Let $P_o$ denote the position with the coordinates $(1,0,0)$ in the Descartes coordinate system $(x,y,z)$.
The point $P_o$ is rotated about the z-axis so that the line $OP_o$ turns directly towards the positive y-axis through an angle $\phi$. The position of the point after this rotation is denoted by $P_1$
$P_1$ is then rotated about the line in the x-y plane perpendicular to $OP_1$ so that the line $OP_1$ turns directly towards the positive z-axis through an angle $\lambda$, giving $P_2$. Find the coordinates of $P_2$
Initially I approached this question using the spherical coordinates:
$$x=rsin(\theta)cos(\phi), y=rsin(\theta)sin(\phi), z=rcos(\theta) \tag{1}$$
where $\theta$ is the polar angle and $\phi$ the azimuthal angle.
Setting $\theta=\frac{\pi}{2}-\lambda$ and $\phi=\phi$ gave me the correct the answer:
$$P_2=(cos(\phi)cos(\lambda),sin(\phi)cos(\lambda),sin(\lambda))\tag{2}$$
However, I then tried an alternative method by rotating the coordinate axes and an incorrect answer was obtained:
I rotated the coordinate axes $(x,y,z)$ by the angle $\phi$ anticlockwise about the z-axis. Denoting the new coordinate axes by $(\bar{x},\bar{y},\bar{z})$, we have
$$x=\bar{x}cos(\phi)-\bar{y}sin(\phi)$$
$$y=\bar{x}sin(\phi)+\bar{y}cos(\phi)$$
$$z=\bar{z} \tag{3}$$
since $$ \left(\begin{matrix} x\\y\\z \end{matrix} \right)=\left(\begin{matrix} cos{(\phi)}&-sin(\phi)&0 \\ sin(\phi)&cos(\phi)&0\\0&0&1\end{matrix}\right) \left(\begin{matrix} \bar{x}\\ \bar{y} \\ \bar{z} \end{matrix}\right) \tag{4}$$
Now in the $(\bar{x},\bar{y},\bar{z})$ coordinate system, $P_1$ has the coordinates $(1,0,0)$.
Rotating $P_1$ through the angle $\lambda$ anticlockwise about $\bar{y}$ gives $P_2$
$$\left(\begin{matrix} cos(\lambda)&0&sin(\lambda)\\0&1&0\\-sin{(\lambda)}&0&cos(\lambda) \end{matrix}\right) \left( \begin{matrix}1\\0\\0 \end{matrix}\right)=\left(\begin{matrix}cos(\lambda)\\0\\-sin(\lambda) \end{matrix}\right)=\left(\begin{matrix}\bar{x}\\\bar{y}\\\bar{z}\end{matrix}\right) \tag{5}$$
Solving for $x,y$ and $z$ via $(3)$ in the original coordinate system yields
$$P_2=(cos(\phi)cos(\lambda),sin(\phi)cos(\lambda),-sin(\lambda))\tag{6}$$
which is not the correct the answer and the problem seems to originate from the $z$ component which has an extra minus sign in front of it.
What conceptual errors are present in my working?
Best Answer
The second rotation must be:
$$\left(\begin{matrix} cos(\lambda)&0&-sin(\lambda)\\0&1&0\\sin{(\lambda)}&0&cos(\lambda) \end{matrix}\right) $$ because the angle $\lambda$ is from the $\bar x$ axis to the $\bar z$ axis