# Eureka Math Precalculus Module 2 Lesson 25 Answer Key

## Engage NY Eureka Math Precalculus Module 2 Lesson 25 Answer Key

### Eureka Math Precalculus Module 2 Lesson 25 Example Answer Key

Example
When three-dimensional objects are projected onto screens with finite dimensions, it often limits the field of view (FOV), or the angle the scene represents. This limiting effect can vary based on the size of the screen and position of the observer.
a. Sketch a diagram that could be used to calculate a viewer’s field of view θ in relation to the horizontal width of the screen w and the distance the viewer is from the screen d.

b. Assume that a person is sitting directly in front of a television screen whose width is 48 inches at a distance of 8 feet from the screen. Use your diagram and right-triangle trigonometry to find the viewer’s horizontal field of view θ.
$$\tan \left(\frac{\theta}{2}\right) = \frac{\frac{w}{2}}{d}$$
$$\tan \left(\frac{\theta}{2}\right) = \frac{\frac{4}{2}}{8}$$ = 0.25
tan-1(0.25) = $$\frac{\theta}{2}$$
14°≈$$\frac{\theta}{2}$$
θ≈28°

c. How far would a viewer need to be from the middle of a computer screen with a width of 15 inches to produce the same field of view as the person in front of the television?
$$\tan \left(\frac{\theta}{2}\right) = \frac{\frac{w}{2}}{d}$$
$$\tan \left(\frac{28^{\circ}}{2}\right) = \frac{\frac{15}{2}}{d}$$
d = $$\frac{\frac{15}{2}}{\tan \left(14^{\circ}\right)}$$ ≈ 30 inches

d. Write a general statement about the relationship between screen size and field of view.
The smaller the screen, the closer a viewer must be to the screen to result in a given horizontal FOV.

### Eureka Math Precalculus Module 2 Lesson 25 Exercise Answer Key

Exercise 1.
In this drawing task, the “eye” or the “camera” is the point, and the shaded figure is the “TV screen.” The cube is in the
3-D universe of the computer game.
By using lines drawn from each vertex of the cube to the point, draw the image of the 3-D cube on the screen. A horizon line and two additional vanishing points have been included to help you. The image point of the first vertex is shown.

Draw line segments connecting the “eye” (V1) with the vertices of the cube.
Draw a vertical line segment on the screen from the given image point (P1‘) to the point where it intersects the line segment from V1 to the vertex beneath it (P2‘).

Draw dashed line segments connecting the two image points to vanishing points 2 and 3.

Plot a point where each of the line segments from V1 to the vertices intersect the line segments connecting the dashed line segments. These are the remaining image points for the vertices of the cube.

Connect the image points to form the projected image of the cube.

Exercise 2.
Let’s assume that the point V1 in our projection diagram is at the origin and the upper-right vertex of the cube is located at $$\left(\begin{array}{l} 5 \\ 8 \\ 4 \end{array}\right)$$. If our screen represents the plane y = 2, use matrix multiplication to determine the vector that represents the line of sight from the observer to the projected point on the screen. Explain your thinking.
The vector from the viewer to the vertex on the actual object is $$\left[\begin{array}{l} 5 \\ 8 \\ 4 \end{array}\right]$$. Since the projected point and the actual vertex are along the same line of sight, the vector from the line of sight of the viewer to the projected point can be found by resizing the vector to the vertex so that the y-coordinate of the projected point is 2. This can be accomplished by rescaling with a factor of 0.25 using the matrix multiplication $$\left[\begin{array}{ccc} 0.25 & 0 & 0 \\ 0 & 0.25 & 0 \\ 0 & 0 & 0.25 \end{array}\right]\left[\begin{array}{l} 5 \\ 8 \\ 4 \end{array}\right] = \left[\begin{array}{c} 1.25 \\ 2 \\ 1 \end{array}\right]$$.

### Eureka Math Precalculus Module 2 Lesson 25 Problem Set Answer Key

Question 1.
Projecting the image of a three-dimensional scene onto a computer screen has the added constraint of the screen size limiting our field of view, or FOV. When we speak of FOV, we wish to know what angle of view the scene represents. Humans have remarkably good peripheral vision. In New York State, the requirement for a driver’s license is a horizontal FOV of no less than 140°. There is no restriction placed on the vertical field of vision, but humans normally have a vertical FOV greater than 120°.
a. Consider the (simulated) distance the camera is from the screen as d, the horizontal distance of the screen as w, and the horizontal FOV as θ. Then use the diagram below and right-triangle trigonometry to help you find θ in terms of w and d.

$$\tan \left(\frac{\theta}{2}\right) = \frac{\frac{w}{2}}{d}$$
$$\frac{\theta}{2} = \tan ^{-1}\left(\frac{\frac{w}{2}}{d}\right)$$
θ = $$2 \tan ^{-1}\left(\frac{\frac{w}{2}}{d}\right)$$

b. Repeat procedures from part (a), but this time let h represent the height of the screen and ψ represent the vertical FOV.

