Rotation Matrix following Right-Hand Rule. In $\Bbb{R}^3$, let $L$ be the line spanned and oriented by $v=(−4,6,−7)$.
Let $R$ the rotation of $\Bbb{R}^3$ through the angle $\frac{\pi}{2}$ about the $v$ oriented line $L$ according to the Right-Hand Rule.
I know what a rotation matrix in $2$ dimensions look like in terms of $\cos \theta$ and $\sin \theta$. But how to find the $R$ for the above? How to use the Right-Hand Rule?

Here is how to find the matrix for any angle $\theta$ about your oriented  axis.Let $$\mathbf c=\frac{1}{\sqrt {16+36+49}} \begin{bmatrix}-4\\6\\-7\end{bmatrix}$$. Take any unit vector $\mathbf a$ orthogonal to $\mathbf a$. We arbitrarily take $$\mathbf a=\frac{1}{\sqrt{36+16}}\begin{bmatrix}6\\4\\0\end{bmatrix}$$. Let $\mathbf {b=c \times a}. $ Let $A$ be the matrix whose first, second and third columns are $\mathbf {c,a} \text { and }\mathbf b$ respectively. Let $B$ be the matrix whose first, second and third columns are $\mathbf c,(\cos \theta)\mathbf a+(\sin \theta)\mathbf b, \text { and } (- \sin \theta)\mathbf a+(\cos \theta)\mathbf b$ respecively. Solve $MA=B$ for $M$, which is not too hard, because $A$ is orthogonal. Then $M$ is the matrix that rotates any point about your oriented axis through the angle $\theta.$ 
