choice is $(C)$.