$$\tan \left(\frac{\psi}{2}\right) = \frac{\frac{h}{2}}{d}$$
$$\frac{\psi}{2} = \tan ^{-1}\left(\frac{\frac{h}{2}}{d}\right)$$
ψ = $$2 \tan ^{-1}\left(\frac{\frac{h}{2}}{d}\right)$$

c. If a particular game uses an aspect ratio of 16:9 as its standard view and treats the camera as though it were 8 units away, find the horizontal and vertical FOVs for this game. Round your answers to the nearest degree.
θ = 2⋅tan-1$$\left(\frac{\frac{16}{2}}{8}\right)$$
= 2 ⋅ tan-1(1)
= 2⋅45
= 90
The horizontal field of view is 90°.
ψ = 2⋅tan-1$$\left(\frac{\frac{9}{2}}{8}\right)$$
= 2 ⋅ tan-1$$\left(\frac{4.5}{8}\right)$$
≈ 2⋅29.36
= 58.72
≈59
So, the vertical field of view is about 59°.

d. When humans sit too close to monitors with FOVs less than what they are used to in real life or in other games, they may grow dizzy and feel sick. Does the game in part (c) run the risk of that? Would you recommend this game be played on a computer or on a television with these FOVs?
The game in part (c) has a smaller field of vision than most humans are probably used to. Since computer games are played much closer and that increases the chance of dizziness and illness, this game would probably be better on a television than a computer.

Question 2.
Computers regularly use polygon meshes to model three-dimensional objects. Most polygon meshes are a collection of triangles that approximate the shape of a three-dimensional object. If we define a face of a polygon mesh to be a triangle connecting three vertices of the shape, how many faces at minimum do the following shapes require?
a. A cube
A cube normally has 6 square faces, but it requires 2 triangles to make a square, so a polygon mesh of a cube would have 12 faces.

b. A pyramid with a square base
The pyramid has 4 triangular faces already, and would require 2 more to make up the base, so 6 faces.

c. A tetrahedron
A tetrahedron is composed of 4 triangles, so 4 faces for its polygon mesh.

d. A rectangular prism
A rectangular prism is the same as a cube: It has 12 faces.

e. A triangular prism
A triangular prism has 2 faces make up the ends, and each side requires 2 more faces, so 2 + 2⋅3, or 8 faces.

f. An octahedron
An octahedron is 8 triangular pieces, so 8 faces.

g. A dodecahedron
A dodecahedron normally has 12 pentagonal faces. Each pentagon requires a minimum of 3 triangles to construct it, so a dodecahedron would have 36 faces in its polygon mesh.

h. An icosahedron
An icosahedron has 20 triangular faces, so its polygon mesh does as well.

i. How many faces should a sphere have?
Answers will vary. An icosahedron is a rough approximation of a sphere, so 20 is a good estimate, especially for graphics in the early years of computers and video games. Some students might suggest 360, which is a good estimate of the number of lines to approximate a circle. The best polygon mesh for a sphere would have as many faces as the computer can handle without slowing down..

Question 3.
In the beginning of 3-D graphics, objects were created only using the wireframes from a polygon mesh without shading or textures. As processing capabilities increased, 3-D models became more advanced, and shading and textures were incorporated into 3-D models. One technique that helps viewers visualize how shading works on a 3-D figure is to include both an “eye” and a “light source.” Vectors are drawn from the eye to the figure, and then reflected to the light (this technique is called ray tracing). See the diagram below.

a. Using this technique, the hue of the object depends on the sum of the magnitudes of the vectors. Assume the “eye” in the picture above is located at the origin, vsis the vector from the eye to the location $$\left(\begin{array}{l} 4 \\ 5 \\ 3 \end{array}\right)$$, and the “light source” is located at $$\left(\begin{array}{l} 5 \\ 6 \\ 8 \end{array}\right)$$. Then find vl, the vector from vsto the light source, and the sum of the magnitudes of the vectors.
vl is the difference between the light source and the object, so vl = $$\left(\begin{array}{l} 5 \\ 6 \\ 8 \end{array}\right)-\left(\begin{array}{l} 4 \\ 5 \\ 3 \end{array}\right)$$ = $$\left(\begin{array}{l} 1 \\ 1 \\ 5 \end{array}\right)$$
‖vl ‖ = $$\sqrt{1^{2} + 1^{2} + 5^{2}} = \sqrt{27} = 3 \sqrt{3}$$
‖vs‖ = $$\sqrt{4^{2} + 5^{2} + 3^{2}} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5 \sqrt{2}$$
‖vl ‖ + ‖v s‖ = $$3 \sqrt{3} + 5 \sqrt{2}$$

b. What direction does light travel in real life, and how does this compare to the computer model portrayed above? Can you think of any reason why the computer only traces the path of vectors that start at the “eye?”
In a computer game, the camera eye is at the origin, and the tip of a dog’s nose has coordinates $$\left(\begin{array}{c} 2 \\ 10 \\ 3 \end{array}\right)$$. If the computer screen represents the plane y = 1, determine the coordinates of the projected point that represents the tip of the dog’s nose.
The vector representing the line of sight from the viewer to the tip of the dog’s nose is $$\left[\begin{array}{c} 2 \\ 10 \\ 3 \end{array}\right]$$ . To resize the vector so that the y-coordinate of the projected point is 1, we need to rescale by a factor of 0.1.
$$\left[\begin{array}{ccc} 0.1 & 0 & 0 \\ 0 & 0.1 & 0 \\ 0 & 0 & 0.1 \end{array}\right]\left[\begin{array}{c} 2 \\ 10 \\ 3 \end{array}\right] = \left[\begin{array}{c} 0.2 \\ 1 \\ 0.3 \end{array}\right]$$, so the projected point is $$\left(\begin{array}{c} 0.2 \\ 1 \\ 0.3 \end{array}\right)$$.