= Solution
A <Givens rotation> equals the identity except in coordinates $i,j$, where its block can be taken as
$$
\begin{pmatrix}c&s\\-s&c\end{pmatrix},\qquad c=\cos\theta,\ s=\sin\theta.
$$
The two block columns are orthonormal because $c^2+s^2=1$; other coordinates are unchanged, so the full matrix is <orthogonal>. If $(v_i,v_j)\ne(0,0)$, take $r=\sqrt{v_i^2+v_j^2}$, $c=v_i/r$ and $s=v_j/r$. The transformed pair is $(r,0)$. If both entries are zero, the identity already suffices.
For the specified vector one possible matrix is
$$
\boxed{Q=\begin{pmatrix}4/5&0&-3/5\\3/5&0&4/5\\0&-1&0\end{pmatrix},\qquad Qv=\begin{pmatrix}0\\5\\0\end{pmatrix}}.
$$
Its rows are orthonormal. It is \b[not unique]: premultiply by any <orthogonal> transformation fixing the second coordinate axis pointwise. This rotates or reflects the other two output coordinates without changing the required image. Equivalently its second row must be $v^T/\|v\|$, but any orthonormal completion of that row works.
Back to article page
