Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 5A / b / i / Solution

The identity in part (a), together with $|a|=1$, givesFor every $n\ge 1$, $x_n$ is perpendicular to $a$, because it is a scalar multiple of $a\times x_{n-1}$. ThereforeAlso $|x_1|=\lambda |a\times x_0|$ and $|x_2|={\lambda }^2|a\times x_0|$. Induction using the two-step recurrence yields