TITLE: If $\vec{a},\vec{b},\vec{c},\vec{d}$are unit vectors, $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ and $\vec{a}.\vec{c}=\frac{1}{2}$,then
QUESTION [5 upvotes]: If $\vec{a},\vec{b},\vec{c},\vec{d}$are unit vectors such that $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ and $\vec{a}.\vec{c}=\frac{1}{2}$,then
$(A)\vec{a},\vec{b},\vec{c}$ are non-coplanar
$(B)\vec{a},\vec{b},\vec{d}$ are non-coplanar
$(C)\vec{b},\vec{d}$ are non-parallel.
$(D)\vec{a},\vec{d}$ are parallel and $\vec{b}$ and $\vec{c}$ are parallel.

My Attempt:
$(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$
$(\vec{a}.\vec{c})(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$
$\frac{1}{2}(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$

Book's Solution which i did not understand:
$(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ is possible only when $|\vec{a}\times\vec{b}|=|\vec{c}\times\vec{d}|=0$ and $(\vec{a}\times\vec{b})$ is parallel to $(\vec{c}\times\vec{d})$ And the correct option given is $(C)$.
I could not solve this problem after some efforts.Please help me.Thanks.

REPLY [4 votes]: While Michael's hint was a bit vague, it did answer the question nicely. This answer is just an elaboration on it.
Since the vectors given are all unit vectors, obviously the maximum value of $|\vec{a}\times\vec{b}|$ and  $|\vec{c}\times\vec{d}|$ is unity. Then it follows that the maximum value of $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})$ is also unity. This maximum clearly occurs when $\vec{a}\perp\vec{b}$ , $\vec{c}\perp\vec{d}$ and $\vec{a}\times\vec{b}\parallel\vec{c}\times\vec{d} $
Now you just have to use your imagination. Since the two planes that are spanned by the pairs of vectors $\vec{a},\vec{b}$ and $\vec{c},\vec{d}$ are parallel, and $\vec{a}$  is not collinear with $\vec{c}$, there is no way for $\vec{b}$ to be collinear with $\vec{d}$.