Induced homomorphism $d\rho:\mathfrak{spin}(V)\rightarrow \mathfrak{so}(V)$ Picture below is from the 72 of Jost's Riemannian geometry and geometric analysis . SO(V) is the special orthogonal group , Spin(V) is subgroup of CL(V)  (Clifford algebra).  In fact ,there are many notation  needed to explain. But they are not important for my question.
First, why $e_i\wedge e_j$ is the tangent vector of rotation from $e_i$ to $e_j$ ?
Second , why product of two reflection about hyperplane orthogonal to $-\cos \theta e_i+\sin \theta e_j$ and $e_i$  is a rotation from $e_i$ to $e_j$ ?



 A: For the first question, use the natural isomorphism $\Lambda^2V\cong\mathfrak{so}(V)$ for a finite dimensional vector space $V$ with nondegenerate bilinear form $(-,-)$.  I think in this book it would be given by $x\wedge y\mapsto y\otimes (x,-)-x\otimes(y,-)\in V\otimes V^*$.  It follows that if $e_1,\dots,e_n$ is a basis of $V$, then $e_i\wedge e_j$ will correspond to the matrix with a 1 in the $(j,i)$th position and a $-1$ in the $(i,j)$th position.  In the ordered basis $(e_i,e_j)$, the one-parameter subgroup of rotations through an angle $\theta$ in the $e_ie_j$ plane from $e_i$ to $e_j$ is $\begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix}$.  Differentiating this subgroup at $\theta=0$ gives $\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$, which is the matrix of $e_i\wedge e_j$ as written.
For the second question, one can write out this transformation explicitly: recall that the reflection through the hyperplane orthogonal to a vector $v$ is given by $w\mapsto w-2\frac{(v,w)}{(v,v)}v$.  In our case, it suffices to look at the two-dimensional subspace spanned by $e_i$ and $e_j$, which we give, as an ordered basis, $(e_i,e_j)$.  Then the matrix of the first reflection is:
$$
\begin{bmatrix}1-2\cos^2\theta & 2\sin\theta\cos\theta\\2\sin\theta\cos\theta&1-2\sin^2\theta\end{bmatrix}=\begin{bmatrix}-\cos(2\theta) & \sin(2\theta)\\\sin(2\theta) & \cos(2\theta)\end{bmatrix}
$$
The matrix of the second reflection is:
$$
\begin{bmatrix}-1 & 0\\0 & 1\end{bmatrix}
$$
Multiplying these gives
$$
\begin{bmatrix}\cos(2\theta) & -\sin(2\theta)\\\sin(2\theta) & \cos(2\theta)\end{bmatrix}
$$
which is exactly the matrix of rotation by the angle $2\theta$ in the plane spanned by $e_i$ and $e_j$.
