## Engage NY Eureka Math Precalculus Module 2 Lesson 27 Answer Key

### Eureka Math Precalculus Module 2 Lesson 27 Example Answer Key

Example

a. Select the Sea Surface as your background scene, and select the setup scene mode. From the Gallery By Class Hierarchy, select Swimmer Classes, then Marine Mammals, and a Dolphin. Name your dolphin. Describe what you see.

Answer:

We have a scene with an ocean surface and a dolphin facing forward in the center.

b. Click on the dolphin, and from the handle style buttons in the top – right corner of the screen, select translation. By selecting the arrows on the dolphin, we can move it to different locations in the screen. Move the dolphin left and right, up and down, forward and backward. Then, move it so that its coordinates are (0, 0, 0).

c. Use a matrix to describe each of the movements of the dolphin from its location at the origin.

i. Move 2 units right.

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{c}

– 2 \\

0 \\

0

\end{array}\right]\)

ii. Move 4 units down.

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{c}

0 \\

– 4 \\

0

\end{array}\right]\)

iii. Move 3 units forward.

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{c}

0 \\

0 \\

– 3

\end{array}\right]\)

d. Click on the rotation button from the handle style buttons, and practice rotating the dolphin about the three axes through its center. Use a matrix to represent the motion of the dolphin described. Assume the center of rotation of the dolphin is at the origin.

i. Rotation counterclockwise one full turn about the z – axis

Answer:

\(\left[\begin{array}{ccc}

\cos \left(360^{\circ}\right) & – \sin \left(360^{\circ}\right) & 0 \\

\sin \left(360^{\circ}\right) & \cos \left(360^{\circ}\right) & 0 \\

0 & 0 & 1

\end{array}\right] = \left[\begin{array}{lll}

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

ii. Rotation counterclockwise one half turn about the x – axis

Answer:

\(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left(180^{\circ}\right) & – \sin \left(180^{\circ}\right) \\

0 & \sin \left(180^{\circ}\right) & \cos \left(180^{\circ}\right)

\end{array}\right] = \left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & – 1 & 0 \\

0 & 0 & – 1

\end{array}\right]\)

e. Select Edit Code from the screen. Drag and drop “this.dolphin turn” from the “Procedures” menu and drop it into the “declare procedure” region on the right. Select from the drop – down menus: this.dolphin, turn: LEFT 0.25, asSeenBy: this, duration: 2.0, animationStyle: BEGIN_AND_END_ABRUPTLY. Then run the program. Describe what you see. Represent the motion using a matrix.

Answer:

The dolphin undergoes a quarter counterclockwise rotation in a horizontal circle about the y – axis.

\(\left[\begin{array}{ccc}

\cos \left(90^{\circ}\right) & 0 & – \sin \left(90^{\circ}\right) \\

0 & 1 & 0 \\

\sin \left(90^{\circ}\right) & 0 & \cos \left(90^{\circ}\right)

\end{array}\right] = \left[\begin{array}{ccc}

0 & 0 & – 1 \\

0 & 1 & 0 \\

1 & 0 & 0

\end{array}\right]\)

f. Drag and drop “this.dolphin roll” from the “Procedures” menu, and drop it into the “declare procedure” region on the right beneath the turn procedure. Select from the drop – down menus: this.dolphin, roll: RIGHT 3.0, asSeenBy: this, duration: 2.0, animationStyle: BEGIN_AND_END_ABRUPTLY. Then run the program. Describe what you see. Represent the motion using a matrix.

Answer:

The dolphin performs three backward counterclockwise somersaults in the vertical plane about the negative z – axis.

\(\left[\begin{array}{ccc}

\cos \left(1080^{\circ}\right) & – \sin \left(1080^{\circ}\right) & 0 \\

\sin \left(1080^{\circ}\right) & \cos \left(1080^{\circ}\right) & 0 \\

0 & 0 & 1

\end{array}\right] = \left[\begin{array}{lll}

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

### Eureka Math Precalculus Module 2 Lesson 27 Exercise Answer Key

Exercises

Exercise 1.

