<s>[INST] let $\mathbf a$ and $\mathbf b$ be 3D vectors. Find a $3\times3$ matrix $\mathbf R$ such that $\mathbf {Ra} = \mathbf a_{\bot \mathbf b}$. Hi as the title says im trying to find this.
let $\mathbf a$ and $\mathbf b$ be 3D vectors. Find a  $3\times3$ matrix $\mathbf R$ such that  $\mathbf {Ra} = \mathbf a_{\bot \mathbf b}$.
according to my exercises the answer is
$$ R = \frac{1}{b^2} 
    \begin{bmatrix}
    b^2_y+b^2_z & -b_xb_y & -b_xb_z \\
    -b_xb_y & b^2_x+b^2_z & -b_yb_z \\
    -b_xb_z & -b_yb_z & b^2_x+b^2_y \\
    \end{bmatrix}
$$
I haven't been able to get to this solution and i've managed to get as far as
$$
 a_{\bot b} = a - a_{||b} = a - \frac{a\cdot b}{b^2}b
$$
and i can substitute $ a_{||b} $ for its expression as a matrix product  $$
 a_{||b} = \frac{1}{b^2}bb^{\mathrm T}a $$
and this is an outer product so it becomes
$$a_{\bot b} = \frac{1}{b^2}\begin{bmatrix}
    b^2_x & b_xb_y & b_xb_z \\
    b_xb_y & b^2_y & b_yb_z \\
    b_xb_z & b_yb_z & b^2_z+b^2_y \\
    \end{bmatrix}\begin{bmatrix}
    a_x  \\
    a_y  \\
    a_z  \\
    \end{bmatrix}$$
from this I can get
$$
Ra = a -  \frac{1}{b^2}\begin{bmatrix}
    b^2_x & b_xb_y & b_xb_z \\
    b_xb_y & b^2_y & b_yb_z \\
    b_xb_z & b_yb_z & b^2_z+b^2_y \\
    \end{bmatrix}\begin{bmatrix}
    a_x  \\
    a_y  \\
    a_z  \\
    \end{bmatrix}
$$
This is as far as i was able to get and im not sure the steps needed to get the last equation to the first.
Thanks for any insight anyone can provide.
 [/INST] The last few steps will be
$$ \begin{align*} Ra &= a - \frac{1}{b^2}\begin{bmatrix}b_x^2 & b_xb_y & b_xb_z\\ b_xb_y & b_y^2 & b_yb_z\\ b_xb_z & b_yb_z & b_z^2\end{bmatrix}a\\ &= \frac{1}{b^2}\Bigg(b^2I -  \begin{bmatrix}b_x^2 & b_xb_y & b_xb_z\\ b_xb_y & b_y^2 & b_yb_z\\ b_xb_z & b_yb_z & b_z^2\end{bmatrix}\Bigg)a \\ \end{align*}$$
Notice that $b^2 = b_x^2 + b_y^2 + b_z^2$. So
$$\begin{align*} Ra &= \frac{1}{b^2}\Bigg(\begin{bmatrix}b_x^2  + b_y^2 + b_z^2& 0 & 0\\ 0 & b_x^2  + b_y^2 + b_z^2 & 0\\ 0 & 0 & b_x^2  + b_y^2 + b_z^2\end{bmatrix}-\begin{bmatrix}b_x^2 & b_xb_y & b_xb_z\\ b_xb_y & b_y^2 & b_yb_z\\ b_xb_z & b_yb_z & b_z^2\end{bmatrix}\Bigg)a \\ &= \frac{1}{b^2}\begin{bmatrix}b_y^2 + b_z^2 & -b_xb_y & -b_xb_z\\ -b_xb_y & b_x^2 + b_z^2 & -b_yb_z\\ -b_xb_z & -b_yb_z & b_x^2 + b_y^2\end{bmatrix}a\end{align*} $$
Hence
$$ R = \frac{1}{b^2}\begin{bmatrix}b_y^2 + b_z^2 & -b_xb_y & -b_xb_z\\ -b_xb_y & b_x^2 + b_z^2 & -b_yb_z\\ -b_xb_z & -b_yb_z & b_x^2 + b_y^2\end{bmatrix}$$
</s>