[QUE/ME-08011] ME-PROBLEM

Find rotation matrix for a rotation by an angle \(\alpha\) about the axis \(1,2,1\) where \(\cos\alpha=\frac{3}{5}, \sin\alpha =\frac{4}{5}\).

The unit vector along the direction \((1,2,1)\) is given by \(\hat{n}=\frac{1}{\sqrt{6}}(1,2,1)\).\\ Under a rotation by an angle \(\alpha\) about axis \(\hat{n}=(n_1,n_2,n_3)\), the new components \(\vec{X}\) are related to old components \(\vec{x}\) by equation \begin{equation} \vec{X} = \vec{x} -\sin \alpha (\hat{n}\times\vec{x}) + (1-\cos\alpha)\hat{n}\times(\hat{n}\times\vec{x})) \end{equation} We compute \[\hat{n}\times\vec{x}=\big({n_2} {x_3}-{n_3} {x_2},{n_3} {x_1}-{n_1} {x_3},{n_1} {x_2}-{n_2} {x_1}\big)\] \begin{eqnarray} (\hat{n}\times(\hat{n}\times\vec{x}))_1&=&{n_2} ({n_1} {x_2}-{n_2} {x_1})-{n_3} ({n_3} {x_1}-{n_1} {x_3})\\ (\hat{n}\times(\hat{n}\times\vec{x}))_2&=&{n_3} ({n_2} {x_3}-{n_3} {x_2})-{n_1} ({n_1} {x_2}-{n_2} {x_1})\\ (\hat{n}\times(\hat{n}\times\vec{x}))_3&=& {n_1} ({n_3} {x_1}-{n_1} {x_3})-{n_2} ({n_2} {x_3}-{n_3} {x_2}) \end{eqnarray} Therefore \begin{eqnarray}\nonumber X_1&=&(1-\cos \alpha ) ({n_2} ({n_1} {x_2}-{n_2} {x_1})-{n_3} ({n_3} {x_1}-{n_1} {x_3}))-\sin \alpha ({n_2} {x_3}-{n_3} {x_2})+{x_1}\\\nonumber X_2&=&(1-\cos \alpha ) ({n_3} ({n_2} {x_3}-{n_3} {x_2})-{n_1} ({n_1} {x_2}-{n_2} {x_1}))-\sin \alpha ({n_3} {x_1}-{n_1} {x_3})+{x_2}\\\nonumber X_3&=& (1-\cos \alpha ) ({n_1} ({n_3} {x_1}-{n_1} {x_3})-{n_2} ({n_2} {x_3}-{n_3} {x_2}))-\sin \alpha ({n_1} {x_2}-{n_2} {x_1})+{x_3} \end{eqnarray} Therefore \begin{equation} \begin{pmatrix} X_1\\X_2\\X_3 \end{pmatrix} = \underline{\Rbb} \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}. \end{equation} where the matrix \(\Rbb\) is given by \begin{eqnarray}\nonumber \begin{pmatrix} -(1-\cos\alpha)(n_2^2+n_3^2) + 1 & (1-\cos\alpha)n_1n_2 + \sin\alpha n_3 & (1-\cos\alpha)n_3n_1 - \sin\alpha n_2\\ (1-\cos\alpha)n_1 n_2 -n_3\sin\alpha & -(1-\cos\alpha)(n_3^2+n_1^2) +1 & (1-\cos\alpha)n_2n_3 + n_1\sin\alpha\\ (1-\cos \alpha)n_1n_3 + n_2 \sin\alpha & (1-\cos\alpha)n_2n_3 -n_1 \sin\alpha & -(1-\cos\alpha)(n_1^2+n_2^2) + 1 \end{pmatrix} \end{eqnarray} Substituting values \begin{equation*} \hat{n}=\frac{1}{\sqrt{6}}(1,2,1),\quad (1-\cos\alpha)=\frac{2}{5}, \quad\sin\alpha=\frac{4}{5} \end{equation*} and simplifying gives % \begin{equation} \left(\begin{array}{ccc} \frac{2}{3} & \frac{2}{15} \left(1+\sqrt{6}\right) & \frac{1}{15} \left(1-4 \sqrt{6}\right) \\[2mm] \frac{2}{15} \left(1-\sqrt{6}\right) & \frac{13}{15} & \frac{2}{15} \left(1+\sqrt{6}\right) \\[2mm] \frac{1}{15} \left(1+4 \sqrt{6}\right) & \frac{2}{15} \left(1-\sqrt{6}\right) & \frac{2}{3} \end{array} \right) \end{equation}