= Solution
The identity in part (a), together with $|a|=1$, gives
$$
\boxed{x_2=\lambda^2\left[a(a\mathbin{\cdot}x_0)-x_0\right]}.
$$
For every $n\geq1$, $x_n$ is perpendicular to $a$, because it is a scalar multiple of $a\times x_{n-1}$. Therefore
$$
\boxed{x_{n+2}=\lambda^2a\times(a\times x_n)=-\lambda^2x_n}.
$$
Also $|x_1|=\lambda|a\times x_0|$ and $|x_2|=\lambda^2|a\times x_0|$. Induction using the two-step recurrence yields
$$
\boxed{|x_n|=\lambda^n|a\times x_0|\qquad(n\geq1)}.
$$
Back to article page
