Verifying that the Rodrigues formula gives the same result as $Rot(x,\phi)$? How to verify that the Rodrigues formula with $x$ as an axis of rotation and $\phi$ the angle of rotation with $u$ a unit vector along $x$ and  $Rot(x,\phi)$ gives the same result?

I only know that
$$Rot_x(\phi)=
\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\phi & -\sin\phi\\
0 & \sin\phi & \cos\phi
\end{pmatrix}$$
I have:
\begin{align*}
\begin{pmatrix}
\cos\phi & 0 & 0\\
0 & \cos\phi & 0\\
0 & 0 & \cos\phi
\end{pmatrix}+
u*u^T(1-\cos\phi)+\hat u\sin\phi
\end{align*}
But I don't know what $u*u^T$ is and not sure of $\hat u$
 A: For $u$ a column vector $u=\left(\begin{array}{c}u_x\\u_y\\u_z\end{array}\right)$ we have $u^T=(u_x\ u_y\ u_z)$, so $$u u^T=\left(\begin{array}{ccc}u_x^2 & u_x u_y& u_x u_z\\u_xu_y&u_y^2&u_y u_z\\u_xu_z&u_yu_z&u_z^2\end{array}\right)$$
$\hat{u}$ is the cross product matrix (see for example https://en.wikipedia.org/wiki/Cross_product#Conversion_to_matrix_multiplication) and has the form
$$\hat{u} =\left(\begin{array}{ccc}0 & -u_z& u_y\\u_z&0&-u_x\\-u_y&u_x&0\end{array}\right)$$
For your problem, $u$ is the rotation axis. For the rotation around $x$ axis, $u=\left(\begin{array}{c}1\\0\\0\end{array}\right)$, $u u^T=\left(\begin{array}{ccc}1 & 0& 0\\0&0&0\\0&0&0\end{array}\right)$, and $\hat{u} =\left(\begin{array}{ccc}0 & 0& 0\\0&0&-1\\0&1&0\end{array}\right)$.
Putting it all together, we get:
$$\left(\begin{array}{ccc}\cos\phi & 0& 0\\0&\cos\phi&0\\0&0&\cos\phi\end{array}\right)+\left(\begin{array}{ccc}1-\cos\phi & 0& 0\\0&0&0\\0&0&0\end{array}\right)+\left(\begin{array}{ccc}0 & 0& 0\\0&0&-\sin\phi\\0&\sin\phi&0\end{array}\right)=\left(\begin{array}{ccc}1 & 0& 0\\0&\cos\phi&-\sin\phi\\0&\sin\phi&\cos\phi\end{array}\right)$$
You can get similar results if $u=\left(\begin{array}{c}0\\1\\0\end{array}\right)$ for rotations around $y$ axis and $u=\left(\begin{array}{c}0\\0\\1\end{array}\right)$ for rotations around $z$ axis
