Question

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

• A. a, b and c are non-coplanar
• B. b, c and d are non-coplanar
• C. b and d are non-parallel
• D. None of these

Answer 1

Answer: (D)

None of these

Given $\parallel a\times b\parallel <=\parallel a\parallel \parallel b\parallel =1$ and $\parallel c\times d\parallel <=\parallel c\parallel \parallel d\parallel =1$ and that$\left|(a\times b).(c\times d)\right|<=\parallel a\times b\parallel \ast \parallel c\times d\parallel =1$, with equality only when they're parallel, 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) are ruled out.

Furthermore: since $\parallel a\times b\parallel <=1$ and $\parallel a\times b\parallel <=1$ and $\parallel a\times b\parallel \ast \parallel c\times d\parallel =1$.we have that $\parallel a\times b\parallel =1$ and $\parallel a\times b\parallel =1$ (otherwise we get a trivial contradiction).

This means 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 n a unit circle.

Now, a.c $1/2$ implies a is at an angle 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.