Open ALICE 3.1. Select a background and three characters to create a scene. Describe the scene, including the coordinates of the pivot point for each character and the direction each character is facing.

Answer:

Scene description: Answers will vary. I created a sea surface scene that contains an adult walrus facing forward, a baby walrus to the left of the adult and facing it (looking to the right), and a dolphin breaching on its back, with its nose pointed up and to the right. The dolphin is behind and to the left of the walruses.

Coordinates of the pivot point for each character:

Answers will vary. For example, dolphin (4.16, 0.5, 4.1); baby walrus (0.6, 0.11, 0.55);

adult walrus ( – 0.46, – 0.03, 1.11).

Exercise 2.

a. Describe the location of a plane x = 5 in your scene from the perspective of the viewer.

Answer:

The plane x = 5 is a vertical plane 5 units to the left of the center of the screen.

b. Determine the coordinates of the pivot points for each character if they were projected onto the plane x = 5.

Answer:

Answers will vary. An example of an acceptable response is shown.

For the dolphin, 4.16t = 5, so t≈1.20 and image = \(\left[\begin{array}{c}

5 \\

6 \\

4.92

\end{array}\right]\).

For the baby walrus, 0.6t = 5, so t = \(\frac{25}{3}\) and image = \(\left[\begin{array}{c}

5 \\

0.92 \\

4.58

\end{array}\right]\).

For the adult walrus, – 0.46t = 5, so t≈ – 10.87 and image = \(\left[\begin{array}{c}

\mathbf{5} \\

\mathbf{0 . 3 3} \\

\mathbf{ – 1 2 . 0 7}

\end{array}\right]\).

Exercise 3.

Create a short scene that includes a one – step turn or roll procedure for each of the characters. The procedures should be unique. Write down the procedures in the space provided. After each procedure, describe what the character did in the context of the scene. Then describe the character’s motion using transformational language. Finally, represent each procedure using matrix operations.

Answer:

Answers will vary. An example of an acceptable response is shown.

Procedure 1, adult walrus: this.adultWalrus turn: FORWARD 4.0, asSeenBy: this, duration: 2.0.

Character motion (in scene): The adult walrus performed four forward somersaults toward the foreground.

Transformational motion: The adult walrus performed a counterclockwise rotation of four full turns about the

x – axis.

Matrix representation of motion: \(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left(1440^{\circ}\right) & – \sin \left(1440^{\circ}\right) \\

0 & \sin \left(1440^{\circ}\right) & \cos \left(1140^{\circ}\right)

\end{array}\right]\) A for any point A on the walrus translated to be centered around the origin.

Procedure 2, baby walrus: this.babyWalrus turn: LEFT 2.0, asSeenBy: this, duration: 2.0.

Character motion (in scene): The baby walrus spun twice in a counterclockwise circle next to the adult.

Transformational motion: The baby walrus rotated about the y – axis counterclockwise.

Matrix representation of motion: \(\left[\begin{array}{ccc}

\cos \left(720^{\circ}\right) & 0 & – \sin \left(720^{\circ}\right) \\

0 & 1 & 0 \\

\sin \left(720^{\circ}\right) & 0 & \cos \left(720^{\circ}\right)

\end{array}\right]\) B for any point B on the walrus translated to be centered around the origin.

Procedure 3, dolphin: this.dolphin roll: LEFT 4.0, asSeenBy: this, duration: 2.0.

Character motion (in scene): The dolphin performed four back somersaults out of the water.

Transformational motion: The dolphin rotated counterclockwise about the z – axis.

Matrix representation of motion:

\(\left[\begin{array}{ccc}

\cos \left(720^{\circ}\right) & – \sin \left(720^{\circ}\right) & 0 \\

\sin \left(720^{\circ}\right) & \cos \left(720^{\circ}\right) & 0 \\

0 & 0 & 1

\end{array}\right]\) C for any point C on the dolphin translated to be centered around the origin.

