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

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

The recurrence $x_{n+2}=-{\lambda }^2{x}_n$ shows that every odd-indexed position vector is a scalar multiple of $x_1$, while every positive even-indexed position vector is a scalar multiple of $x_2$. HenceMoreover $x_2=\lambda a\times x_1$, so these two straight lines through the origin are perpendicular.

Let $d_n=x_{n+1}-x_n$ be the direction vector of the segment $X_n{X}_{n+1}$. Consecutive vectors $x_n,x_{n+1}$ are perpendicular and $|x_{n+1}|=\lambda |x_n|$. Using $x_{n+2}=-{\lambda }^2{x}_n$,Thus adjacent segments are perpendicular. AlsosoIf $0<\lambda <1$, the norm formula in part (i) gives $x_n\to 0$. If $\lambda =1$, then $x_{n+2}=-x_n$: the points $X_1,X_2,-X_1,-X_2$ repeat cyclically and form a square centered at the origin.

Put $L=|x_n|$ and choose perpendicular unit vectors $e,f$ such that $x_n=Le$ and $x_{n+1}=\lambda Lf$. Direction vectors for the two lines areandTheir scalar product and norms giveTherefore the acute angle between the straight lines satisfies