Quadratic form(?) of Rotation [closed]

As pointed out by @preferred_anon,

$R = R_y R_x R_z$ is a rotation matrix about an axis $\mathbf{a}$ that can be identified from the Rodriguez formula, and also the angle of rotation $\theta_0$ can be identified from the same formula.

Hence the image of a vector $v_1$ is

$v_2 = R v_1$

where the coordinates of $v_1$ and $v_2$ are in the world coordinate system. Now define a new basis for $\mathbb{R}^3$ with the first vector being the unit vector along the axis of rotation $a$, and the second and third vectors making up a right handed orthonormal basis of $\mathbb{R}^3$

i.e. define

$ R' = [a, u_1, u_2] $

where $u_1 \perp a$ and $u_2 = a \times u_1 $.

Vector $v_1$ can be expressed as

$v_1 = R' v'_1$

and,

$v_2 = R' v'_2$

where $v'_1$ and $v'_2$ are the coordinate vectors in the new basis.

Hence,

$v'_2 = {R'^{-1}}{ R}{ R'} v'_1 = {R'^T}{ R}{ R'} v'_1$

Since $v_2$ is the image of $v_1$ under a rotation about axis $\mathbf{a}$, then $v'_2$ is the image of $v'_1$ under a rotation about the $x'$ axis (the first vector of $R'$)

Thus

${R'^T} {R }{R'} = R_x(\theta_0)$

From which,

$R = {R'} {R_x(\theta_0)}{R'^T} $