### Eureka Math Precalculus Module 2 Lesson 27 Problem Set Answer Key

Question 1.

For the following commands, describe a matrix you can use to get the desired result. Assume the character is centered at the origin, facing in the negative z direction with positive x on its right for each command. Let c be a nonzero real number.

a. move LEFT c

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{c}

– c \\

0 \\

0

\end{array}\right]\)

b. move RIGHT c

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{l}

c \\

0 \\

0

\end{array}\right]\)

c. move UP c

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{l}

0 \\

c \\

0

\end{array}\right]\)

d. move DOWN c

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{c}

0 \\

– c \\

0

\end{array}\right]\)

e. move FORWARD c

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{c}

0 \\

0 \\

– c

\end{array}\right]\)

f. move BACKWARD c

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{l}

0 \\

0 \\

c

\end{array}\right]\)

g. turn LEFT c

Answer:

\(\left[\begin{array}{ccc}

\cos \left(360 c^{\circ}\right) & 0 & – \sin \left(360 c^{\circ}\right) \\

0 & 1 & 0 \\

\sin \left(360 c^{\circ}\right) & 0 & \cos \left(360 c^{\circ}\right)

\end{array}\right]\)

h. turn RIGHT c

Answer:

\(\left[\begin{array}{ccc}

\cos \left( – 360 c^{\circ}\right) & 0 & – \sin \left( – 360 c^{\circ}\right) \\

0 & 1 & 0 \\

\sin \left( – 360 c^{\circ}\right) & 0 & \cos \left( – 360 c^{\circ}\right)

\end{array}\right]\)

i. turn FORWARD c

Answer:

\(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left( – 360 c^{\circ}\right) & – \sin \left( – 360 c^{\circ}\right) \\

0 & \sin \left( – 360 c^{\circ}\right) & \cos \left( – 360 c^{\circ}\right)

\end{array}\right]\)

j. turn BACKWARD c

Answer:

\(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left(360 c^{\circ}\right) & – \sin \left(360 c^{\circ}\right) \\

0 & \sin \left(360 c^{\circ}\right) & \cos \left(360 c^{\circ}\right)

\end{array}\right]\)

k. roll LEFT c

Answer:

\(\left[\begin{array}{ccc}

\cos \left( – 360 c^{\circ}\right) & – \sin \left( – 360 c^{\circ}\right) & 0 \\

\sin \left( – 360 c^{\circ}\right) & \cos \left( – 360 c^{\circ}\right) & 0 \\

0 & 0 & 1

\end{array}\right]\)

l. roll RIGHT c

Answer:

\(\left[\begin{array}{ccc}

\cos \left(360 c^{\circ}\right) & – \sin \left(360 c^{\circ}\right) & 0 \\

\sin \left(360 c^{\circ}\right) & \cos \left(360 c^{\circ}\right) & 0 \\

0 & 0 & 1

\end{array}\right]\)

m. resize c

Answer:

\(\left[\begin{array}{lll}

c & 0 & 0 \\

0 & c & 0 \\

0 & 0 & c

\end{array}\right]\)

n. resizeWidth c

Answer:

\(\left[\begin{array}{lll}

\boldsymbol{c} & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right]\)

o. resizeHeight c

Answer:

\(\left[\begin{array}{lll}

1 & 0 & 0 \\

0 & c & 0 \\

0 & 0 & 1

\end{array}\right]\)

p. resizeDepth c

Answer:

\(\left[\begin{array}{lll}

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & c

\end{array}\right]\)

Question 2.

