# How to find the matrix of rotation in $\mathbb{R}^3$ by the angle a around the vector column $(1, 2, 3)^T$

Find the matrix of rotation in $\mathbb{R}^3$ by the angle $\theta$ around the column vector $(1, 2, 3)^T$. We assume that rotation is counterclockwise if we sit at the tip of the vector and look at the origin.

Could I use the product of three 3x3 matrices to rotate vectors about the x,y,and z points individually? If so, what would they look like?

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If you have a question, then please ask a question, don't hand out assignments or give orders. Your post is in imperative mode, which is used to give orders, not ask questions. – Arturo Magidin Feb 15 '11 at 21:48

Do you know anything about quaternions? In particular, in the standard representation of rotations by quaternions, a rotation of $\theta$ degrees about the vector $\vec{v} = (v_0, v_1, v_2) = v_0{\bf i}+v_1{\bf j}+v_2{\bf k}$ is represented by the quaternion ${\bf q} = \mathrm{cos}(\theta/2) + \mathrm{sin}(\theta/2)\hat{v}$, where $\hat{v}$ is the normalization of $\vec{v}$. From there it's easy to find the rotation matrix; see the Wikipedia page on quaternions and rotation for more details.
First normalize the column vector to a unit vector $\textbf{u}$. Then the rotation matrix $R$ is given by the following:
$$R = \textbf{u} \otimes \textbf{u}+ \cos \theta(I- \textbf{u} \otimes \textbf{u})+\sin \theta[\textbf{u}]_{\times}$$
where $[\textbf{u}]_{\times}$ is the skew-symmetric form of $\textbf{u}$, $\otimes$ is the tensor product, and $I$ is the identity matrix.