Find the normalization transformation that maps a window whose lower left corner is at (1,1) and upper right corner is at (3, 5) onto a viewport that is the entire normalized device screen.

The correct answer is option 1.

Given Data:

Window to viewport mapping

FORMULA:

Window to viewport mapping

S_x = \(V_xmax-V_xmin \over W_xmax-W_xmin\)

S_y = \(V_ymax-V_ymin \over W_ymax-W_ymin\)

Normalization transformation N = \(\begin{bmatrix} S_x & 0 & -S_x \times W_xmin+V_xmin\\ 0 & S_y & -S_y \times W_ymin+V_ymin\\ 0 & 0 & 1 \end{bmatrix}\)

CAlCULATION:

S_x = \(1-0 \over 3-1\)

S_y = \(1 \over 2\)

S_y = \(1-0 \over 5-1\)

S_y = \(1 \over 4\)

\(-S_x \times W_xmin+V_xmin\) = \({-1 \over 2 }\times(1)+0\) = \(-1 \over 2\)

\(-S_y \times W_ymin+V_ymin\) = \({-1 \over 4 }\times(1)+0\) = \(-1 \over 4\)

Hence the correct answer is

\(\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0&{ - \frac{1}{2}}\\ 0&{\frac{1}{4}}&{ - \frac{1}{4}}\\ 0&0&1 \end{array}} \right)\)