= Solution
The transformed components are $a'=e^{i\alpha_3/2}a$, $b'=e^{-i\alpha_3/2}b$. Hence $(a')^*b'=e^{-i\alpha_3}a^*b$, while the two <moduli> are unchanged. Taking real and imaginary parts in the expression from part (ii) gives
$$
\boxed{\begin{pmatrix}x'_1\\x'_2\\x'_3\end{pmatrix}=\begin{pmatrix}\cos\alpha_3&\sin\alpha_3&0\\-\sin\alpha_3&\cos\alpha_3&0\\0&0&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}.}
$$
This is a right-handed active <rotation> through \b[$-\alpha_3$ about the third axis]. The minus sign follows from the positive sign in $U=e^{+iT/2}$; changing that convention reverses the rotation. A $2\pi$ spatial <rotation> changes the <spinor> sign while leaving the three-vector unchanged.
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