Rotating base frame thus one of the child frames points to a specific location Lately I've been studying with frames and transformation matrices. I have a problem that looks simple yet I couldn't wrap my head around it. Here is my problem with images.
Let's say we have two frames and a point (white sphere in picture). Frame A and Frame B. Frame B is child of Frame A.
Frame A, Frame B and a white sphere as a point.
I want to rotate Frame A thus the x-axis of Frame B points to my point.
Desired situation.
I know transformation matrices and location of point. Which math should I do exactly to calculate how much rotation do I need to rotate Frame A ?
 A: Let frame $A$ be specified by vector $d_1$ and rotation matrix $R_1$ such that
$p_0 = d_1 + R_1 p_1 \hspace{48pt} (1)$
where $p_0$ is the world coordinate vector, and $p_1$ is the coordinate vector with respect to frame $A$.  Frame $B$ is the child of frame $A$, thus
$ p_1 = d_2 + R_2 p_2 \hspace{48pt}(2)$
where $p_2$ is the coordinate vector with respect to frame $B$.
Let $P$ be the world coordinate of the given fixed point, then
$P = d_1 + R_1 (d_2 + R_2 p_2 ) \hspace{48pt}(3)$
Since $p_2$ is along the $x$-axis of the child frame $B$, then
$p_2 = t \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = t v \hspace{48pt}(4)$
Plug this into $(3)$:
$P = d_1 + R_1 d_2 + t R_1 R_2 v \hspace{48pt}(5) $
Here $R_1$ is the desired rotation matrix not the starting one. $t$ and $R_1$ are unknown in $(5)$.
From $(5)$ it follows that
$ P - d_1 = R_1 ( d_2 + t R_2 v ) \hspace{48pt}(6) $
Since $R_1$ is a rotation matrix, then
$ \| P - d_1 \| = \| d_2 + t R_2 v \| \hspace{48pt}(7) $
This can be used to determine $t$ explicitly using dot product as follows:
$ (P - d_1) \cdot (P - d_1) = (d_2 + t R_2 v) \cdot (d_2 + t R_2 v)  \hspace{48pt} (8) $
And taking the positive root of this quadratic equation.  Now define
$ q_1 = d_2 + t R_2 v \hspace{48pt} (9a)$
$ q_2 = P - d_1\hspace{48pt} (9b)$
then,
$ q_2 = R_1 q_1 \hspace{48pt}(9c) $
where both $q_1$ and $q_2$ are known vectors.
The solution for $R_1$ satisfying $(9c)$ is not unique; there is an infinite number of possible rotation matrices.  To generate the $R_1$ with minimum rotation angle, we take the axis of its rotation to be
$ a = \dfrac{q_1 \times q_2 }{\| q_1 \times q_2 \|} \hspace{48pt}(10) $
and the angle of rotation is
$ \theta = \cos^{-1} ( q_1 \cdot q_2 )\hspace{48pt}(11) $
Now we can use the Rodrigues' rotation matrix formula, which is
$ R_1 = {a a}^T + (I - {aa}^T) \cos \theta + S_a \sin \theta \hspace{48pt}(12) $
Matrix $S_a = \begin{bmatrix} 0 && - a_z && a_y \\ a_z && 0 && -a_x \\ -a_y && a_x && 0 \end{bmatrix} \hspace{48pt}(13) $
We now have one possible desired rotation matrix.  Finally, let's say we started with the original rotation matrix for the frame as $R_{1O}$, and we want to apply rotation $R^*$ with respect to the world coordinate system to $R_{1O}$ that modifies it to the $R_1$ that we derived, the relation between them is
$ R_1 = R^* R_{1O} \hspace{48pt} (14) $
Hence, the required rotation is
$R^* = R_1 R_{1O}^T \hspace{48pt} (15)$
