<s>[INST] How do I show the direction angle of a vector in 3d space? Given a vector $P$ in $3\text D$ space, I understand that the rotation matrices around:
the $x$-axis ($yz$ plane) would be:
$$\begin{pmatrix} 1&0&0\\ 0  & \cosα &  \sinα\\ 0 & -\sinα  & \cosα\\ \end{pmatrix}\begin{pmatrix}x\\y\\z \end{pmatrix}=\begin{pmatrix}x'\\y'\\z' \end{pmatrix}$$
the $y$-axis ($xz$ plane) would be:
$$\begin{pmatrix} \cos \alpha&0&-\sin \alpha\\ 0  &1 & 0\\ \sin \alpha & 0  & \cosα\\ \end{pmatrix}\begin{pmatrix}x\\y\\z \end{pmatrix}=\begin{pmatrix}x'\\y'\\z' \end{pmatrix}$$
the $z$-axis ($xy$ plane) would be:
$$\begin{pmatrix} \cos \alpha&\sin \alpha&0\\ -\sin \alpha  & \cosα &  0\\ 0 & 0  & 1\\ \end{pmatrix}\begin{pmatrix}x\\y\\z \end{pmatrix}=\begin{pmatrix}x'\\y'\\z' \end{pmatrix}$$
I however cannot understand neither can I correctly visualize the exact position of the direction angle $α$ (see diagram) in $3\text D$ space. 

N/B: I have drawn (see diagram) the rotation around the $y$-axis to show the illustration of the math. 

 [/INST] Hint.-Look at your first figure. You have
$$(|\vec v|\cos \alpha)^2=(|\vec v|\cos \theta_x)^2+(|\vec v|\cos \theta_y)^2$$ so you have
$$\cos^2(\alpha)=
\left(\frac{x}{\sqrt{x^2+y^2+z^2}}\right)^2+\left(\frac{y}{\sqrt{x^2+y^2+z^2}}\right)^2$$
$$\cos(\alpha)=\sqrt{\frac{x^2+y^2}{x^2+y^2+z^2}}$$
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