Posons \(k =\frac{1}{\sqrt{2}}(e_1+e_2)\). Soit \(u=xe_1+ye_2+ze_3\in E\). On sait que

\[\begin{align*}\ensuremath r(u)&=(\cos\theta)u+(1-\cos\theta)\left(u|k\right)k+(\sin\theta)k\wedge u=\frac{1}{2}u +\frac{1}{4}(u|(e_1+e_2))(e_1+e_2)+\frac{\sqrt{6}}{4}(e_1+e_2)\wedge u\\ &\text{\og}=\text{\fg}\;\frac{1}{2}\left( \begin{array}{c} x\\ y\\ z \end{array} \right)+\frac{1}{4}(x+y)\left( \begin{array}{c} 1\\ 1\\ 0 \end{array} \right)+\frac{\sqrt{6}}{4}\left( \begin{array}{c} 1\\ 1\\ 0 \end{array} \right)\wedge\left( \begin{array}{c} x\\ y\\ z \end{array} \right)\\ &=\frac{1}{2}\left( \begin{array}{c} x\\ y\\ z \end{array} \right)+\frac{1}{4}(x+y)\left( \begin{array}{c} 1\\ 1\\ 0 \end{array} \right)+\frac{\sqrt{6}}{4}\left( \begin{array}{c} z\\ -z\\ -x+y \end{array} \right)\\ &=\left( \begin{array}{c} \frac{3}{4}x+\frac{1}{4}y+\frac{\sqrt{6}}{4}z\\ \rule[-4mm]{0mm}{11mm}\frac{1}{4}x+\frac{3}{4}y-\frac{\sqrt{6}}{4}z\\ -\frac{\sqrt{6}}{4}x+\frac{\sqrt{6}}{4}y+\frac{1}{2}z \end{array} \right)=\left( \begin{array}{cccc} \frac{3}{4}&\frac{1}{4}&\frac{\sqrt{6}}{4}\\ \rule[-4mm]{0mm}{11mm}\frac{1}{4}&\frac{3}{4}&-\frac{\sqrt{6}}{4}\\ -\frac{\sqrt{6}}{4}&\frac{\sqrt{6}}{4}&\frac{1}{2} \end{array} \right)\left( \begin{array}{c} x\\ y\\ z \end{array} \right) \end{align*}\]

et la matrice cherchée est

\(M=\left( \begin{array}{cccc} \frac{3}{4}&\frac{1}{4}&\frac{\sqrt{6}}{4}\\ \rule[-4mm]{0mm}{11mm}\frac{1}{4}&\frac{3}{4}&-\frac{\sqrt{6}}{4}\\ -\frac{\sqrt{6}}{4}&\frac{\sqrt{6}}{4}&\frac{1}{2} \end{array} \right)\).

exo7 5793