Mumbai University > Mechanical Engineering > Sem 7 > CAD CAM CAE

Marks: 8 Marks

Year: Dec 2016

Normal Vector N=I+J+K

$n_1=1, n_2=1, n_3=1 \\ |N|=\sqrt{3} \ \ and \ \ \lambda = \sqrt{2} \\ R_{xy}=\begin{bmatrix}\dfrac{\sqrt2}{\sqrt3}&0&\dfrac{1}{\sqrt3}&0 \\ -\dfrac{1}{\sqrt6}&\dfrac{1}{\sqrt2}&\dfrac{1}{\sqrt3}&0 \\ -\dfrac{1}{\sqrt6}&-\dfrac{1}{\sqrt2}&\dfrac{1}{\sqrt3}&0 \\ 0&0&0&1\end{bmatrix} \\ R_{xy}^{-1}=\begin{bmatrix}\sqrt{\dfrac{2}{3}}&-\dfrac{1}{\sqrt6}&-\dfrac{1}{\sqrt6}&0 \\ 0&\dfrac{1}{\sqrt2}&-\dfrac{1}{\sqrt2}&0 \\ +\dfrac{1}{\sqrt3}&\dfrac{1}{\sqrt3}&\dfrac{1}{\sqrt3}&0 \\ 0&0&0&1\end{bmatrix} \\ M=\begin{bmatrix}1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \\ \therefore \text{Reflection R Matrix in given by} \\ RT=R_{xy} \times M \times R_{xy} \\ \therefore RT=\begin{bmatrix}\dfrac{1}{3}&-\dfrac{2}{3}&-\dfrac{2}{3}&0 \\ -\dfrac{2}{3}&\dfrac{1}{3}&-\dfrac{2}{3}&0 \\ -\dfrac{2}{3}&-\dfrac{2}{3}&\dfrac{1}{3}&0 \\ 0&0&0&1\end{bmatrix}$