In each of the transformations above, we have assumed that each animation will take one second of time. If T is the number of seconds an animation takes and t is the current running time of the animation, then rewrite the following commands as a function of t.

a. move RIGHT c duration T

Answer:

\(\left[\begin{array}{l}

0 \\

0 \\

0

\end{array}\right] + \left[\begin{array}{l}

c \\

0 \\

0

\end{array}\right] \frac{t}{T}\)

b. turn FORWARD c duration T

Answer:

\(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos \left( – 360 c^{\circ}\right) & – \sin \left( – 360 c^{\circ}\right) \\

0 & \sin \left( – 360 c^{\circ}\right) & \cos \left( – 360 c^{\circ}\right)

\end{array}\right] \frac{t}{T}\)

c. resize c duration T

Answer:

\(\left[\begin{array}{lll}

c & 0 & 0 \\

0 & c & 0 \\

0 & 0 & c

\end{array}\right] \frac{t}{T}\)

Question 3.

For computational simplicity, we have been assuming that the pivot points of our characters occur at the origin.

a. If we apply a rotation matrix like we have been when the pivot point is at the origin, what happens to characters that are not located around the origin?

Answer:

The characters trace out a circular path much like we saw in previous lessons when we were rotating points instead of objects.

b. Let x = \(\left[\begin{array}{l}

\chi \\

y \\

Z

\end{array}\right]\) be the pivot point of any three – dimensional object and A = \(\left[\begin{array}{l}

a \\

b \\

c

\end{array}\right]\) a point on the surface of the object. Moving the pivot point to the origin has what effect on A? Find A’, the image of A after moving the object so that its pivot point is the origin.

Answer:

A’ = A – x

= \(\left[\begin{array}{l}

a \\

b \\

c

\end{array}\right] – \left[\begin{array}{l}

x \\

y \\

z

\end{array}\right]\)

= \(\left[\begin{array}{l}

a – x \\

b – y \\

c – z

\end{array}\right]\)

c. Apply a rotation of θ about the x – axis to A’. Does this transformation cause a pivot or what you described in part (a)?

Answer:

\(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & \cos (\theta) & – \sin (\theta) \\

0 & \sin (\theta) & \cos (\theta)

\end{array}\right]\left[\begin{array}{c}

a – x \\

b – y \\

c – z

\end{array}\right] = \left[\begin{array}{c}

a – x \\

\cos (\theta)(b – y) – \sin (\theta)(c – z) \\

\sin (\theta)(b – y) + \cos (\theta)(c – z)

\end{array}\right]\)

Since the pivot point has been shifted to the origin, the rotation rotates the object about itself instead of through space.

d. After applying the rotation, translate the object so that its pivot point returns to x = \(\left[\begin{array}{l}

\chi \\

y \\

Z

\end{array}\right]\). Was the

x – coordinate affected by the rotation? Was the pivot point?

Answer:

\(\left[\begin{array}{c}

a – x \\

\cos (\theta)(b – y) – \sin (\theta)(c – z) \\

\sin (\theta)(b – y) + \cos (\theta)(c – z)

\end{array}\right] + \left[\begin{array}{l}

x \\

y \\

z

\end{array}\right] = \left[\begin{array}{c}

a \\

\cos (\theta)(b – y) – \sin (\theta)(c – z) + y \\

\sin (\theta)(b – y) + \cos (\theta)(c – z) + z

\end{array}\right]\)

No, the x – coordinate stayed the same once we shifted it back to the starting point. The pivot point would also have stayed the same since we shifted it to the origin (the origin does not change during rotations) and then shifted it back in the opposite direction.

e. Summarize what you found in parts (a)–(d).

Answer:

To pivot points that are not centered on the origin, we have to translate the object so that its pivot point is at the origin, then perform the rotation, and then shift the object back. If R is the rotation matrix, A is the point on the surface of the object, and x is the pivot point, then we do the following:

R(A – x) + x

for each point A on the surface of the object.

Extension:

Question 4.

In first – person computer games, we think of the camera moving left – right, forward – backward, and up – down. For computational simplicity, the camera and screen stays fixed and the objects in the game world move in the opposite direction instead. Let the camera be located at (0, 0, 0), the screen be located at z = 1, and the point v = \(\left[\begin{array}{c}

10 \\

6 \\

5

\end{array}\right]\) represent the center of an object in the game world. If a character in ALICE is the camera, then answer the following questions.

a. What are the coordinates of the projection of v on the screen?

