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Let's just consider the lines in $\mathbb{R}^3$. For any line $l$, suppose it is determined by $(x_1:x_2:x_3)$. For any point $(a,b,c)$, after the rotation it is mapped to(as we can compute) $(a',b',c')=(2\lambda x_1-a,2\lambda x_2-b,2\lambda x_3-c)$, where $\lambda=\frac{ax_1+bx_2+cx_3}{x_1^2+x_2^2+x_3^2}$. I computed by using the condition that their middle point is on the line $(x_1:x_2:x_3)$, and $\overline{xx'}$ is orthogonal to the line $l$.

And we can of course compute the matrix of this transformation. But I have trouble checking it indeed satisfies the condition for rotation matrix $\det(A)=1, A^T=A^{-1}$.

I am wondering are their any more algebraic way of finding the matrix? Also, I wish someone could verify whether my answer is correct. Thanks!

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    Consider the orthogonal complement of the line $L$, which is a plane $P$. Find the orthogonal projection of the point $x$ onto the line, $x_L$, and onto the plane, $x_P$ (hint: if you can find one, you get the other for free via $x=x_L+x_P$). A rotation by $\pi$ keeps $x_L$ the same and negates $x_P$. – 2017-08-29

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Consider a vector $v=(a,b,c)$ and associate to it the skew matrix $A$ given by\begin{pmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a &0 \end{pmatrix} Then $R=e^A$ is rotation about the direction determined by $v$ and angle determined by $\theta=\lVert v\rVert$. Because the powers of $A$ can be computed in terms of $A$, you can find a simple expression for $R$. See here.

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    Somehow I think it is a quite different problem.... In my problem, it asks us to find maps for each line to the matrix that rotate vectors about the line by $\pi$. I don't see how the link can be used in this problem... – 2017-01-29