Question

If a, b, c, d are unit vectors such that (axb).(cxd)=1 and a.c=1/2 then

a) a, b, c are non-coplanar

,

b) b, c, d are non-coplanar

,

c) b, d are non-parallel

,

d) a, dare parallel and b, c are parallel

.

Answer 1

Given ||a x b|| <= ||a|| ||b|| (since, a, b are unit vectors)
<= 1
and ||c x d|| <= ||c|| ||d||
<= 1
and that | (a x b) . (c x d) | <= ||a x b|| * || c x d||
<= 1
( since , we know modulus of cross product will be <= 1 if the vectors are unit vectors).
with equality only when they are parallel,
from the question we have that a x b and c x d are parallel. (more specifically co-directional since their dot product is positive).

But this means that a,b define the very same plane as c,d ,Since
their cross products are perpendicular to their respective plane.

Therefore a,b,c and d are all co=planar unit vectors. So 1) and 2) options are wrong.

Furthermore , since || a x b|| <=1 and ||c x d||<=1 and ||a x b||*||c x d|| = 1, we have that ||a x b|| = 1 and ||c x d|| = 1 (otherwise we get a trivial contradiction).

This mans that a is perpendicular to b and c is perpendicular to d.
And as shown above , all four are coplanar unit vectors. Therefore we can visualise them in a unit circle.

Now, a.c = 1/2 implies a is at anangle of 60 degrees with c.
Since b and d are perpendicular to a and c respectively, this means the angle between b and d is either 60 or 120 degrees and therefore b and d are guaranteed not to be parallel.

Therefore 3) is correct.