Answer:

v’ = \(\left[\begin{array}{c}

2 \\

1.2 \\

1

\end{array}\right]\)

b. What is the value of the image of v as the character moves 4 units closer to v in the x direction?

Answer:

v’ = \(\left[\begin{array}{l}

6 \\

6 \\

5

\end{array}\right]\)

c. What are the coordinates of the projection of v’?

Answer:

v’ = \(\left[\begin{array}{c}

1.2 \\

1.2 \\

1

\end{array}\right]\)

d. If the character jumps up 6 units, then where does the image of v move on the screen?

Answer:

v moves to the apparent position \(\left[\begin{array}{c}

10 \\

0 \\

5

\end{array}\right]\), which appears to be located at \(\left[\begin{array}{l}

2 \\

0 \\

1

\end{array}\right]\) on the screen.

e. In Lesson 26, you learned that if the camera is not in the standard orientation, that rotation matrices need to be applied to the camera first. In Lesson 27, you learned that rotation matrices only pivot an object if that object is located at the origin. If a camera is in a nonstandard orientation and not located at the origin, then should the rotation matrices be applied first or a translation to the origin?

Answer:

The translation to the origin needs to take place before the rotation or else the camera will be tracing out a circular path around the axes instead of rotating to the correct position.

### Eureka Math Precalculus Module 2 Lesson 27 Exit Ticket Answer Key

Question 1.

Consider the following set of code for the ALICE program featuring a bluebird whose center is located at \(\).

this.bluebird resize 2.0

this.bluebird resizeHeight 0.5

this.bluebird turn RIGHT 0.25

this.bluebird move FORWARD 1.0

For an arbitrary point x on this bluebird, write the four matrices that represent the code above, and state where the point ends after the program runs.

Answer:

Resize 2.0: \(\left[\begin{array}{lll}

2 & 0 & 0 \\

0 & 2 & 0 \\

0 & 0 & 2

\end{array}\right]\)

ResizeHeight 0.5: \(\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & 0.5 & 0 \\

0 & 0 & 1

\end{array}\right]\)

Turn RIGHT 0.25: turning right is a rotation about the y – axis clockwise, so we get

\(\left[\begin{array}{ccc}

\cos \left( – 90^{\circ}\right) & 0 & – \sin \left( – 90^{\circ}\right) \\

0 & 1 & 0 \\

\sin \left( – 90^{\circ}\right) & 0 & \cos \left(90^{\circ}\right)

\end{array}\right] = \left[\begin{array}{ccc}

0 & 0 & 1 \\

0 & 1 & 0 \\

– 1 & 0 & 0

\end{array}\right]\)

Move forward is a translation by 1 unit in the positive x direction after the turn right. There is no linear transformation we can use, but we can write this as x’ + \(\left[\begin{array}{l}

1 \\

0 \\

0

\end{array}\right]\).

x would end at

\(\left[\begin{array}{ccc}

0 & 0 & 1 \\

0 & 1 & 0 \\

– 1 & 0 & 0

\end{array}\right]\left[\begin{array}{ccc}

1 & 0 & 0 \\

0 & 0.5 & 0 \\

0 & 0 & 1

\end{array}\right]\left[\begin{array}{lll}

2 & 0 & 0 \\

0 & 2 & 0 \\

0 & 0 & 2

\end{array}\right] x + \left[\begin{array}{l}

1 \\

0 \\

0

\end{array}\right]\)

\(\left[\begin{array}{ccc}

0 & 0 & 2 \\

0 & 1 & 0 \\

– 2 & 0 & 0

\end{array}\right] x + \left[\begin{array}{l}

1 \\

0 \\

0

\end{array}\right]\)

If x = \(\left[\begin{array}{l}

a \\

b \\

c

\end{array}\right]\), then the image of x at the end of the program would be \(\left[\begin{array}{c}

2 c + 1 \\

b \\

– 2 a

\end{array}\right]\).