\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
The coordinates of \(P_1\) are \[ P_1 (\cos \phi , \sin \phi , 0), \] and the coordinates of \(Q_1\) are \[ Q_1 (- \sin \phi , \cos \phi , 0). \]
Since the rotation is about \(z\)-axis, the position of \(R\) remains unchanged \[ R_1 (0, 0, 1). \]
This rotation axis is precisely \(OQ_1\), since it is contained in the \(x\)-\(y\) plane, and is perpendicular to \(OP_1\). Hence, the position of \(Q\) remains unchanged, and hence \[ Q_2 (- \sin \phi , \cos \phi , 0). \]
If we drop a perpendicular from \(P_2\) to the line \(OP_1\), and call the intersection be \(P'\). We can see from trigonometry that \[ P_2 P' = \sin \lambda , \] and \[ O P' = \cos \lambda . \]
Hence, the \(x\)-coordinate of \(P_2\) is \(\cos \lambda \cos \phi \), and the \(y\)-coordinate of \(P_2\) is \(\cos \lambda \sin \phi \). The \(z\)-coordinate of \(P_2\) is \(\sin \lambda \), and hence \[ P_2 (\cos \phi \cos \lambda , \sin \phi \cos \lambda , \sin \lambda ). \]
The relative positions of \(P, Q\) and \(R\) remains unchanged under rotation, and hence \begin {align*} \vect {r}_{R_2} & = \vect {r}_{P_2} \times \vect {r}_{Q_2} \\ & = \begin {pmatrix} \cos \phi \cos \lambda \\ \sin \phi \cos \lambda \\ \sin \lambda \end {pmatrix} \times \begin {pmatrix} - \sin \phi \\ \cos \phi \\ 0 \end {pmatrix} \\ & = \begin {vmatrix} \ihat & \jhat & \khat \\ \cos \phi \cos \lambda & \sin \phi \cos \lambda & \sin \lambda \\ - \sin \phi & \cos \phi & 0 \end {vmatrix} \\ & = \begin {pmatrix} \sin \phi \cos \lambda \cdot 0 - \sin \lambda \cos \phi \\ - (\cos \phi \cos \lambda \cdot 0 + \sin \lambda \cdot \sin \phi ) \\ \cos \phi \cos \lambda \cdot \cos \phi + \sin \phi \cos \lambda \cdot \sin \phi \end {pmatrix} \\ & = \begin {pmatrix} - \sin \lambda \cos \phi \\ - \sin \lambda \sin \phi \\ \cos ^2 \phi \cos \lambda + \sin ^2 \phi \cos \lambda \end {pmatrix} \\ & = \begin {pmatrix} - \sin \lambda \cos \phi \\ - \sin \lambda \sin \phi \\ \cos \lambda \end {pmatrix}, \end {align*}
and hence \[ R_2 (- \sin \lambda \cos \phi , - \sin \lambda \sin \phi , \cos \lambda ). \]
The angle of rotation is the angle between \(O P_0\) and \(O P_2\), and hence \begin {align*} \cos \theta & = \frac {\bvect {O P_0} \cdot \bvect {O P_2}}{\abs *{\bvect {O P_0}} \cdot \abs *{\bvect {O P_2}}} \\ & = \begin {pmatrix} 1 \\ 0 \\ 0 \end {pmatrix} \cdot \begin {pmatrix} \cos \phi \cos \lambda \\ \sin \phi \cos \lambda \\ \sin \lambda \end {pmatrix} \\ & = \cos \phi \cos \lambda , \end {align*}
as desired.
The axis of this rotation must be perpendicular to both \(O P_1\) and \(O P_2\), and hence their cross product \begin {align*} \bvect {O P_0} \times \bvect {O P_2} & = \begin {pmatrix} 1 \\ 0 \\ 0 \end {pmatrix} \times \begin {pmatrix} \cos \phi \cos \lambda \\ \sin \phi \cos \lambda \\ \sin \lambda \end {pmatrix} \\ & = \begin {vmatrix} \ihat & \jhat & \khat \\ 1 & 0 & 0 \\ \cos \phi \cos \lambda & \sin \phi \cos \lambda & \sin \lambda \end {vmatrix} \\ & = \begin {pmatrix} 0 \\ - \sin \lambda \\ \sin \phi \cos \lambda \end {pmatrix} \end {align*}
is a vector in the direction of the axis.