Given two distinct nonzero vectors $v_{1}and{v}_2$ in 3 dimensions, define a sequence of vectors by  $v_{n+2}=v_n\times v_{n+1}\left(so{v}_3=v_1\times v_2,v_4=v_2\times v_{3}andsoon\right)$

Let,  $S=\{v_n/n=1,2,\mathrm{.....}\}andU=\{\frac{v_n}{\left|v_n\right|}/n=1,2,\mathrm{.....}\}.$

(Note: Here $\times$ denotes the cross product of vectors and  $\left|v\right|$ denotes the magnitude of the vector v. The vector 0 with 0 magnitude, if it occurs in S, is counted. But in that case of course the 0 vector is not considered while listing elements of U. number of elements in set $S$  is called cardinality of $S$ ) which of the following statement(s) is/are True

It is easiest to do this geometrically, remembering that the cross product $p\times q$ of vectors p and q is perpendicular to both of them and $|p\times q|=\left|p\right|\left|q\right|$ sin (angle between n p and q) $=\left|p\right|\left|q\right|$ if p and q are perpendicular. The cross product of nonzero vectors is zero if and only if the vectors are collinear. It is see that the only way the zero vector is S is if $v_3$ is zero, which will happen only when the nonzero vectors  $v_1$ and  $v_2$ are collinear, and in that case the sequence is zero all the way from $v_3$  onwards.
As the staring vectors  $v_1$ and $v_2$  are distinct and nonzero, the third vector  $v_3=v_1\times v_2$, being perpendicular to both $v_1$ and $v_2$ is distinct from them. This is true even if $v_3$ is 0 due to $v_1$ and $v_2$  being collinear. So (a) false.