---
geometry: margin=2cm
output: pdf_document
title: Exam 2
subtitle: CSCI 5607
date: \today

author: |
  | Michael Zhang
  | zhan4854@umn.edu $\cdot$ ID: 5289259
---

\renewcommand{\c}[1]{\textcolor{gray}{#1}}

## Reflection and Refraction

1. Consider a sphere $S$ made of solid glass ($\eta$ = 1.5) that has radius $r =
   3$ and is centered at the location $s = (2, 2, 10)$ in a vaccum ($\eta =
   1.0$). If a ray emanating from the point $e = (0, 0, 0)$ intersects $S$ at a
   point $p = (1, 4, 8)$:

   a. (2 points) What is the angle of incidence $\theta_i$ ?

   First, the normal at the point $(1, 4, 8)$ is determined by subtracting
   that point from the center $(2, 2, 10)$, which gets us $N = (2 - 1, 2 - 4,
      10 - 8) = (1, -2, 2)$. Then, to determine the angle between

   b. (1 points) What is the angle of reflection $\theta_r$ ?

   c. (3 points) What is the direction of the reflected ray?
   d. (3 points) What is the angle of transmission $\theta_t$ ?
   e. (4 points) What is the direction of the transmitted ray?

   Using Snell's law, we know that $\eta_1 \sin \theta_1 = \eta_2 \sin
      \theta_2$. In this case, let material 1 be the vacuum, and material 2 be
   the glass. Then, we have $1.0 \times \sin \theta_1$

## Geometric Transformations

2. \c{(8 points) Consider the airplane model below, defined in object
   coordinates with its center at $(0, 0, 0)$, its wings aligned with the $\pm
   x$ axis, its tail pointing upwards in the $+y$ direction and its nose facing
   in the $+z$ direction. Derive a sequence of model transformation matrices
   that can be applied to the vertices of the airplane to position it in space
   at the location $p = (4, 4, 7)$, with a direction of flight $w = (2, 1, –2)$
   and the wings aligned with the direction $d = (–2, 2, –1)$.}

   The translation matrix is

   $$
   \begin{bmatrix}
     1 & 0 & 0 & x \\
     0 & 1 & 0 & y \\
     0 & 0 & 1 & z \\
     0 & 0 & 0 & 1 \\
   \end{bmatrix}
   =
   \begin{bmatrix}
     1 & 0 & 0 & 4 \\
     0 & 1 & 0 & 4 \\
     0 & 0 & 1 & 7 \\
     0 & 0 & 0 & 1 \\
   \end{bmatrix}
   $$

   Since the direction of flight was originally $(0, 0, 1)$, we have to
   transform it to $(2, 1, -2)$.

## Clipping

9. \c{Consider the triangle whose vertex positions, after the viewport
   transformation, lie in the centers of the pixels: $p_0 = (3, 3), p_1 = (9,
   5), p_2 = (11, 11)$.}

   Starting at $p_0$, the three vectors are:

   - $v_0 = p_1 - p_0 = (9 - 3, 5 - 3) = (6, 2)$
   - $v_1 = p_2 - p_1 = (11 - 9, 11 - 5) = (2, 6)$
   - $v_2 = p_0 - p_2 = (3 - 11, 3 - 11) = (-8, -8)$

   The first edge vector $e$ would be $(6, 2)$, and the edge normal would be
   that rotated by $90^\circ$.

   a. \c{(6 points) Define the edge equations and tests that would be applied,
   during the rasterization process, to each pixel $(x, y)$ within the bounding
   rectangle $3 \le x \le 11, 3 \le y \le 11$ to determine if that pixel is
   inside the triangle or not.}

   b. \c{(3 points) Consider the three pixels $p_4 = (6, 4), p_5 = (7, 7)$, and
   $p_6 = (10, 8)$. Which of these would be considered to lie inside the
   triangle, according to the methods taught in class?}