Question

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

• A. $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar
• B. $\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ are non coplanar
• C. $\overrightarrow{b},\overrightarrow{d}$ are non parallel
• D. $\overrightarrow{a},\overrightarrow{d}$ are parallel and $\overrightarrow{b},\overrightarrow{c}$ are parallel

Answer 1

The correct option is C $\overrightarrow{b},\overrightarrow{d}$ are non parallel

Given ||a x b|| <= ||a|| ||b|| = 1
and ||c x d|| <= ||c|| ||d|| = 1
and that | (a x b) . (c x d) | <= ||a x b|| * ||c x d|| = 1, with adequation alone if they're parallel, we accept that a x b and c x d are alongside (more accurately co-directional back their dot artefact is positive).
But this agency that a, b ascertain the actual aforementioned even as c, d, back their cantankerous articles are erect to their corresponding plane.
Therefore a, b, c and d are all co-planar assemblage vectors. So 1) and 2) are disqualified out.
Furthermore, back ||a x b|| <= 1 and ||c x d|| <= 1 and ||a x b|| * ||c x d|| = 1, we accept that ||a x b|| = 1 and ||c x d|| = 1 (otherwise we get a atomic contradiction).
This agency that a is erect to b and c is erect to d.
And as apparent above, all four are coplanar assemblage vectors. Accordingly we can visualise them in a assemblage circle.
Now, a.c = 1/2 implies a is at an bend of 60 degrees with c
Since b and d are erect to a and c respectively, this agency the bend amid b and d is either 60 or 120 degrees and accordingly b and d are affirmed not to be parallel.